Nicholson - Introduction to Abstract Algebra - Chapter 2 Solutions

Section 2.1

Problem 2.1.1.

Solution: Part (a) Not associative or commutative.

(a ∗ b) ∗ c = (a− b)− c 6= a− (b− c) = a− b+ c,

a ∗ b = a− b 6= b− a = b ∗ a.There is no unity because the operator is not commutative for any element. Therefore there are no units in Z with this

operator.Part (b) Associative and commutative.

(a ∗ b) ∗ c =1

2

(1

2ab

)c =

1

2a

(1

2bc

)= a ∗ (b ∗ c),

a ∗ b =1

2ab =

1

2ba = b ∗ a.

Two is unity since 2 ∗ a = a ∗ 2 = (2 ∗ a)/2 = a. Each non-zero element q ∈ Q has unity 4/q since q ∗ (4/q) =(q ∗ 4/q)/2 = 2. Zero has no inverse.

Part (c) Associative and commutative.

(a ∗ b) ∗ c = (a+ b− ab) + c− (a+ b− ab)c = a+ b+ c− ab− ac− bc+ abc,

a ∗ (b ∗ c) = a+ (b+ c− bc)− a(b+ c− bc) = a+ b+ c− ab− ac− bc+ abc,

so (a ∗ b) ∗ c = a ∗ (b ∗ c). Further:a ∗ b = a+ b− ab = b+ a− ba = b ∗ a.

Zero is unity since a ∗ 0 = a+ 0− a(0) = a = 0 + a− 0(a) = 0 ∗ a. Each a ∈ R where a 6= 1 is a unit whose inverse isa/(1− a). If a ∗ b = 0, then:

a ∗ b = a+ b− ab = a+ b(1− a) = 0 =⇒ b =−a

1− a=

a

a− 1.

A simple calculation shows a ∗ a/(a− 1) = 0. One is not a unit.Part (d) Associative but not commutative.

(a ∗ b) ∗ c = b ∗ c = c = a ∗ c = a ∗ (b ∗ c).

a ∗ b = b 6= a = b ∗ a.There is no unity. If a is the unity and b ∈ M where a 6= b, then a ∗ b = b 6= a (and b cannot be unity since unity is

unique). Since there is no unity, there are no units.Part (e) Associative but not commutative.

((p, q) ∗ (p′, q′)) ∗ (p′′, q′′) = (p, q′) ∗ (p′′, q′′) = (p, q′′) = (p, q) ∗ (p′, q′′) = (p, q) ∗ ((p′, q′) ∗ (p′′, q′′)),

(p, q) ∗ (p′, q′) = (p, q′) 6= (p′, q) = (p′, q′) ∗ (p, q).

There is no unity or units for the same reason as in part (d).Part (f) Associative and commutative. It is trivial to show that max(max(a, b), c) = max(a, b, c). As a result:

(a ∗ b) ∗ c = max(a, b) ∗ c = max(a, b, c) = a ∗max(b, c) = a ∗ (b ∗ c),

a ∗ b = max(a, b) = max(b, a) = b ∗ a.Zero is unity for given any a ∈ Z≥0, it follows that a ≥ 0, so a ∗ 0 = 0 ∗ a = max(0, a) = a. There are no units other

than 0 (obviously 0 ∗ 0 = 1). For if a ∈ N and a > 0, then given any b ∈ Z≥0 it follows that a ∗ b = b ∗ a = max(a, b) > 0.Part (g) This binary operator is associative, as easily shown by application of Proposition 1 below. The operator is

also commutative since gcd(a, b) = gcd(b, a).Unity is 1 since 1 ∗ a = a ∗ 1 = gcd(a, 1) = a. Each a ∈ N is a unit with the inverse equal to any element of N that is

relatively prime to a.

Proposition 1. If a, b, c ∈ Z, then gcd(a, gcd(b, c)) = gcd(gcd(a, b), c).

Proof. Designate:m0 = gcd(a, b) and m1 = gcd(b, c),m2 = gcd(m0, c) and m3 = gcd(a,m1).

By definition, m2 | c and m2 | m0, so m2 divides both a and b. Accordingly, m2 | m1, from which it follows that m2 | m3.Similarly, m3 | a and m3 | m1, so m3 divides both b and c. Therefore m3 | m0, so m3 | m2. Since m2 and m3 divide eachother, they are equal.

Part (h) The operator is associative:

[(x, y) ∗ (x′, y′)] ∗ (x′′, y′′) = (xx′, xy′) ∗ (x′′, y′′) = (xx′x′′, xx′y′′),(x, y) ∗ [(x′, y′) ∗ (x′′, y′′)] = (x, y) ∗ (x′x′′, x′y′′) = (xx′x′′, xx′y′′).

The operator is not commutative:

(x, y) ∗ (x′, y′) = (xx′, xy′) 6= (xx′, x′y) = (x′, y′) ∗ (x, y).

There is no unity. Suppose (a0, a1) ∈ R2 were the unity. Choose distinct (x, y), (x, y′) ∈ R2 where y, y′ 6= 0and y 6= y′. If (x, y) ∗ (a0, a1) = (a0x, a1x) = (x, y), then a1x = y (and clearly a1 6= 0 since y 6= 0). But then(x, y′) ∗ (a0, a1) = (a0x, a1x), so it must be that a1x = y′. But this implies that y = y′, contradicting that y and y′ aredistinct. Hence there is no unity. Because there’s no unity, there are no units. Since there is no unity, there are no units.

Problem 2.1.2.

Solution: Part (a) Suppose x, y, or z equals 1. We may assume without loss of generality that z = 1 because theunity element is commutative; therefore xyz = xzy = zxy (by the Theorem of General Associativity), so this assumptionaccounts for either x or y being equal to 1, as well. Accordingly, (xy)z = (xy)1 = xy = x(y1) = x(yz).

Part (b) Since M is a monoid, 1 ∈M , so there is a single element a in M not equal to 1. Because the binary operationis closed, a2 = a or a2 = 1. Therefore there are two possible monoids with two elements:

1 a1 1 aa a a

1 a1 1 aa a 1

Part (c) First, the operation on M is associative. If a, b, c ∈ S, then by hypothesis (ab)c = a(bc). By hypothesis,1(bc) = bc = (1b)c, a(1c) = ac = (a1)c, and a(b1) = ab = (ab)1. Second, 1 is the unity of M . If s ∈ S, then1s = s1 = s for any s ∈ S. In addition, we will show that 1 · 1 = 1. If 1 · 1 = t ∈ S, then given any u ∈ S we have(u1)1 = u1 = u = u(1 · 1) = ut2 and (1(1u) = 1u = u = (1 · 1)u = t2u. Thus t2 would be the unity of S, which doesnot exist by hypothesis. Accordingly, 1 · 1 = 1, so 1 is M ’s unity. Finally, the operation is closed. Clearly ab ∈ S ⊆ M byvirtue that the operator is closed in S. By defintion, 1a = a1 = a ∈ S ⊆M . Finally, 1 · 1 ∈M .

We conclude that M is a monoid.

Problem 2.1.3.

Solution: Part (a) From the table as filled in, (ab)b = bb = a, so a(bb) = aa = a. We then infer that (bb)b = ab = b, sob(bb) = ba = b. The Cayley table is then:

a ba a bb b a

Clearly the operator is commutative.Part (b) We can follow the same logic as in part (a) to find that there are three possible Cayley tables:

A =a b

a b ab a b

B =a b

a a ab a b

C =a b

a b ab b b

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A and B are commutative with unity b. C is not commutative and has no unity.

Problem 2.1.4.

Solution: Suppose M is a monoid and M is the set of all non-empty subsets of M . Given X,Y ∈M , define the operationXY = {xy : x ∈ X and y ∈ Y }.

The set M is a monoid. First we will prove that the operation is closed. Since x, y ∈M , which is a monoid, xy ∈M .Since X and Y are non-empty, XY is a subset of M and contained in M .

Next, we will show that the operation is associative. We have (XY )Z = {αy : α ∈ XY and z ∈ Z}. DesignateU = {xyz : x ∈ X and y ∈ Y and z ∈ Z}. If αz ∈ (XY )Z, then α ∈ XY , so α = xy where x ∈ X and y ∈ Y .Therefore αz = xyz where x ∈ X, y ∈ Y , and z ∈ Z, so αz ∈ U . Accordingly, (XY )Z ⊆ U . Conversely, if xyz ∈ U , thenx ∈ X, y ∈ Y , and z ∈ Z, so xy ∈ XY and hence xyz ∈ (XY )Z. We conclude that (XY )Z = U . On the other hand,X(Y Z) = {xβ : x ∈ X and β ∈ Y Z}. Following the same argument, we see that X(Y Z) = U , so (XY )Z = X(Y Z).The operation is associative.

Finally, we will show that there is some unity {1} in M . Designate 1 as the unity of M (which must exist since M is amonoid). Given X ∈ M , we have {1} ·X = {1x : x ∈ X} = {x : x ∈ X} = X = {x1 : x ∈ X} = X · {1}. We concludethat {1} is the unity of M .

Units of M . Define V = {m ∈ M : m is a unit in M}. Accordingly, given m ∈ V , there is some n ∈ V such thatmn = nm = 1. Therefore {m} · {n} = {mn} = {1} = {nm} = {n} · {m}, so every set {m} where m ∈ V is a unit inM . By contrast, clearly if m′ ∈ M\V , then m′ has no inverse, so {m′} cannot be a unit in M . In addition, is |N | > 1where N ∈ M , then given any P ∈ M , there is some np ∈ NP where p is not the inverse of n, so np 6= 1 and thereforeNP 6= {1}. Hence {{m}}m∈V is the set of units in M .

Commutativity of M . If M is commutative, then ab = ba for any a, b ∈M . If X,Y ∈M , then XY = {xy : x ∈ Xand y ∈ Y } = {yx : x ∈ X and Y ∈ Y } = Y X. Therefore M is commutative.

Problem 2.1.5.

Solution: Proving that W is a monoid with unity λ is straightforward. If |A| = 1, then W is a commutative monoidbecause there is only one letter in A, which we will designate a, so the juxtaposition of a word of N a’s with a wordof M a’s is the same irrespective of order. If |A| > 1, then the commutativity of the operator in W follows from thecommutativity of the operator in A. The only unit of W is λ since if AB ∈W and AB = λ, then A = B = λ.

Problem 2.1.6.

Solution: Given σ, τ ∈ F and x ∈ X, we have σ(x)·τ(x) ∈M since the binary operator in M is closed. Define α : X →Mwhere α(x) = σ(x) · τ(x) for each x ∈ X. Since α is a map from X to M , it belongs to F . Accordingly, F is closed.

The associativity of F ’s operator follows from the associativity of M ’s operator.Denote 1 ∈M as the unity of M . Define ε : X →M where ε(x) = 1 for all x ∈ X. As a map from X to M , it follows

that ε ∈ F . We then have for any σ ∈ F and x ∈ X:

σ(x) · ε(x) = σ(x) · 1 = σ(x) = 1 · σ(x) = ε(x) · σ(x)

Hence ε is F ’s unity.Let U be the set of units in M . Define β : X →M where the image X under β is a subset of U . As a result, [β(x)]−1

exists for all x. Then define γ : X →M where γ(x) = [β(x)]−1 for each x ∈ X. Accordingly:

β(x) · γ(x) = β(x) · [β(x)]−1 = 1 = [β(x)]−1 · β(x) = γ(x) · β(x).

As maps from X to F , both β and γ are contained in F . Accordingly β is a unit of F . Now let δ ∈ F where δ(X) is not asubset of U . Therefore there is some x ∈ X where δ(x) is not a unit. Consequently, for every ζ ∈ F , either ζ(x) · δ(x) 6= 1or δ(x) · ζ(x) 6= 1; therefore δ cannot be a unit of F . As a result, the units of F are {β : X →M : β(X) ⊆ U}.

Finally, suppose M ’s operator commutes. As a result, σ(x) · τ(x) = τ(x) · σ(x), so F ’s operator commutes.

Problem 2.1.7.

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Solution: The proofs are straightforward. If 1M is the unity of M and 1N is the unity of N , then (1M , 1N ) is theunity of M × N . The monoid M × N is commutative if both M and N are commutative. The units of M × N are{(m,n) ∈M ×N : m is a unit of M and n is a unit of N}.

Problem 2.1.8.

Solution: Part (a)Part (b)

Problem 2.1.9.

Solution: Part (a)(a5)5 = a25 = (a12)2a = (b5)5 = b25 = (b12)2b = (a12)2b

Using left-cancellation, we infer that a = b.Part (b) Suppose m,n ∈ N where m and n are relatively prime. If m = 1 or n = 1, by hypothesis am = a = bm = b, so

we’re done. Assume that both m and n are greater than one. The linear combination 1 = mx+ ny exists where x, y ∈ Z.It must be that exactly one of x and y is less than zero (otherwise, the linear combination could not equal one). We mayassume without loss of generality that y < 0, so −y > 0. The follows that mx = 1 + n(−y). As a result:

(am)x = amx = a1+n(−y) = (an)−ya = (bm)x = bmx = b1+n(−y) = (bn)−yb = (an)−yb.

Applying left-cancellation, we find that a = b.

Problem 2.1.11.

Solution: Suppose e0 and e1 are distinct left unities of an operator. Given any r, we have e0r = r and e1r = r, soe0r = e1r. Because e0 6= e1, it cannot be that r is a right unity. We conclude that the operator has no right unity.

Problem 2.1.12.

Solution: Part (a) Let u be a unit of monoid M , whose inverse we’ll designate v. If au = bu, then:

a = a1 = a(uv) = (au)v = (bu)v = b(uv) = b.

Part (b) Suppose M = {a1, . . . , an} is a finite monoid where if au = bu then a = b. Let aj , ak ∈ M . If aju = aku,then aj = ak. It follows that if aj 6= ak, then aju 6= aku. Consequently, a1u, . . . , anu are distinct.

Since M is finite and all aju are distinct, its follows that the map φu : M → M where φu(a) = au is injective.Therefore φu must be surjective, as well, or the image of two distinct aj , ak under φu would be equal. As a result, thereexists am = φ−1u ({1}), so amu = 1. To establish that u is a unit, all that’s left to show is that uam = 1. We have(uam)u = u(amu) = u1 = 1u, so uam = 1 by hypothesis. Therefore u is a unit with inverse am.

Problem 2.1.13.

Solution: Suppose uv is a unit of monoid M where if av = bv then a = b. Designate ξ as the inverse of uv. We have:

ξ(uv) = (ξu)v = 1, and [v(ξu)]v = v[(ξu)v] = v1 = 1v,

so by hypothesis v(ξu) = 1. Thus v−1 = ξu. Further

(uv)ξ = u(vξ) = 1, and (vξ)u = v(ξu) = vv−1 = 1.

Accordingly, u−1 = vξ, so both u and v are units in M .

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Problem 2.1.14.

Solution: Suppose u has a left inverse v and right inverse w, so uv = wu = 1. Therefore w(uv) = (wu)v = 1v = v =w1 = w. As a result, w is the inverse of u, so u is a unit.

Problem 2.1.15.

Solution: Part (a) Given a monoid M and u ∈ M , define σ : M → M where σ(a) = ua for all a ∈ M . Suppose σ is abijection. There is some a0 ∈ M where σ(a0) = ua0 = 1. In addition, since σ(1) = u, it follows that if σ(a) = 1, thena = 1. Because σ(a0u) = uaou = u, it follows that a0u = 1. Thus u is a unit with inverse a0.

Conversely, suppose u is a unit. Given a ∈ M , we have σ(a) = u. If σ(a) = σ(b), then ua = ub, so u−1(ua) = a =u−1(ub) = b; hence σ is injective. If c ∈ M , there is some a2 ∈ M where u−1c = a2. Therefore ua2 = u(u−1c) = c, soσ(a2) = c; thus σ is surjective. We conclude that σ is a bijection.

Part (b) The inverse is σ−1(a) = u−1a because given a ∈M :

σ−1 ◦ σ(a) = σ−1(ua) = u−1(ua) = (u−1u)a = 1a = a.

Problem 2.1.16.

Solution: This is easily shown by mathematical induction on n.

Problem 2.1.17.

Solution: Suppose u and v are units in M .Part (a) If u−1 = v−1, then (u−1)−1 = u = (v−1)−1 = v by Theorem 2.1.8.Part (b) Let a ∈M and ua = au. It follows that:

u−1(ua)u−1 = (u−1u)au−1 = au−1 = u−1(au)u−1 = (u−1a)(uu−1) = u−1a.

Part (c) If uv = vu, then:(uv)−1 = v−1u−1 = (vu)−1 = u−1v−1.

Problem 2.1.18.

Solution: (1) ⇒ (2). Assume if ab is a unit in monoid M then both a and b are units. Suppose ab = 1. It follows thatab(1) = (1)ab = 1, so ab is a unit with inverse 1. Accordingly, a and b are units, so (ab)−1 = b−1a−1 = 1−1 = 1 byTheorem 2.1.5(1) and (3). We then have:

b(b−1a−1)a = (bb−1)(a−1a) = 1 = b1a = ba.

(2) ⇒ (1). Assume if ab = 1 then ab = 1. Suppose ab is a unit with inverse ξ. Therefore ab(ξ) = a(bξ) = 1. Sincebξ ∈M , then (bξ)a = 1, so a is a unit. Similarly, ξ(ab) = (ξa)b = 1, so by the same argument b is a unit.

Problem 2.1.20.

Solution: Part (a) Let a, b, c ∈ M be given. Since 1 ∈ M is a unit, a = a1, so a ∼ a. If a ∼ b, then a = bu0 forsome unit u0, so au−10 = (bu0)u−10 = b(u−10 u0) = b. Because u−10 is a unit, b ∼ a. Finally, if a ∼ b and b ∼ c, there areunits u1 and u2 such that a = bu1 and b = cu2. As a result, au−11 u−12 = bu1(u−11 u−12 ) = bu−12 and bu−12 = cu2u

−12 = c.

Accordingly, au−11 u−12 = c, so a ∼ c. We conclude that ∼ is an equivalence on M .Part (b) For a, b ∈M , define the operation ab = ab. To show that this operation is well-defined, we must establish that

it results in exactly one equivalence class ab contained in M . We have a = {α ∈ M : α ∼ a} and b = {β ∈ M : β ∼ b}.

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Since ab ∈M because M is a monoid, it follows that ab is contained in M . Further, since ab results in a single element inM , it follows that ab results in a single equivalence class in M . The operation is well-defined.

Part (c) We will show that M is associative. Let a, b, c ∈ M . It follows that (ab)c = abc = (ab)c . Therefore(ab)c = {β ∈ M : (ab)c ∼ β}. Conversely, a(bc) = abc = a(bc) . Therefore a(bc) = {β ∈ M : a(bc) ∼ β}. Because(ab)c = a(bc), it follows that (ab)c = a(bc). The operator is associative.

Next, we will establish that 1 is M ’s unity. If a ∈ M , the 1a = 1a = a = a1 = a1. By Theorem 2.1.5(1), 1 is also aunit.

Now suppose that b ∈ M is a unit. Observe that if x ∈ 1, then x ∼ 1, so x = 1 · u = u where u is a unit, so x is aunit. Further if u is a unit, then u = 1 · u, so u ∈ 1. Therefore 1 is exactly the set of all units in M and thus containsb by hypothesis. But b is also contained in b (since b ∼ b). Because equivalence classes are either disjoint or equal (seeTheorem 1 below), we conclude that 1 = b. Therefore 1 is the only unit in M .

Finally, we will show that M is commutative. Since ab = ba because M is commutative, ab = ab = ba = ba.

Theorem 1. Given any equivalence relation, two equivalence classes a and b are either disjoint or equal.

Proof. Suppose a and b are not disjoint, in which case there is some x ∈ a ∩ b. Therefore x ∼ a and x ∼ b. If y ∈ a, theny ∼ x (with the order of operands allowed by commutativity), so y ∼ b by transitivity, from which it follows that y ∈ b.Similarly, if z ∈ b, then z ∼ x, so x ∼ a and x ∈ a. We conclude that a equals b if the two equivalence classes are notdisjoint.

Problem 2.1.21.

Solution: Part (a) Let α, β ∈ E(M). Given x, y ∈M :

αβ(xy) = α ◦ β(xy) = α(β(x) · y) = α(β(x)) · y,

which is an element of E(M). Therefore the operation over E(M) of composition of mappings is closed.Since function composition is associative, the operation over E(M) is associative.Finally, the identity map 1M : M → M where 1M (x) = x for all x ∈ M is the unity of E(M). Given α ∈ E(M) and

x ∈M :1Mα(x) = 1M ◦ α(x) = 1M (α(x)) = α(x) = α(1M (x)) = α ◦ 1M (x) = α1M (x).

Accordingly, E(M) is a monoid.Part (b) For α ∈M define αa : M →M as αa(x) = ax for x ∈M . Given x, y ∈M :

αa(xy) = a(xy) = (ax) · y = αa(x) · y,

which is contained in E(M).Part (c) We will show that θ is surjective. Designate T = {αa : a ∈ M}. In part (b), we showed that T ⊆ E(M).

Now we will show that the two sets are equal. Given α ∈ E(M), let a = α(1). For x ∈ M , we have α(x) = α(1 · x) =α(1) · x = ax = αa(x). Accordingly α ∈ T , so E(M) is a subset of T . We conclude that E(M) = T .

Now let α ∈ E(M). There is some a ∈M where α = αa. It follows that θ(a) = αa = α, so θ is surjective.The map θ is also injective. If θ(a) = θ(b), then αa = αb. Therefore αa(1) = a = αb(1) = b, so a = b. Observe that

θ is a bijection.We have θ(1) = α1 where α1(x) = x for x ∈M . But 1M (x) = x, so α1 = 1M .Finally, we will show that θ(ab) = θ(a) · θ(b). Let a, b ∈ M . We have θ(ab) = αab where αab(x) = (ab)x. Further,

θ(a) = αa and θ(b) = αb where αa(x) = ax and αb(x) = bx. We thus have αa(x) · αb(x) = αa ◦ αb(x) = αa(bx) =a(bx) = (ab)x = αab(x). Hence θ(ab) = θ(a) · θ(b).

Section 2.2

Problem 2.2.1.

Solution: We’ll sketch answers here.Part (a) Not a group. 0 is the unity of G. Given x ∈ Z>0, for any y ∈ Z≥0 we have x + y ≥ x > 0, so x has no

inverse.

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Part (b) This is a group. 0 is the unity.Part (c) This is a group. The unity is −1, and the inverse of any a ∈ G is −2− a.Part (d) Not a group. 0 is the unity. For any a, the inverse must be of the form a/a− 1, so a = 1 has no inverse.Part (e) Not a group because the operation is not closed. Observe that (1 2)(1 3) = (1 3 2) /∈ G. However, note that

ε is G’s unity and any non-identity element of G is a transposition and therefore is its own inverse.Part (f) This is a group. 0 is the unity, and every element has an inverse.Part (g) This is a group. Observe that 16 is the unity and every element has an inverse.Part (h) This is a group.Part (i) Not a group because not every element has an inverse. Define α : Z≥0 → Z≥0 as α(n) = n+ 1. Clearly α is

injective, so α ∈ G. Assume β : Z≥0 → Z≥0 is α’s inverse. Since β must belong to G, the map is injective. For n > 0, itmust be that β(n) = n− 1. As a result, for n ∈ Z≥0:

β ◦ α(n) = β(n+ 1) = (n+ 1)− 1 = n.

But then if m = β(0) ∈ Z≥0, then β(m+ 1) = m and m+ 1 6= 0, which contradicts that β is injective. Therefore α hasno inverse.

Part (j) Not a group. The element a appears twice in the d column, contracting the corollary to Theorem 2.1.1 thateach column of the Cayley table contains each element of a group exactly once.

Problem 2.2.2.

Solution: This is a straighforward proof. Let a, b, c ∈ Gop. It follows that a, b, c ∈ G. Since G is a group, we can leverageits properties as a group.

• Operation Is Closed. We have a ◦ b = ba ∈ G because G is a group by hypothesis. Since every element of G is anelement of Gop, the ◦ operation is closed in Gop.

• Operation Is Associative. We also have:

(a ◦ b) ◦ c = (ba) ◦ c = cba = a ◦ (cb) = a ◦ (b ◦ c),

so the operation is also associative.

• 1 is Unity of Gop. Designate 1 as the unity of G, in which case 1 ∈ Gop. Therefore a ◦ 1 = 1a = a = a1 = 1 ◦ a,so 1 is the unity of Gop.

• Everything Element Has an Inverse. Each a ∈ Gop has an inverse a−1 in G because G is a group. Thereforea ◦ a−1 = a−1a = 1 = aa−1 = a−1 ◦ a, so a−1 is the inverse of a in Gop.

We conclude that Gop is a group.

Problem 2.2.3.

Solution: Part (a) The complete table is:

1 a b c d1 1 a b c da a d 1 b cb b 1 c d ac c b d a 1d d c a 1 b

Here are the chains of deduction:

1. Since ab = 1, it follows that ba = 1 because a and b are inverses.

2. ad 6= d because that would imply that a = 1 (see Theorem 2.2.6 for cancellation laws in groups; this rule is usedthrough the analysis). The only remaining option is that ad = c, from which it follows that aa = d by the corollaryto Theorem 2.2.6, which requires that every row and column have each element exactly once.

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3. ca 6= c, so a = 1, from which we infer that ca = b and da = c.

4. cb 6= c, so cb ∈ {a, d}. Further, bb 6= b, so bb ∈ {c, d}, and db 6= d, so db ∈ {a, c}. But da = c, so by the corollaryto Theorem 2.2.5 db = a. It follows that cb = d and bb = c.

5. Every element other than c and d are inverses. Since c and d’s inverses are unique, their inverses cannot be 1, a, orb. It follows that c and d are inverses of each other, so cd = dc = 1.

6. We can fill in the rest of the table by checking which elements are missing from a row or column. We find thatdd = b, cc = a, bc = d, and bd = a.

Part (b) The complete table is:

1 a b c d1 1 a b c da a d 1 b cb b 1 c d ac c b d a 1d d c a 1 b

Here are the chains of deduction:

1. ba 6= c (since then a would equal 1), so ba = 1 (and hence ab = 1 by symmetry of inverses).

2. bd = a by the corollary to Theorem 2.2.5

3. ac 6= c, so ac ∈ {b, d}. Since bc = d by the corollary to Theorem 2.2.5, ac = b.

4. Since a and b are inverses of each other and 1 is its own inverse, it follows by the uniqueness of inverses that c andd are inverses of each other. Therefore cd = dc = 1.

5. By the corollary to Theorem 2.2.5, cc = a.

6. Since cb 6= b, it follows that cb = d. Then ca = b.

7. ad 6= d, so ad = c and cb = d.

8. Looking at what is remaining, w see that da = c, db = a, and dd = b by the corollary to Theorem 2.2.5.

Problem 2.2.4.

Solution: The empty set is not a group. A group must have some identity element; axiom G3 is clear that this elementmust exist). Since the empty set has no elements, it cannot have an identity element, hence it is not a group.

Problem 2.2.5.

Solution: Each row should have exactly one 1, and no two rows should have a 1 in the same column.

Problem 2.2.6.

Solution: First, let’s change the expression of the result of the operation ⊕ to a slightly different form. We will show thatA ⊕ B = (A ∪ B)\(A ∩ B). If u ∈ A ⊕ B = (A\B) ∪ (B\A), then either x ∈ A but x /∈ B or x ∈ B but x /∈ A. As aresult, x ∈ A ∪ B but x /∈ A ∩ B, so x ∈ (A ∪ B)\(A ∩ B). Conversely, if x′ ∈ (A ∪ B)\(A ∩ B), then x ∈ A ∪ B butx /∈ A ∩B. Therefore x is in A or B but cannot be in both, so x ∈ A\B or x ∈ B\A. Accordingly x ∈ (A\B) ∪ (B\A).We conclude that (A\B) ∪ (B\A) = (A ∪B)\(A ∩B).

Let A,B,C ∈ G. We have A ⊕ B = (A ∪ B)\(A ∩ B), which is a subset of U ; hence the operation in closed in G.The operation is also associative:

(A⊕B)⊕ C = [(A ∪B)\(A ∩B)]⊕ C = ((A ∪B) ∪ C)\((A ∩B) ∩ C)

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= (A ∪ (B ∪ C))\(A ∩ (B ∩ C)) = A⊕ [(B ∪ C)\(B ∩ C)] = A⊕ (B ⊕ C).

The operator ⊕ is also commutative:

A⊕B = (A ∪B)\(A ∩B) = (B ∪A)\(B ∩A) = B ⊕A.

The empty set is an element of G because the empty set is a subset of any set. We have:

A⊕ ∅ = (A ∪ ∅)\(A ∩ ∅) = A\∅ = A = (∅ ∪A)\(∅ ∩A) = ∅ ⊕A.

Therefore ∅ is the unity of G.Finally, given A ∈ G:

A⊕A = (A ∪A)\(A ∩A) = A\A = ∅.

Thus every element of G is its own inverse. We conclude that G is an abelian group.

Problem 2.2.7.

Solution: Throughout this proof, we designate A,B,C ∈ G as:

A =

1 a0 a10 1 a20 0 1

, B =

1 b0 b10 1 b20 0 1

, C =

1 c0 c10 1 c20 0 1

.

Each matrix is an element of R3×3.First we will show that the operation of matrix multiplication in this group is closed. We have:

AB =

1 a0 + b0 b1 + a0b2 + a10 1 a2 + b20 0 1

.

This is an upper diagonal matrix with 1’s along the diagonal and all real entries; therefore AB ∈ G.Second, the operation is associative because matrix multiplication is associative.Third, G has the unity element I (the identity matrix of R3×3). As true for any matrix, AI = IA = A.Finally, every matrix in G has an inverse. For any A, define D as:

D =

1 −a0 a0a2 − a10 1 −a20 0 1

.

A simple calculation shows that AD = DA = I, so D is the inverse of A. We conclude that G is a group.

Problem 2.2.8.

Solution: Part (a) Let σ = (1 2)(3 4), τ = (1 3)(2 4), and µ = (1 4)(2 3). The Cayley table is:

ε σ τ µε ε σ τ µσ σ ε µ ττ τ µ ε σµ µ τ σ ε

Obviously the operation is closed, ε is the unity, and every element has an inverse. The operation inherits associativityfrom the associativity of the product of permutations. G is a group.

ε2, σ2, τ2, and µ2 all equal ε.Part (b) Let σ = (1 2 3 4), τ = (1 3)(2 4), and µ = (1 4 3 2). The Cayley table is:

ε σ τ µε ε σ τ µσ σ τ µ ετ τ µ ε σµ µ ε σ τ

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The operation is closed, ε is the unity, and every element has an inverse. The operation inherits associativity from theassociativity of permutation products. It is a group.

ε2 = τ2 = ε.

Problem 2.2.9.

Solution: Observe that σ6 = ε. The set G with the operation is an abelian group:

1. The operation is closed. Given τ ∈ G, it follows that τε = ετ = τ . Given σj , σk ∈ G where j, k ∈ {1, 2, 3, 4, 5},it follows that σjσk = σj+k. By the division algorithm, j + k = 6q + r where q ∈ Z and 0 ≤ r ≤ 5. If r = 0, thenσj+k = σ6q = (σ6)q = εq = ε. On the other hand, if r 6= 0, then σj+k = σ6qσr = (σ6)qσr = εσr = σr. Thereforethe operation over any two elements of G results in an element of G.

2. The operation is associative. It inherits this property from the associativity of the product of permuations.

3. The operation is commutative. The operation of any element of G with ε is commutative. Given σj , σk ∈ Gwhere j, k ∈ {1, 2, 3, 4, 5}, we have σjσk = σj+k = σk+j = σkσj .

4. G has a unity element. Obviously ε is the unity.

5. Every element has an inverse. The inverse of ε is ε. Given σj ∈ G where j ∈ {1, 2, 3, 4, 5}, it follows thatσjσ6−j = σ6 = ε = σ6−jσj . Since 1 ≤ 6− j ≤ 5, it must be that σ6−j is contained in G and is σj ’s inverse.

If τ ∈ {ε, σ3}, then τ2 = ε. If τ ∈ {ε, σ2, σ4}, then τ3 = ε.

Problem 2.2.10.

Solution: Part (a) Suppose a4 = 1 and ab = ba2 in a group. We have:

(a3)ab = a4b = b = a3(ba2) = a3ba2 = a2(ab)a2 = a2ba4 = a2b.

By the Cancellation Law, a2 = 1. Therefore:

ab = ba2 = b1 = b.

Again applying the Cancellation Law, a = 1.Part (b) Suppose a6 = 1 and ab = ba3 in a group. We have:

(ab)a3 = aba3 = a(ab) = a2b = (ba3)a3 = ba6 = b.

By the Cancellation Law, a2 = 1. Using this result, we have:

ab = ba3 = ba(a2) = ba1 = ba.

Part (c) Suppose a6 = 1 and ab = ba2. We have:

(ab)a4 = aba4 = (ba2)a4 = ba6 = b.

Further:aba4 = a(ba2)a2 = a(ab)a2 = a2ba2 = a2(ab) = a3b,

so a3b = b. By the Cancellation Law, a3 = 1. Using this result:

(ab)a = aba = (ba2)a = ba3 = b1 = b.

Problem 2.2.11.

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Solution: Part (a) Suppose (ab)n = 1 in some group for n ≥ 0. The proposition is trivial for n = 0. We will prove bymathematical induction that (ba)n = 1 for n ≥ 1. We have (ab)1 = ab = 1, so (ab)a = a(ba) = 1a = a1. By the Law ofCancellation, ba = 1. Now assume the inductive hypothesis holds for some n ∈ N. We have:

(ab)n+1 = (ab)nab = (ba)nab = (ba)nba = (ba)n+1 = 1,

where we used the inductive hypothesis and the fact that ba = 1 = ab. The proposition holds for all n ≥ 0.Part (b) Let n < 0. Using the result in part (a) and Theorem 2.2.3(5):

(ba)−n = [(ba)n]−1 = 1−1 = 1.

Problem 2.2.12.

Solution: Let G = {1, a, b, c} be a group where a2 = b2 = 1. Since 12 = 1 as well, it must be that c2 = 1 (otherwise, cwould be an inverse of a, b, or 1, which is impossible because their inverses are unique). It cannot be that ab = b or ab = b(since then b = 1 or a = 1, respectively), so ab = c. By the same logic, ba = c. We can then fill in the Cayley table forthe single missing elements in each row and column, so ac = b, ca = b, cb = a, and bc = a.

Since c = ab, the Cayley table is:

1 a b ab1 1 a b aba a 1 ab bb b ab 1 aab ab b a 1

Note that G is commutative and every element is its own inverse.

Problem 2.2.13.

Solution: Let α : G→ G where α(g) = g−1 for g ∈ G. If α(g0) = α(g1), then g−10 = g−11 , so g0g−10 g1 = g1 = g0g

−11 g1 =

g0, hence α is injective. Given g2 ∈ G, it follows that α(g−12 ) = (g−12 )−1 = g2 by Theorem 2.2.3(2). Thus α is surjective.

Problem 2.2.14.

Solution: Let G be a group and a, b, c ∈ G. Suppose a−1x0b = a−1x1b = c for some x0, x1 ∈ G. We have:

a(a−1x0b)b−1 = (aa−1)x0(bb−1) = x0 = a(a−1x1b)b

−1 = (aa−1)x1(bb−1) = x1,

so x0 = x1. The solution x to a−1xb = c is unique.

Problem 2.2.15.

Solution: Let G be a group with some element a, and let X be a finite subset of G. Define Xa = {xa : x ∈ X} (notethat Xa is the left coset of X generated by a). We will show that there is a bijection from X to Xa. Define φa : X → Xasuch that φa(x) = xa for x ∈ X. Given y ∈ Xa, by definition there is some x ∈ X such that xa = y. Thereforeφa(x) = xa = y, so φa is surjective. Now suppose φa(x0) = φa(x1) for some x0, x1 ∈ X. Accordingly, x0a = x1a, so byright cancellation x0 = x1, and φa is injective. It follows that φa is a bijection, and by Theorem 2, X and Xa have thesame cardinality.

Theorem 2. Given finite sets A and B, if there is a bijection ζ : A→ B then A and B have the same cardinality.

Proof. Let |A| = M and |B| = N . Therefore there is are bijections α : A → SM and β : B → SN where Sq = {n ∈N : n ≤ q} (i.e., a section of natural numbers). If there is a bijection ζ : A → B, then β ◦ ζ : A → SN is a bijection, so|A| = N . Consequently, |A| = |B|, so the two sets have the same cardinality..

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Problem 2.2.16.

Solution: Suppose fgh = 1 in some group. Therefore:

(fgh)f = f(ghf) = 1f = f.

By left cancellation, ghf = 1.It is not the case that fgh = 1 implies that gfh = 1. We know from exercise 2.2.7 that U as the subset of R3×3

consisting of only upper triangular matrices with 1’s along the diagonal is a group under matrix multiplication with unityI. Define F,G,H ∈ U as:

F =

1 2 20 1 20 0 1

, G =

1 5 60 1 −30 0 1

, H =

1 −7 −90 1 10 0 1

.

Observe that GH is the inverse of F . Therefore FGH = GHF = I. However:

GFH =

1 0 160 1 00 0 1

6= I.

However, if a group is abelian, then the proposition holds, for if fgh = 1, then gfh = 1 by the commutativity of theoperator. Obviously this doesn’t work for U because matrix multiplicative is not commutative.

Problem 2.2.17.

Solution: Let G be a group and suppose e ∈ G where e2 = e. This is equivalent to e2 = ee = e1. By the Law ofCancellation, e = 1, so 1 is the only idempotent in G.

Problem 2.2.18.

Solution: Let G be a group and g, h ∈ G. If gh = hg, then (gh)−1 = h−1g−1 = (hg)−1 = g−1h−1 by Theorem 2.2.3(3).Conversely, if g−1h−1 = h−1g−1, then (hg)−1 = (gh)−1 by Theorem 2.2.3(3), so ((hg)−1)−1 = hg = ((gh)−1)−1 = gh.

Problem 2.2.19.

Solution: Given a group G, suppose G is abelian. If g, h ∈ G, then gh = hg, so (gh)−1 = h−1g−1 = (hg)−1 = g−1h−1.Conversely, suppose if g, h ∈ G then (gh)−1 = g−1h−1. It follows that (gh)−1 = (hg)−1 by Theorem 2.2.3(3). Therefore((gh)−1)−1 = gh = ((hg)−1)−1 = hg, so G is abelian.

Problem 2.2.20.

Solution: Suppose G is a group where x2 = 1 for any x ∈ G. Given g, h ∈ G:

h1h = hg2h = (hg)(gh) = h2 = 1.

Multiplying both sides by hg and observing that (hg)2 = 1 by hypothesis:

(hg)(hg)(gh) = (hg)2(gh) = 1(gh) = gh = (hg)1 = hg.

We conclude that G is abelian.The converse is not true. The set Z under the operation of addition is an abelian group (it is closed, addition is

associative and commutative, 0 is the unity, and the inverse of any n is −n). However, 1 ∈ Z but 1 + 1 = 2 6= 0.

Problem 2.2.21.

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Solution: Let G be a group and gh ∈ G. If G is abelian, then (gh)2 = (gh)(gh) = g(hg)h = g(gh)h = g2h2. Conversely,if (gh)2 = g2h2, then (gh)2 = g(hgh) = g(ghh), so by left cancellation hgh = (hg)h = ghh = (gh)h. Applying rightcancellation, hg = gh. Therefore G is abelian.

Problem 2.2.23.

Solution: Part (a) Let G be a group with element g. If g2 = 1, then g2g−1 = g = 1g−1 = g−1. Conversely, if g = g−1,then gg = g2 = g−1g = 1.

Part (b) Suppose G is a group with a finite, even cardinality |G| = 2k where k ∈ N (since G cannot be empty, seeexercise 2.2.4). Assume every g 6= 1 has an inverse where g−1 6= g. We know that if h 6= g then g−1 cannot be the inverseof h, for if hg−1 = 1, then hg−1g = h = g. As a result, each element of G\{1} may be paired up with exactly other oneelement of G\{1} that is its inverse, and each element of G\{1} belongs to exactly one pair. This requires an even numberof elements, but |G\{1}| is odd. Therefore it cannot be that every g 6= 1 has an inverse g−1 6= g. We conclude that thereis some element ξ 6= 1 where ξ = ξ−1. From part (a), ξ2 = 1.

Problem 2.2.24.

Solution: Let a and b be elements of a group G. Obviously (aba−1)0 = 1 = ab0a−1 = aa−1. We will prove bymathematical induction on k ∈ N that (aba−1)k = abka−1. For k = 1, we have (aba−1)1 = ab1a−1. If the inductivehypothesis holds for some k ∈ N, then:

(aba−1)k+1 = (aba−1)k(aba−1) = (abka−1)(aba−1) = abk(a−1a)(ba−1) = abkba−1 = abk+1a−1.

Hence the inductive hypothesis holds for all k ∈ N.Given k ∈ Z<0, let j = −k so j ∈ N, in which case:

(aba−1)k = (aba−1)−j =[(aba−1)j

]−1= (abja−1)−1 = (a−1)−1b−ja−1 = ab−ja−1 = abka−1.

The proposition holds for all k ∈ Z.

Problem 2.2.26.

Solution: Let Cn be a cyclic group of order n were Cn = 〈a〉 where an = 1. Given b, c ∈ Cn, it follows that b = aj andc = ak where j, k ∈ {0, 1, . . . , n− 1}. Therefore:

bc = ajak = aj+k = ak+j = akaj = cb.

The cyclic group G is abelian.

Problem 2.2.27.

Solution: From example 2.2.6, we know that Z/nZ under addition is an abelian group with unity 0. If a ∈ Z/nZ, thena ∈ {0, . . . , n− 1}. If a 6= 0, then

∑aj=1 1 = a. If a = 0, then

∑nj=1 1 = n = 0. Therefore every element of Z/nZ is a

”power” of 1, hence 1 is its generator. It follows that Z/nZ is cyclic.

Problem 2.2.28.

Solution: Let G be a group with elements a and b, and suppose that an = bn and am = bm where gcd(m,n) = 1. FromTheorem 1.2.4, there exist x, y ∈ Z such that 1 = mx+ ny. Accordingly:

a = a1 = amx+ny, and b = b1 = bmx+ny.

We have:a = amx+ny = amxany = (am)x(an)y = (bm)x(bn)y = bmxbny = bmx+ny = b,

so a = b.

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Problem 2.2.29.

Solution: Part (a) Suppose G has a left unity and every element has a left inverse. First we will establish left cancellation.Let f, g, h ∈ G and k be the left inverse of h. If gf = gh, then:

k(gf) = (kg)f = ef = f = k(gh) = (kg)h = eh = h.

Any g ∈ G has a left inverse h, so:

(hg)e = h(ge) = e · e = e(hg) = (eh)g = hg.

By left cancellation, ge = g, so e is the unity of G.Further:

(hg)h = h(gh) = eh = h = he.

By left cancellation, gh = e, so h is the inverse of g. We could that G is a group.Part (b) Suppose G is finite and both left and right cancellation hold. Given g ∈ G, it cannot be that each gn for

n ∈ N is distinct because G is finite. Therefore there is some m ∈ N such that gm+n = gmgn = gn. By right cancellation,since gm must equal some element of G, it follows that gm must equal some eg ∈ G such that egg = g (i.e., eg is a leftunity of g). Analogously, gm+n = gngm = gn, so by left cancellation gm must equal some element e′g ∈ G such thatge′g = g (i.e., e′g is a right unity of g). We infer that e′g = eg, and we will designate this element e.

Now we must show that e is a left and right unity of every element in G. Let h ∈ G. We have hgm = hgn = h(egn) =(he)gn. By right cancellation, he = h. Similarly, gmh = gnh = (gne)h = gn(eh), so by left cancellation h = eh. Weconclude that e is the unity of G.

Because G has an associative operator that is closed by hypothesis and has a unity, G is a monoid. Let u ∈ G be given.Because right cancellation holds in G, the equality au = bu implies that a = b. Applying exercise 2.1.12(b), u is a unit.Hence every element of G has an inverse. We conclude that G is a group.

Part (c) Suppose for all g, h ∈ G the equations gx = h and xg = h are solvable. It follows that there is some eg ∈ Gsuch that geg = g. Further, for some j ∈ G, it must be that eg = jg = eg, so jg = jgeg = e2g. Thus eg is idempotent.Further, given h ∈ G, it follows that there is some x ∈ G such that egx = h. Therefore:

egh = eg(egx) = e2gx = egx = h,

so egh = h for any h ∈ G. Thus eg is the left unity for h, which we will designate e.Following a similar argument, there is some x′ ∈ G such that h = x′e, so:

he = (x′e)e = x′e2 = x′e = h,

so he = h, so e is the right unity for any h ∈ G. We conclude that e is the unity of G.Given an h ∈ G, there is some j ∈ G such that hj = e, so j is the right inverse of h. Similarly, there is some j′ ∈ G

such that j′h = e. Then:(hj)j = ej = j = (j′h)j = j′(hj) = j′e = j′.

Consequently, j′ = j, so for every h ∈ G there is a j such that hj = jh = e; hence every element of h has an inverse. Weconclude that G is a group.

Part (d) Suppose gx = h has a unique solution for all g, h ∈ G. As an initial matter, we follow the same argument asin the beginning of the solution for part (c) to find that there is a left unity e ∈ G such that eg = g for all g ∈ G.

Now the argument departs from that in part (c). Given g ∈ G, by hypothesis there is some x ∈ G that uniquely solvesex = g. Therefore:

g · g = g2 = (ex)2 = (ex)(ex) = e(xe)x = (ex)x = gx.

Therefore gx = g2. But gg = g2, and since the solution to this equation is unique, it follows that x = g. Therefore eg = g,so e is a right inverse. Thus e is the unity of G.

Now let h ∈ G. There is some y ∈ G that uniquely solves hy = e. We then have:

(hy)h = h(yh) = eh = he.

Since he = h and h(yh) = h, we infer that yh = e because the solution is unique. Therefore hy = yh = e, so y is theinverse of h. Hence every element of G has an inverse, so G is a group.

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Section 2.3

Problem 2.3.1.

Solution: Part (a) Not a subgroup because the operation is not closed (e.g., 1 + 1 = 2 /∈ H).Part (b) H is a subgroup of G. Under multiplication 1 is the unity of G and is contained in H. Simple calculations

show multiplication is closed in H. Finally, for every element in H its inverse in G is contained in H.Part (c) Not a subgroup because the operation is not closed (e.g., 3 · 3 = 9 /∈ H).Part (d) Not a subgroup because the operation is not closed (e.g., (1 2 3)(1 2 3) = (1 3 2) /∈ H).Part (e) Not a subgroup because the operation is not closed (e.g., [(1 2)(3 4)][(1 3)(2 4)] = (1 4)(2 3) /∈ H).Part (f) H is a subgroup of G. The unity of G (the identity matrix I) is contained in H. Straightforward calculations

show that the product of each element of H is in H. Finally, the inverse in G of every element in H is contained in H.Part (g) H is a subgroup of G. 0 = 6 is the unity of Z/6Z and is contained in H. Calculations show H is closed under

addition. The inverses in G of each element in H are contained in H: 2−1

= 4, 4−1

= 2, 6−1

= 6.Part (h) Not a subgroup because the operation is not closed. Given any x ∈ Z>0, if y ∈ Z≥0 then x+ y ≥ x 6= 0.Part (i) H is a subgrou of G. Note the the operation in G and H is to add the components of two tuples element by

element (e.g., (1, 2) · (3, 4) = (1 + 3, 2 + 4) = (4, 6)).The element (0, 0) is the unity of G. Because 0 + 0 is even, (0, 0) ∈ H. Let a = (a0, a1), b = (b0, b1) ∈ H. We will

show that either both a0 and a1 are even (in which case we will say a is even) or both a0 and a1 are odd (in which casewe will say a is odd), with the same for (b0, b1). Obviously the sum of two even numbers or the sum of two odd numbersis even. If a0 were even and a1 where odd, then a0 + a1 = 2k0 + (2k1 + 1) = 2(k0 + k1) + 1 for some k0, k1 ∈ Z, which isodd, so (a0, a1) /∈ H. As a result there are three possible cases for ab = (a0, a1) · (b0, b1):

1. a and b are both even. It follows that a0 = 2k0, a1 = 2k1, b0 = 2l0 and b1 = 2l1 for some k0, k1, l0, l1 ∈ Z.Therefore (a0 + b0) + (a1 + b1) = 2(k0 + k1 + l0 + l1), which is even.

2. a and b are both odd. Since a0 = 2k0 + 1, a1 = 2k1 + 1, b0 = 2l0 + 1 and b1 = 2l1 + 1 for some k0, k1, l0, l1 ∈ Z,then (a0 + b0) + (a1 + b1) = 2(k0 + k1 + l0 + l1 + 2), which is even.

3. a is even and b is odd. Since a0 = 2k0, a1 = 2k1, b0 = 2l0 + 1 and b1 = 2l1 + 1 for some k0, k1, l0, l1 ∈ Z, then(a0 + b0) + (a1 + b1) = 2(k0 + k1 + l0 + l1 + 1), which is even. The case where a is odd and b is even is analogous.

It follows that ab ∈ H for all a, b ∈ H. The set H is closed.Finally, given (a0, a1) ∈ H, its inverse in G is (−a0,−a1). Since a0 + a1 is even by virtue of its inclusion in H, it

follows that (−a0) + (−a1) = −(a0 + a1) is also even, so (−a0,−a1) ∈ H. By the Subgroup Test, H is a subgroup of G.

Problem 2.3.2.

Solution: Let H be a subset of group G. Suppose H is a subgroup of G. It cannot be that H is empty because then itwould not contain a unity, which is necessary for any group. If a, b ∈ H, then b has an inverse b−1 in H, so by closureab−1 ∈ H. Conversely, suppose H is nonempty and a, b−1 ∈ H for any a, b ∈ H. Therefore aa−1 = 1 ∈ H, so H containsthe unity of G. In addition, given any b ∈ H, it follows that 1b−1 = b−1 ∈ H, so H contains the inverse in G of everyelement in H. Finally, given a, b ∈ H, since b−1 ∈ H we have a(b−1)−1 = ab ∈ H, so H is closed. By the Subgroup Test,H is a subgroup of G.

Problem 2.3.3.

Solution: K is a subgroup of G. The subgroup K must contains the unity 1H of H, which serves as K’s unity. Thesubgroup H contains the unity 1G of G, which serves as H’s unity. Since unities are unique under Theorem 2.1.1, 1G = 1H ,so K contains 1G, which is K’s unity. The subgroup K is closed under its operation by definition. Finally, if k ∈ K, thenits inverse k−1 in H is contained in K, and since k ∈ H, it follows that k−1 is k’s inverse in G. By the Subgroup Test, Kis a subgroup of G.

Problem 2.3.4.

Page 15

Solution: The group SX is the set of all permutations of the elements of R\{0, 1}. The operation is composition of thepermutation functions. The unity ε of SX is an element of G. The operation in G is closed. For example:

µ2λ1 =1

1−x1

1−x − 1=

1

x= µ1.

Here is the Cayley table:

ε λ1 λ2 µ1 µ2 µ3

ε ε λ1 λ2 µ1 µ2 µ3

λ1 λ1 λ2 ε µ2 µ3 µ1

λ2 λ2 ε λ1 µ3 µ1 µ2

µ1 µ1 µ3 µ2 ε λ2 λ1µ2 µ2 µ1 µ3 λ1 ε λ2µ3 µ3 µ2 µ1 λ2 λ1 ε

Finally, every element in G has an inverse, and that inverse is the same as in SX .

Problem 2.3.5.

Solution: Part (a) Suppose G is an abelian group and H = {a ∈ G : a2 = 1}. Since 12 = 1, it follows that 1 ∈ H. Givena, b ∈ G, it must be that a2 = b2 = 1. Therefore:

(ab)2 = (ab)(ab) = a(ba)b = a(ab)a = a2b2 = 12 = 1,

where we relied upon the commutativity of the operation in G. Thus ab ∈ G. Finally, given c ∈ H, since c2 = 1, we havec2c−2 = 1 = c−2 = (c−1)2, so c−1 ∈ H. By the Subgroup Test, H is a subgroup of G.

Part (b) Note that S3 is a non-abelian group. Let H = {ε, (1 2), (1 3) (2 3) }, which is a subset of S3. Since Hconsists of the identity permutation and single transpositions, it follows that the square of any element in H equals theidentiy permutation. But (1 2)(2 3) = (1 2 3) /∈ H, so H is not closed. Therefore H is not a subgroup of S3.

Problem 2.3.6.

Solution: Part (a) Suppose G is an abelian group and H = {g2 : g ∈ G}. Any element of H is of the form x2 where x ∈ G.Obviously 12G (the unity of G) belongs to H. Given a2, b2 ∈ H, we have a2b2 = (aa)(bb) = a(ab)b = a(ba)a = (ab)2.Since ab ∈ G, it follows that (ab)2 = a2b2 ∈ H. Finally, if d2 ∈ H, then in G the element d−2 is its inverse (sinced2d−2 = d0 = 1G). Since (d2)−1 = d−2 = (d−1)2 and d−1 ∈ G, it follows that (d−1)2 ∈ H. By the Subgroup Test, H isa subgroup of G.

Part (b) Let G = Z with the operation of addition, which is an abelian group. Then H = {n2 : n ∈ Z}. It followsthat 12 and 22 are contained in H, but 12 + 22 = 5 /∈ H because

√5 /∈ Z. We conclude that H is not a subgroup of G.

Part (c) A4 = {ε, (1 2 3), (1 2 4), (1 3 4), (2 3 4), (1 3 2), (1 4 2), (1 4 3), (2 4 3), (1 2)(3 4), (1 3)(2 4), (1 4)(23) }.

H = {ε, (1 3 2), (1 4 2), (1 4 3), (2 4 3), (1 2 3), (1 2 4), (1 3 4), (2 3 4) } (the square of the product of disjointtranspositions collapses to ε). Observe that (1 3 2)(1 4 2) = (1 4)(2 3), which is not contained in H. Therefore H is nota subgroup of A4.

Problem 2.3.7.

Solution: Part (a) Let G be a group and 〈g〉 = {gk : k ∈ Z}. Since g0 = 1, the unity of G is contained in 〈g〉. Anyelement of 〈g〉 is of the form gn where n ∈ Z. Given gj , gk ∈ 〈g〉, we have gjgk = gj+k. Since j + k ∈ Z, it follows thatgjgk ∈ 〈g〉. Finally, the inverse of gk in G is g−k; since −k ∈ Z, it must be that g−k ∈ 〈g〉. By the Subgroup Test, 〈g〉 isa subgroup of G.

Part (b) Suppose G is a finite group and H = {gk : k ∈ N} for some g ∈ G. Given gj , gk ∈ H, we have gjgk = gj+k,so gjgk ∈ H, showing that H is closed. Further, H must be finite because it is a subset of finite set G: Given gl ∈ H,since g ∈ G it must be that gl ∈ G by closure of G. By Theorem 2.3.2, H is a subgroup of G.

Page 16

Problem 2.3.8.

Solution: Part (a) Suppose X is a non-empty subset of group G and:

〈X〉 = {xk11 xk22 · · ·xkmm : m ≥ 1, xi ∈ X and ki ∈ Z for each i}.

First, a few observations. Note that X may be infinite, or even uncountable. The elements of 〈X〉, by contrast, consistsonly of finite products of the elements of X, and may include the same terms at different positions in the product.

Let x ∈ X, in which case x = x1, a finite product of elements of X. Therefore x ∈ 〈X〉, so X ⊆ 〈X〉.Now we will show that 〈X〉 is a subgroup of G. Choose x ∈ X. Since x0 = 1 ∈ 〈X〉, the unity of G belongs to 〈X〉.

Given a, b ∈ 〈X〉, we may express each as:

a =∏mj=1 x

αj

j , and b =∏m+nj=m+1 x

αj

j , (1)

where m,n ≥ 1, xj ∈ X, and αj ∈ Z. Their product is:

ab =

m∏j=1

xαj

j

m+n∏j=m+1

xαj

j

=

m+n∏j=1

xαj

j .

Since m+ n ≥ 2, we infer that ab ∈ 〈X〉. Finally, given a ∈ 〈X〉, its inverse in G is:

a−1 =

m∏j=1

x−αm−j+1

m−j+1 ,

which is easily verified. Since −αm−j+1 ∈ Z, it follows that a−1 ∈ 〈X〉. Hence 〈X〉 is a subgroup of G.Part (b) Suppose H is a subgroup of G, and H contains X. If a ∈ 〈X〉, then we can express a as in equation (1)

above. Since each xj ∈ H by hypothesis, it follows that xαj

j ∈ H by closure of H, and the product of all xαj

j (in anyorder!) is contained in H. Therefore a ∈ H, so 〈X〉 is a subset of H.

Problem 2.3.9.

Solution: Given group G and g ∈ G, define C(g) = {z ∈ G : zg = gz}. Clearly 1 ∈ C(g) by the definition of G’s unity.If a, b ∈ G, then ag = ga and bg = gb, Therefore:

(ab)g = a(bg) = a(gb) = (ag)b = (ga)b = g(ab),

so ab ∈ C(g). Finally, if c ∈ C(g), then:

c−1(cg)c−1 = (c−1c)(gc−1) = gc−1 = c−1(gc)c−1 = (c−1g)(cc−1) = c−1g,

so c−1 ∈ C(g). By the Subgroup Test, C(g) is a subgroup of G.

Problem 2.3.10.

Solution: Suppose X = {1, 2, . . . , n} for some n ∈ N and H = {σ ∈ Sn : σk = k for all k ∈ X}. It must be thatvarepsilon ∈ H because εk = k for all k ∈ X. Given τ, µ ∈ H, since τk = k and µk = k for all k ∈ X, it follows thatτµk = τk = k, so τµ ∈ H. Finally, given v ∈ H, its inverse in Sn is v−1; we have v−1(vk) = (v−1v)k = εk = k = v−1k,so v−1 ∈ H. Thus H is a subgroup of Sn.

Problem 2.3.11.

Solution: Define:

G = {(a b0 a

)∈ GL2(R) : a, b ∈ R, a 6= 0}.

Page 17

Obviously the identity matrix I (which is the unity of GL2(R)) belongs to G as it is an upper diagonal matrix witha = 1 and b = 0. If A,B ∈ G, then:

AB =

(a0 a10 a0

)(b0 b10 b0

)=

(a0b0 a0b1 + a1b0

0 a0b0

),

where a0, b0 6= 0. Further, a0b1 + a1b0 ∈ R, hence AB ∈ G. Thus G is closed under matrix multiplication.A simple calculation shows that for C ∈ G, in GL2(R) its inverse is C−1 is:

C−1 =

(1/a0 −a1/a20

0 1/a0

).

This is a valid matrix because a0 6= 0. Since 1/a0 6= 0, it follows that C−1 ∈ H. Thus G is a subgroup of GL2(R).

Problem 2.3.12.

Solution: Define:

G = {(

1 b0 1

)∈ GL2(R) : b ∈ R}.

The argument that G is a subgroup of GL2(R) is analogous the argument used in exercise 2.3.11. Given any A ∈ G,the inverse in GL2(R) is:

A−1 = {(

1 −a0 1

),

where a is the entry in the first row, second column of A. Obviously A−1 ∈ G. Therefore G is a subgroup of GL2(R).

Problem 2.3.13.

Solution: Part (a) Let G be a group and H = {(g, g) : g ∈ G}. The element (1, 1) is the unity of G × G. Since 1 isthe unity of the G, it follows that (1, 1) ∈ H. Given a, b ∈ H, we may express each as a = (α, α) and b = (β, β) whereα, β ∈ G. Their product is ab = (αβ, αβ). Since G is closed, αβ ∈ G, so ab ∈ H. Moreover, α has the inverse α−1 in G,so (α−1, α−1) ∈ G×G and is the inverse of (α, α) because:

(α, α)(α−1, α−1) = (αα−1, αα−1) = (1, 1) = (α−1α, α−1α) = (α−1, α−1)(α, α).

It follows that (α−1, α−1) ∈ H. By the Subgroup Test, H is a subgroup of G×G.Part (b) The group G must be an abelian group. Define H = {(g, g−1) : g ∈ G}. Obviously (1, 1) ∈ H for any group

G. Given a, b ∈ H where a = (α, α−1) and b = (β, β−1) for some α, β ∈ G, we have:

(α, α−1)(β, β−1) = (αβ, α−1β−1).

To be an element of H, it must be that (αβ)−1 = α−1β−1 = (βα)−1. Taking the inverse of both sides, αβ = βα;hence G must be abelian.

Finally, it’s obvious that (α−1, α) is the inverse of (α, α−1) in G × G and is contained in H for any group G. Weconclude that the only requirement is that G be abelian.

Problem 2.3.14.

Solution: Let X be an infinite set and G be the set of permutations σ of X where σx = x for all but a finite number ofx ∈ X. The identity permutation ε in SX belongs to G because εx = x for all x ∈ X (hence the relation is true for all buta finite number (zero) of x ∈ XX).

If σ0, σ1 ∈ G, let U0 = {x ∈ X : σ0x 6= x} and U1 = {x ∈ X : σ1x 6= x}. By hypothesis, both U0 and U1 are finite withM0 and M1 elements, respectively. We will show that if σ0σ1x 6= 0 then σ0x 6= x or σ1x 6= x. If σ0x = x and σ1x = x,then σ0σ1x = σ0x = x. The contrapositive of the foregoing proves the proposition. Therefore U2 = {x ∈ X : σ0σ1x 6= x}is a subset of U0 ∪ U1. Since U0 ∪ U1 has at most M0 + M1 elements, it follows that U2 is finite. We conclude thatσ0σ1 ∈ G.

Page 18

Finally, if τ ∈ G, then its inverse in SX is τ−1. Since τ is a bijection, τ−1x 6= x if and only if τx 6= x. Thereforeτ−1x = x for all but a finite number of x ∈ X, hence τ−1 ∈ G. Hence G is a group of SX .

Problem 2.3.15.

Solution: We’ll provide just the answers without proof. Reaching the answer only requires some simple logic.Part (a) Only the trivial subgroup is a subgroup of C4.

C4

{1}

Part (b)

C6

{1, c2, c4} {1, c3}

{1}

Part (c)

S3

{1, σ, σ2} {1, τ} {1, τσ} {1, τσ2}

{1}

Part (d)

(Z/8Z)×

{1, 3} {1, 5} {1, 7}

{1}

Problem 2.3.16.

Solution: Part (a) Suppose the H and K are subgroups of group G. Since 1 ∈ H and 1 ∈ K, it follows that 1 ∈ H ∩K.Given a, b ∈ H ∩K, both elements are in both H and K, so by the closure of these subgroups ab ∈ H and ab ∈ K, henceab ∈ H ∩K. Further, c−1 ∈ H and c−1 ∈ K by the definition of subgroups, so c−1 ∈ H ∩K. By the Subgroup Test,H ∩K is a subgroup of G.

Part (b) Let L be a subgroup of H and a subgroup of K, which are both contained in group G. If a ∈ L, then a ∈ Hand a ∈ K because L is a subset of both. Therefore a ∈ H ∩K, so L is a subset of H ∩K. Since every subgroup of bothH and K is contained in H ∩K, we conclude that H ∩K is the largest subgroup of both H and K.

Problem 2.3.17.

Page 19

Solution: Let H and K be subgroups of group G. Suppose H ∪K is a subgroup of group G. If H = K, then H ⊆ Kand K ⊆ H, so we’re done. On the other hand, if H 6= K we may assume without loss of generality that there is someh ∈ H that does not belong to K. Given any k ∈ K, it must be that hk = c ∈ H ∪K by closure of the subgroup. Sinceh−1 ∈ H by the Subgroup Test, h−1hk = k = h−1c. If c ∈ K, then kc−1 = h−1 belongs to K, from which it follows thath ∈ K; this contradicts that h /∈ K. Accordingly, c ∈ H, so since h−1 ∈ H as well, we infer that k ∈ H. Therefore H is asubset of K.

Conversely, if either H ⊆ K or K ⊆ H, the H ∪K = H or H ∪K = K, either of which is a subgroup of G.

Problem 2.3.18.

Solution: Note that each τa,b is an affine function of R. For fixed a, b ∈ R, τa,b is a bijection, so each τa,b ∈ SR. Theunity ε of SR is contained in G because ε = τ1,0 = 1x + 0 = x. Given τa,b, τc,d ∈ G, it must be that c, d 6= 0. For anyx ∈ R:

τa,bτc,d(x) = τa,b(cx+ d) = a(cx+ d) + b = (ac)x+ (ad+ b).

Since ac 6= 0 and ac, ad+ b ∈ R, it must be that τa,bτc,d ∈ G.The inverse of τa,b in SR is:

τ−1a,b (x) = x/a+ (−b/a) = τ1/a,−b/a,

(for which we can easily show that τa,bτ1/a,−b/a(x) = x = τ1/a,−b/aτa,b for all x ∈ R). Since 1/a 6= 0 (and exists since

a 6= 0) and 1/a,−b/a ∈ R, it follows that τ−1a,b ∈ H. By the Subgroup Test, G is a subgroup of SR.

Problem 2.3.19.

Solution: Let H and K be subgroups of a group G with g ∈ G.Part (a) Suppose G is abelian. Given g ∈ G, if u ∈ gHg−1, then u = ghg−1 for some h ∈ H, so u = ghg−1 =

gg−1h = h, so gHG ⊆ H. Moreover, given h ∈ H, it follows that ghg−1 = h for the same reason, so H ⊆ gHg−1;therefore for any g ∈ G we have gHG = H. Thus H is self-conjugate.

Part (b) Suppose u ∈ (gHg−1) ∩ (gKg−1), so there is an h ∈ H and k ∈ K such that g−1hg = g−1kg. Thereforeg(g−1ghg)g−1 = h = g(g−1kg)−1 = k. Therefore h ∈ H and h ∈ K, so h ∈ H∩K. It follows that ghg−1 ∈ g(H∩K)g−1.

Conversely, suppose v ∈ g(H ∩K)g−1, so there is some l ∈ H and l ∈ K such that glg−1 = v. Therefore glg−1 ∈gHg−1 and glg−1 ∈ gKg−1, so glg ∈ (gHg−1) ∩ (gKg−1). We conclude that (gHg−1) ∩ (gKg−1) = g(H ∩K)g−1.

Problem 2.3.20.

Solution: Part (a) Suppose H is a subgroup of G and a subset of Z(G). Let g ∈ G. If v ∈ gHg−1, then u = ghg−1 forsome h ∈ H. Since h ∈ Z(G), it follows that gh = hg, so u = ghg−1 = hgg−1 = h, so gHg−1 ⊆ H. On the other hand,if h ∈ H, then ghg−1 ∈ gHg−1. Further, ghg−1 = hgg−1 = h, so H ⊆ gHg−1. Therefore gHg−1 = H. Because this istrue for every g ∈ G, we conclude that H is self-conjugate.

Part (b) A straightforward calculation shows that H = {ε, σ, σ2} is self-conjugate:

εHε−1 = H,σHσ−1 = σ2Hσ = H,

(σ2)H(σ2)−1 = σ2Hσ = H,τHτ−1 = τHτ = H,

τσH(τσ)−1 = τσHτσ = H,τσ2H(τσ2)−1 = τσ2Hτσ2 = H.

Problem 2.3.21.

Solution: Define:

G = {(a b0 c

)∈ GL2(R) : a, b, c ∈ R, a, c 6= 0}.

Page 20

Given A,X ∈ G where:

A =

(a b0 c

), and X =

(x0 x10 x2

),

we must find A such that AX = XA. Therefore:

AX =

(ax0 ax1 + bx20 cx2

),

and

XA =

(ax0 bx0 + cx10 cx2

).

The only entry that may differ between the two products is their first row, second column, so it must be that ax1+bx2 =bx0 + cx1. If we set x1 = 1 and x0 = x2 = 0, then we find that a = c. Therefore any A ∈ Z(G) is of the form of adiagonal matrix with equal elements along the diagonal. This makes sense because for some α ∈ R it follows that A = αI,so given X we have AX = α(IX) = αX = α(XI) = X(αI) = XA. Therefore:

Z(G) = {(α 00 α

): α ∈ R}.

Problem 2.3.22.

Solution: We must find the set of 2× 2 real-valued matrices that commute with every element of GL2(R). Given matricesA and B such that:

A =

(a0 a1a2 a3

), and B =

(b0 b1b2 b3

),

we have:

AB =

(a0b0 + a1b2 a0b1 + a1b3a2b0 + a3b2 a2b1 + a3b3

),

and

BA =

(a0b0 + a2b1 a1b0 + a3b1a0b2 + a2b3 a1b2 + a3b3

).

Because it must be the AB = BA, we must find the appropriate aj ’s such that:

a0b0 + a1b2 = a0b0 + a2b1,a0b1 + a1b3 = a1b0 + a3b1,a2b0 + a3b2 = a0b2 + a2b3,a2b1 + a3b3 = a1b2 + a3b3.

(2)

We may find the appropriate elements of A by successively setting B to different basis matrices, which we can take tothe be set of {Bj}3j=0 where the matrix Bj has bj = 1 and all other elements equal to zero.

For AB0 (i.e., b0 = 1 and all other elements are zero), the non-trivial results of equations (2) are:

a1 = 0,a2 = 0,

For AB1, the non-trivial, new result is:a0 = a3.

We conclude that any element of A ∈ Z[GL2(R)] must be such that a1 = a2 = 0, and a0 = a3, i.e.,:

A =

(α 00 α

).

where α ∈ R. Note that A = αI, so given any B ∈ GL2(R), it follows that:

AB = (αI)B = α(IB) = αB = α(BI) = B(αI) = BA.

In conclusion:

Z[GL2(R)] = {(α 00 α

)∈ GL2(R) : α ∈ R}.

Page 21

Problem 2.3.23.

Solution: Yes, a group G may have an abelian subgroup not contained in Z(G). For instance, there is no µ ∈ S3\{ε}such that µν = νµ for all ν ∈ S3, so Z(S3) = {ε}. However, {ε, σ, σ2} is an abelian subgroup of S3.

Problem 2.3.24.

Solution: Suppose ab = ba in some group G containing a and b. Define H = {g ∈ G : agb = bga}. Obviously 1 ∈ Hsince a1b = ab = ba = b1.

To show that H is closed, we will first establish some simple results. Observe that:

ab−1 = (b−1b)(ab−1) = b−1(ba)b−1 = b−1(ab)b−1 = (b−1a)(bb−1) = b−1a.

We can similarly show that a−1b = ba−1.Let u, v ∈ H be given. Observe that since aub = bua, we have:

u = a−1buab−1.

Using the previous results, we have:

a(uv)b = (a)u(vb) = a(a−1buab−1)(vb) = (bu)(ab−1)(vb) = (bu)(b−1a)(vb) = (bub−1)(avb)

= (bub−1)(bva) = (bu)(b−1b)(va) = b(uv)a.

Accordingly, uv ∈ H, so H is closed.Finally, given w ∈ H, its inverse in G is w−1. Since awb = bwa, we have:

a−1w−1b−1 = (bwa)−1 = (awb)−1 = b−1w−1a−1.

We then have:(ba)(a−1w−1b−1)(ba) = b(a−1a)w−1(b−1b)a = bw−1a = (ba)(b−1w−1a−1)(ba)

= (ab)(b−1w−1a−1)(ab) = aw−1b.

Therefore w−1 ∈ H. By the Subgroup Test, H is a subgroup of G.

Problem 2.3.25.

Solution: Let H and K be subgroups of a group G. Define HK = {hk : h ∈ H and k ∈ K} and KH = {kh : h ∈ Hand k ∈ K}. Suppose KH is a subset of HK. Since 1 ∈ H and 1 ∈ K, it follows that 1 ∈ HK. If u0, v0 ∈ HK, thenthere are h0, h1 ∈ H and k0, k1 ∈ K such that u0 = h0k0 and u1 = h1k1. Observe that k0h1 ∈ KH, so k0h1 ∈ HK, sothere is some h2 ∈ H and k2 ∈ K such that k0h1 = h2k2. With this result, we find:

u0u1 = (h0k0)(h1k1) = h0(k0h1)k1 = h0(h2k2)k1 = (h0h2)(k2k1).

By closure of H and K, h0h2 ∈ H and k2k1 ∈ K, so u0u1 ∈ HK. Further, u−10 is the inverse of u0 in G, so u−10 = k−10 h−10 ;it follows that u−10 ∈ KH. By hypothesis, u−10 ∈ HK. By the Subgroup Test, HK is a subgroup of G.

Conversely, suppose HK is a subgroup of G. Let kh ∈ KH be given. By definition hk ∈ HK. It must be thath−1 ∈ HK (since h · 1· ∈ HK) and k−1 ∈ HK (since 1 · k· ∈ HK). In G we may express kh = (h−1k−1)−1. Sinceh−1 ∈ H and k−1 ∈ K, it follows h−1k−1 ∈ HK. But since HK is a subgroup, the inverse of any element in HK is alsoin HK. Therefore kh = (h−1k−1)−1 ∈ HK. We conclude that KH is a subset of HK.

Section 2.1

Problem 2.4.1.

Page 22

Solution: We can use Theorem 2.4.8 to determine all the generators of G where 〈gk〉 = G if and only if gcd(k, o(g)) = 1.Part (a) g, g2, g3, g4.Part (b) g, g3, g7, g9.Part (c) g, g3, g5, g7, g9, g11, g13, g15.Part (d) g, g3, g7, g9, g11, g13, g17, g19.

Problem 2.4.2.

Solution: We can again use Theorem 2.4.8 to determine all the generators Z/nZ. Note that 〈1〉 = Z/nZ where o(1) = n.

Therefore if k ∈ Z/nZ then k = k · 1 = 1k (with the former in additive notation and the latter in multiplicative notation).Hence k is a generator if gcd(k, o(1)) = 1.

Part (a) 1, 2, 3, 4.Part (b) 1, 3, 7, 9.Part (c) 1, 3, 5, 7, 9, 11, 13, 15.Part (d) 1, 3, 7, 9, 11, 13, 17, 19.Unsurprisingly, the results are the same as in exercise 2.4.1 with g = 1.

Problem 2.4.3.

Solution: Part (a) g−1 is a generator of G for if h ∈ 〈g〉 then h = gα for some α ∈ Z. Therefore (g−1)−α = gα. We cansimilarly show that any h′ ∈ 〈g−1〉 is contained in 〈g〉.

Are there any other generators? Since o(g) =∞, there can be no γ ∈ Z\{1} such that gγ = g (otherwise, gγ−1 = 1,contradicting that g is of infinite order). Therefore gγ = g if and only if γ = 1. Since g ∈ G, if (gγ)δ = gγδ = g, thenγδ = 1, so γ = ±1. Thus g and g−1 are the only generators of G.

Part (b) Obviously 1 and −1 generate Z. If k ∈ Z, then k = k · 1 = (−k) · (−1). If u ∈ Z\{−1, 1}, then u - u + 1because, by the division algorithm, u = u(1) + 1 where the non-zero remainder is unique. Thus u cannot generate Z.

Problem 2.4.4.

Solution: Part (a) (Z/7Z)× = {1, 2, 3, 4, 5, 6} is cyclic. A simple calculation shows that o(3) = 6, so 〈6〉 = (Z/7Z)×.Part (b) (Z/12Z)× = {1, 5, 7, 11} is not cyclic because given u ∈ (Z/12Z)× where u 6= 1, we have o(u) = 2 (and

therefore uk ∈ {1, y} for any k ∈ Z).Part (c) (Z/16Z)× = {1, 3, 5, 7, 9, 11, 13, 15} is not cyclic because given u ∈ (Z/16Z)× where u 6= 1, we have

o(u) ∈ {2, 4}.Part (d) (Z/11Z)× = {k : 1 ≤ k ≤ 10} is cyclic with generator 2.

Problem 2.4.5.

Solution: Part (a) Q∗ is not cyclic. Note that this is a multiplicative group. Suppose u ∈ Q is a generator of Q∗, whichwe can express as α/β where α, β ∈ Z\{0} and gcd(α, β) = 1. If a/b ∈ Q∗ where gcd(a, b) = 1 and a, b > 0, thenuk = αk/βk = a/b for some k ∈ Z. But −a/b ∈ Q∗ as well, so αj/βj = −a/b for some j ∈ Z. It follows that:

αj

βjβk

αk=αj−k

βj−k=a

b

(− ba

)= −1.

It follows that α/β = −1 (and, of course, j−k must be odd), but this cannot be the generator of Q∗ because −1j = ±1for all j ∈ Z. Therefore there is no generator of Q∗, so it cannot be cyclic.

Part (b) Q is not cyclic. Note that this is an additive group. Suppose v ∈ Q is its generator, which we can expressas v = α/β where α, β ∈ Z. It cannot be that v = 0 (then kv = 0 for all k ∈ Z), so α, β 6= 0. We have some a′/b′ ∈ Qwhere gcd(a′, b′) = 1 and a′, b′ > 0. Then jv = a′/b′ for some j ∈ Z. But a′/(2b′) ∈ Q as well, so there is some m ∈ Zsuch that mα/β = a′/(2b′), so m = 1/2, contradicting that m is an integer. Hence Q is not cyclic.

Problem 2.4.6.

Page 23

Solution: If a ∈ 〈g〉 then a = gk for some k ∈ Z. It follows that (g−1)−k = gk ∈ 〈g−1〉. Conversely, if b ∈ 〈g−1〉, thenb = (g−1)j for some j ∈ Z, so g−j = (g−1)j ∈ 〈g〉. Hence 〈g〉 = 〈g−1〉.

Problem 2.4.7.

Solution: Part (a) By Theorem 2.4,5, o(g2) = 20/2 = 10.Part (b) By Theorem 2.4.2(a), (g8)k = g8k = 1 if and only if 20 | 8k. Since the order of g8 is the smallest power of g8

that equals one, it follows that 8k must be the least common multiple of 20 and 8, which here is 40. Therefore o(g8) = 40.Part (c) By Theorem 2.4.5, o(g5) = 20/5 = 4.Part (d) Using Theorem 2.4.2(a), o(g3) = 60 (the least common multiple of 3 and 20).

Problem 2.4.8.

Solution: Part (a) By Theorem 2.4.4, σ ∈ S5 can be factored into disjoint cycles γ1, . . . , γr, and |σ| = lcm(o(γ1), . . . , o(γr)),).Out of the possible factorizations of σ, some σ that may be factored into disjoint cycles of 3 and 2 will have o(σ) = 6,whereas any other possible factorization has a smaller order, e.g.:

Factorization(Disjoint Cycles) Cycle Orders Total Order

5 5 54 4 4

3,2 3,2 62,2 2,2 23 3 32 2 2ε 1 1

An example of such a permutation is σ = (1 2 3)(4 5).Part (b) We use the same logic as with part (a). The orders of the various factorizations of elements of S7 are:

Factorization(Disjoint Cycles) Cycle Orders Total Order

7 7 76 6 6

5,2 5,2 105 5 5

4,3 4,3 124,2 4,2 84 4 4

3,3 3,3 33,2 3,2 63 3 3

2,2,2 2,2,2 22,2 2,2 22 2 2ε 1 1

A permuation factorizable into disjoint cycles of 3 and 4 have the highest order. An example of such a permutation isσ = (1 2 3)(4 5 6 7).

Problem 2.4.9.

Solution: By Theorem 2.4.9, the subgroups of a cyclic group G = 〈g〉 with o(g) = n are exactly those 〈gk〉 where k | n.We’ll also use the next result, which is given without proof in example 2.4.14.

Page 24

Proposition 2. Let G = 〈g〉 with o(g) = n. If 〈gj〉 and 〈gk〉 are subgroups of G and j | k, then 〈gk〉 is a subgroup of 〈gj〉.

Proof. By Theorem 2.4.5 o(gj) = n/j. Further, by Theorem 2.4.9 j = n/q and k = n/r for some q, r ∈ N. Since j | k,it follows that js = k for some s ∈ N. As a result, js = k = n/r, so j = n/(rs) and rs = n/j. Therefore r | n/j, so byTheorem 2.4.9(2) 〈(gj)(n/j)/r〉 is a subgroup of 〈gk〉. But we see that:

〈(gj)(n/j)/r〉 = 〈(gj)n/(jr)〉 = 〈(gj)s〉 = 〈gjs〉 = 〈gk〉

Hence 〈gk〉 is a subgroup of 〈gj〉, as we sought to show.

With this in mind, the problem is straightforward. We’ll only provide the Hesse diagrams.Part (a)

〈g〉

〈g2〉

〈g4〉

{1}

Part (b)

〈g〉

〈g2〉 〈g5〉

{1}

Part (c)

〈g〉

〈g2〉 〈g3〉

〈g6〉 〈g9〉

{1}

Part (d)

〈g〉

〈gp〉

〈gp2〉

{1}

Part (e)

〈g〉

〈gp〉 〈gq〉

{1}

Page 25

Part (f)

〈g〉

〈gp〉 〈gq〉

〈gp2〉 〈gpq〉

{1}

Problem 2.4.10.

Solution: Part (a) Suppose g and h are elements of G and hg = gh, o(g) = n, and o(h) = m. It follows that if j ∈ Zthen (gh)j = gjhj by Theorem 2.1.6. Let r = lcm(n,m), in which case nq = r = nq′ for some q, q′ ∈ Z. Therefore:

(gh)r = grhr = (gn)q(hm)q′

= 1q1q′

= 1.

As a result, 1 ≤ o(gh) ≤ r, so the order of gh is finite.Part (b) Designate the matrices as follows:

g =

(0 −11 0

)and h =

(0 1−1 −1

).

A simple calculation shows gh 6= hg. We will show by induction that for k ∈ N:

(gh)k =

(1 k0 1

).

For k = 1, the result is obvious. If the inductive hypothesis holds for some k ∈ N, then:

(gh)k+1 = (gh)k(gh) =

(1 k0 1

)(1 10 1

)=

(1 k + 10 1

),

proving the result for all k ∈ N.Since k + 1 increases monotonically and is never equal to zero for any k, there is no k ∈ N such that (gh)j = I. We

then infer that there is no j ∈ Z<0 such that (gh)j = I; if there were, the (gh)−j = I (possible because g and h areinvertible) where −j ≥ 1, contradicting the result above. Therefore o(gh) =∞.

Problem 2.4.11.

Solution: Part (a) Let G be a cyclic group of order n where G = 〈g〉 for some g ∈ G. If h ∈ G, then h = gk for somek ∈ Z. By the corollary to Theorem 2.4.2, |G| = n = o(g), so gn = 1. Therefore hn = (gk)n = (gn)k = 1k = 1.

Part (b) Suppose hm = 1 where gcd(m,n) = 1. It follows that mx+ ny = 1 for some x, y ∈ Z. As a result:

hmx = 1x = 1 = h1−ny = hh−ny = h(hn)−y = h1−y = h,

where we used the result of part (a) that hn = 1. Hence h = 1.

Problem 2.4.12.

Solution: Let g = e2πi/n. Clearly gn = 1. We will show that any positive power k such that gk = 1 must be greater thanor equal to n. If gk = 1 where k ≥ 1, then gk = e2πik/n = 1. Accordingly, 2πi(k/n) = 2πmi for some m ∈ Z; thereforek = nm, so k is a multiple of n. We conclude that o(g) = n.

Problem 2.4.13.

Page 26

Solution: Part (a) Suppose G = {g1, g2, . . . , gr} is an abelian group and a = g1g2 · · · gr. We have:

a2 = (g1g2 · · · gr)(g1g2 · · · gr) = g21g22 · · · g2r ,

where we relied upon the commutativity of the terms of the product. Because the product includes every element of Gtwice, every element gj of G has some inverse gk in the product. If gj 6= gk, then (using commutativity) gjgk = 1 appearstwice in the product and cancels out. If gj = gk, then g2j = 1. Since this is true of every term in the product, a2 = 1.

Part (b) Observe that (Z/pZ)× = {a : 1 ≤ a ≤ p− 1} since gcd(a, p) = 1 for all 1 ≤ a ≤ p− 1. By exercise 1.3.26,the only self-inverse elements of (Z/pZ)× are 1 and −1 = p− 1. Therefore:

(p− 1)! =

p−1∏k=1

k = (1)(p− 1)

p−2∏k=2

k = (p− 1)

p−2∏k=2

k.

Because the product on the right-hand side includes all the non-self-invertible elements of (Z/pZ)× (all of which arecommutative), the product equals 1. Therefore:

(p− 1)! = p− 1 = −1.

We conclude that (p− 1)! ≡ −1 (mod p).

Problem 2.4.14.

Solution: Suppose G is a group for which {1} and G are the only subgroups. If G = {1}, then it is finite and cyclic(G = 〈1〉) and we’re done. Now assume instead that G 6= {1}, in which case there is some g 6= 1 contained in G. ByTheorem 2.4.1, 〈g〉 is a subgroup of G, and it cannot equal {1} since g 6= 1 but is contained in G. By hypothesis, 〈g〉 = G,so G is cyclic. If o(g) =∞, then all gk for k ∈ Z are distinct by Theorem 2.4.3(3). Therefore g2 6= 1 (since g0 = 1), and〈g2〉 is a subgroup of G not equal to {1}, and therefore G = 〈g2〉 = 〈g〉. But then g /∈ 〈g2〉, a contradiction. It followsthat o(g) is finite.

Finally, we will show that n = o(g) is either one or prime. By Theorem 2.4.9(2), if d ∈ N is any divisor of n, then 〈gd〉is a subgroup of G of order n/d. Since the only subgroups of G are of order 1 or n, it follows that d ∈ {1, n}, from whichit follows that n is prime.

Problem 2.4.15.

Solution: We can use Theorem 2.4.10 to find the solution. Remember that if the generators X of a group 〈X〉 arecontained into another group H, then 〈X〉 is a subgroup of H.

Observe that a−1 = (a)−1 ∈ 〈a, b〉 and b−1 = (b)−1 ∈ 〈a, b〉, so by Theorem 2.4.10 〈a−1, b−1〉 ⊆ 〈a, b〉. By ananalogous argument 〈a, b〉 ⊆ 〈a−1, b−1〉, so the two sets are equal.

Moreover a ∈ 〈a−1, b−1〉 and ab = (a−1)−1(b−1)−1 ∈ 〈a−1, b−1〉, so 〈a, ab〉 ⊆ 〈a−1, b−1〉. Conversely, a−1 ∈ 〈a, ab〉and b−1 = ab)−1a ∈ 〈a, ab〉, so 〈a−1, b−1〉 ⊆ 〈a, ab〉.

We conclude that 〈a, b〉 = 〈a, ab〉 = 〈a−1, b−1〉.

Problem 2.4.16.

Solution: We make liberal use of Theorem 2.4.10 here.Part (a) a = a4a−3 = xy−1 ∈ H, so G ⊆ H. Conversely x = a4, y = a3 ∈ G, so H ⊆ G. Accordingly H = G.Part (b) gcd(6, 8) = 2 and 2 = 6(−5) + 8(4), so a2 = (a6)−5(a8)4 ∈ H, hence 〈a2〉 ⊆ H. Conversely, a6, a8 ∈ 〈a2〉,

so H ⊆ 〈a2〉. Therefore H = 〈a2〉.Part (c) Since gcd(m, k) = d, there are x, y ∈ Z such that d = mx+ ky. As a result, ad = amx+ky = (am)x(ak)y ∈

〈x, y〉; thus 〈ad〉 ⊆ H. On the other hand, m = dq and k = dq′ for some q, q′ ∈ Z, so am = (ad)q, ak = (ad)q′ ∈ 〈ad〉, so

H ⊆ 〈ad〉. Therefore H = 〈ad〉.Part (d) From example 2.2.8, note that S3 = 〈σ, τ〉 where σ = (1 2 3) and τ = (1 2). We have σ =(1 2)(2 3) ∈ H

and τ ∈ H, so S3 ⊆ H. Conversely, (2 3) = τσ ∈ S3, so H ⊆ S3. We conclude that H = S3.Part (e) Skipped.Part (f) Skipped.

Page 27

Problem 2.4.17.

Solution: Part (a) Suppose X ⊆ Y where Y is contained in a group. Assume 〈X〉 is non-empty. If x ∈ X, then x ∈ 〈X〉and x ∈ Y , so x ∈ 〈Y 〉. Therefore 〈X〉 ⊆ 〈Y 〉.

Part (b) Let X be a non-empty subset of some group G. Suppose X is a subgroup of G. Since X as a subgroupobviously contains X, Theorem 2.4.10(2) implies that 〈X〉 ⊆ X. On the other hand, by Theorem 2.4.10(1) X ⊆ 〈X〉. Weconclude that X = 〈X〉.

Problem 2.4.18.

Solution: Let G = 〈g〉 and H = 〈h〉 for some g ∈ G and h ∈ H. Define K = 〈(g, 1), (1, h)〉. If u ∈ G × H,then u = (gα, hβ) for some α, β ∈ Z. By the definition of multiplication of elements of a direct product, (gα, hβ) =(gα, 1)(1, hβ) = (g, 1)α(1, h)β ∈ K, so G ⊆ K. Conversely, if v ∈ K, then v = (g, 1)γ(1, h)δ = (gγ , hδ) ∈ G for someγ, δ ∈ Z. Thus G×H = 〈(g, 1), (1, h)〉.

Problem 2.4.19.

Solution: Suppose G = 〈X〉 and xy = yx for all x, y ∈ X. Given u, v ∈ G, we may express each as:

u =∏Nj=1 x

αj

j and v =∏N+Mj=N+1 x

αj

j ,

where xj ∈ X and N,M ≥ 1. If 1 ≤ j, k ≤ N + M where j 6= k, then xαj

j = a and xαk

k = b for some a, b ∈ X. Sinceab = ba, we have:

xαj

j xαk

k = xαk

k xαj

j

Therefore all elements in the products that constitute u and v commute. It follows that:

uv =

N∏j=1

xαj

j

N+M∏j=N+1

xαj

j

=

N+M∏j=N+1

xαj

j

N∏j=1

xαj

j

= vu,

hence G is abelian.

Problem 2.4.20.

Solution: Part (a) There exist a ∈ C6 and b ∈ C15 such that C6 = 〈a〉 and C15 = 〈b〉. The orders of (a, b), (a, b2), and(a4, b7) are all thirty. We will show that no element of C6 × C15 can be of greater order. For given any aj ∈ C6 andbk ∈ C15, we have:

(aj , bk)30 = ((aj)30, (bk)30) = ((a6)5j , (b15)2k = (1, 1).

Since every element of C6×C15 must have an order less than or equal to thirty, it follows that order thirty is the maximumof the group.

Part (b) Designate H = Cm×Cn. There are g ∈ Cm and h ∈ Cn such that Cm = 〈g〉 and Cn = 〈h〉, and o(g) = m ando(h) = n. Let r = lcm(m,n), so r = mα0 = nα1 for some α0, α1 ∈ Z. It follows that (g, h)r = ((gm)α0 , (hn)α1) = (1, 1),so 1 ≤ o((g, h)) ≤ r. We will show that the order is precisely r. For if s ≥ 1 and (g, h)s = (1, 1), then gs = 1 and hs = 1.By Theorem 2.4.2, m | s and n | s. Since s is a common multiple of m and n, it follows from the definition of the leastcommon multiple that r | s, hence r ≤ s. We conclude o((g, h)) = r.

Now we will show no element of H can be of greater order than r. Given (u, v) ∈ H, we can express u = ga and v = hb

for some a, b ∈ Z. It follows that:(ga)r = (ga)mα0 = (gm)aα0 = 1,

and(hb)r = (hb)nα1 = (hn)bα1 = 1,

Thus (u, v)r = (1, 1), so any element of H cannot have an order greater than r. We conclude that (g, h) is an elementof H with maximum order.

Problem 2.4.22.

Page 28

Solution: Suppose σ ∈ Sn and o(σ) = p where p is prime. It cannot be that σ = ε because o(ε) = 1. By the cycledecomposition theorem, σ may be decomposed into a product of one or more disjoint cycles each of length 2 or greater.Assume there are N such disjoint cycles, each designated γj . By Theorem 2.4.4, o(σ) = lcm(γ1, γ2, . . . , γN ) = p. Thereforeeach o(γj) divides p. Since o(γj) ≥ 2, we infer that o(γj) = p, which also corresponds to the length of cycle γj . Weconclude that σ equals the product of disjoint p-cycles.

Problem 2.4.23.

Solution: Part (a) Let g, h ∈ G be given and n = o(h). It is easily seen that for k ∈ N :

(ghg−1)k =

k times︷ ︸︸ ︷(ghg−1)(ghg−1) · · · (ghg−1) = ghkg−1.

Consequently, (ghg−1)n = ghng−1 = gg−1 = 1. Now suppose there is some j ∈ N such that (ghg−1)j = 1. It followsthat:

g−1(ghg−1)jg = g−1(ghjg−1)g = hj = g−1g = 1.

Accordingly n | j, so j ≥ n. We conclude that o(ghg−1) = n.Part (b) Let g, h ∈ G and n = o(gh). Given k ∈ N where (gh)k = 1, we have:

(hg)k =

k times︷ ︸︸ ︷(hg)(hg) · · · (hg) = h(gh)k−1g = 1.

We then have:h−1(h(gh)k−1g))g−1 = (gh)k−1 = h−1g−1 = (gh)−1,

from which it follows that (gh)k = 1. Therefore if (gh)k = 1 then (hg)k) = 1. Conversely, if (hg)j = 1, then by thesame argument (gh)j = g(hg)j−1h = 1, so (hg)j = 1. We conclude that given k ∈ N we have (gh)k = 1 if and only if(hg)k = 1. The smallest positive m such that (hg)m = 1 is therefore n, so o(hg) = n.

Problem 2.4.25.

Solution: I’m not sure how the hint in the text proves the result, so I instead use the Chinese remainder theorem.Let G and H be cyclic where |G| = m and |H| = n such that gcd(m,n) = 1. We may define G = 〈g〉 and H = 〈h〉

for some g ∈ G and h ∈ H. Obviously if j ∈ Z then (g, h)j = (gj , hj) ∈ G×H, so 〈(g, h)〉 ⊆ G×H. If u ∈ G×H, thenu = (gr, hs) for some r, s ∈ Z. Because m and n are relatively prime, by the Chinese remainder theorem there is someα ∈ Z such that α ≡ r(modm) and α ≡ s(modn). By Theorem 2.4.2, gα = gr and hα = hs. Therefore (g, h)α = (gr, gs),so G×H ⊆ 〈(g, h)〉. We conclude that G×H = 〈(g, h)〉, so G×H is cyclic.

Problem 2.4.26.

Solution: Let g, h ∈ G and o(g) = m and o(h) = n where m and n are relatively prime. Part (a) We have:

(gh)mn = gmnhmn = (gm)n(hn)m = 1n · 1m = 1.

Since mn ≥ 1, the order of gh is finite, so designate r = o(gh). By Theorem 2.4.2(1), r | mn. If we show that mn | r,then o(gh) = r. Since m and n are relatively prime, it suffices to the show that m | r and n | r by Theorem 1.2.5. Since(gh)d = gdhd = 1, it follows that gd = h−d. Therefore h−d ∈ 〈g〉 and gd ∈ 〈h〉, so gd ∈ 〈g〉 ∩ 〈h〉. But since m and n arerelatively prime, if hd

As a result, gd = 1 and h−d = 1, so hd = 1. Therefore m | d and n | d. We conclude that mn = d, so o(gh) = mn.I’m not clear on the second part of the question as to what condition is being dropped. If gh 6= hg, it is not necessarily

the case that o(gh) = lcm(o(g), o(g)). In S5 for example, if σ = (1 2 3) and τ = (2 4), then στ 6= τσ and o(σ) = 3 ando(τ) = 2. But o(στ) = 3 6= lcm(2, 3). Separately, if m and n are not relatively prime, then it could be that g = h−1 (inwhich case o(g) = o(h) = s where s could be greater than 1). However, o(gh) = 1, so the result does not follow there,either.

Page 29

Part (b) Suppose there is a ∈ G such that o(a) = mn where m and n are relatively prime. Theorem 2.4.9 impliesthat o(〈an〉) is a subgroup of 〈a〉 of order m and o(〈am〉) is a subgroup of 〈a〉 of order n. Because m and n are relativelyprime, there are x, y ∈ Z such that 1 = mx+ ny. As a result:

a = amx+ny = amxany = anyamx.

Designate g = any and h = amx, so a = gh = hg. We will show that 〈g〉 is of order m and 〈h〉 is of order n. Since1 = mx + ny, it follows that gcd(y,m) = 1. Since m = o(〈an〉), it follows from Theorem 2.4.8 that 〈any〉 = 〈an〉, soo(〈any〉) = o(〈g〉) = m. By the same argument, 〈amx〉 = 〈am〉, so o(〈amx〉) = o(〈h〉) = n.

Problem 2.4.28.

Solution: Let H be a subgroup of G with a ∈ G and o(a) = n. Suppose m is the smallest positive integer such thatam ∈ H. It must be that 1 ≤ m ≤ n. By the division algorithm, n = mq + r where q ∈ Z and 0 ≤ r ≤ m − 1. Thusn− r = mq, so amq = an−r = ana−r = a−r. Multiplying by am, we have am(q+1) = am−r, where 1 ≤ m− r ≤ m. Sincethe only possibility is m− r = m, it follows that r = 0, hence m | n.

Problem 2.4.29.

Solution: Suppose o(g) = n. Let k ∈ Z be given, and designate d = gcd(n, k). Thus d | n and d | k. It follows that(gk)n/d = (gk/d)n = (gn)k/d = 1. Now suppose j ∈ N such that (gk)j = gjk = 1. By Theorem 2.4.2, n | kj, so nq = kjfor some q ∈ Z. Further, (n/d)q = (k/d)j, so n/d | (k/d)j where n/d and k/d are relatively prime. By Theorem 1.2.5,n/d | j. Since n/d, j ≥ 1, we conclude that n/d ≤ j. As a result, o(gk) = n/d where d = gcd(n, k).

Problem 2.4.30.

Solution: Let G = 〈g〉 for some g ∈ G and o(g) = n. Given gk ∈ G, denote d = gcd(k, n). For some q ∈ Z we havegk = (gd)q ∈ 〈gd〉, so 〈gk〉 ⊆ 〈gd〉. On the other hand, by Theorem 1.2.3 there are x, y ∈ Z such that d = kx+ ny. As aresult:

gd = gkx+ny = (gk)x(gn)y = (gk)x · 1 = (gk)x ∈ 〈gk〉,

so 〈gd〉 ⊆ 〈gk〉. Therefore 〈gd〉 = 〈gk〉.

Problem 2.4.31.

Solution: Let G = 〈g〉 be a cyclic group and A = 〈ga〉 and B = 〈gb〉 for a, b ∈ Z. Let m = lcm(a, b).Part (a) Suppose o(g) =∞. If u ∈ A ∩ B, then (ga)α = (gb)β for some α, β ∈ Z. By Theorem 2.4.3, aα = bβ = c,

so a | c and b | c. By the definition of the least common multiple, m | c, so (ga)α = (gb)β = (gm)γ ∈ 〈gm〉 for someγ ∈ Z. Therefore 〈ga〉 ∩ 〈gb〉 ⊆ 〈gm〉.

Conversely, since ad0 = m and bd1 = m for some d0, d1 ∈ Z, we have gm = (ga)d0 ∈ 〈ga〉 and gm = (gb)d1 ∈ 〈gb〉, sogm ∈ 〈ga〉 ∩ 〈gb〉. Theorem 2.4.10 implies that 〈gm〉 ⊆ 〈ga〉 ∩ 〈gb〉. We conclude that 〈gm〉 = 〈ga〉 ∩ 〈gb〉.

Part (b) Suppose o(g) = n and a | n and b | n. By the definition of the least common multiple, m | n. Further, sincethere are α, β ∈ Z such that aα = bβ = n, it follows that α | n and β | n and n/α = a and n/β = b. As subgroups of G,Theorem 2.4.9(1) implies that 〈ga〉 = 〈gn/α〉 and 〈gb〉 = 〈gn/β〉. If v ∈ 〈ga〉 ∩ 〈gb〉, then u = (gn/α)c = (gn/β)d for somec, d ∈ Z. Since n = mm′ for some m′ ∈ Z, we have:

v = (gn/α)c = (gmm′/α)c = (gm)m

′c/α ∈ 〈gm〉 and v = (gn/β)d = (gmm′/β)d = (gm)m

′d/β ∈ 〈gm〉.

We will show that m′c/α and m′d/β in the expressions above are integers. Since gmm′c/α = gmm

′d/β , it follows thatmm′c/α ≡ mm′d/β (mod n). Since m | n, by exercise 1.4.7 mm′c/α ≡ mm′d/β (mod m), so there is an x such thatmx = m(m′c/α−m′d/β). It follows that x = m′c/α−m′d/β, so m′c/α and m′d/β are integers, as we sought to show.Accordingly, v ∈ 〈gm〉, so 〈ga〉 ∩ 〈gb〉 ⊆ 〈gm〉.

Conversely, gm = (ga)c0 ∈ 〈ga〉 and gm = (gb)c0 ∈ 〈gb〉 for some c0, c1 ∈ Z, so by Theorem 2.4.10 〈gm〉 ⊆ 〈ga〉 ∩ 〈gb〉.Therefore 〈gm〉 = 〈ga〉 ∩ 〈gb〉.

Page 30

Problem 2.4.32.

Solution: Let G be a finite group.(1) ⇒ (2) Suppose G is cyclic and |G| = pn where p is prime and n ≥ 0. If n = 0, then G = {1}, so the only subgroup

is itself, and the desired result is trivial. Assume n ≥ 1. Given subgroups H and K of G, by Theorem 2.4.9 H = 〈gj〉 andK = 〈gk〉 for some j, k ∈ Z where j | pn and k | pn. It follows that j, k ∈ {pm : 0 ≤ m ≤ n}. If j ≤ k, then j | k, so forsome l ∈ Z we have gk = (gj)l ∈ H; hence K ⊆ H. On the other hand, if k ≤ j, by the same argument H ⊆ K. Weconclude that one of the subgroups H and K is a subset of the other.

(2) ⇒ (1) Suppose if H and K are subgroups of G, then H ⊆ K or K ⊆ H. First we will show that G is cyclic.Choose g0 ∈ G. If G = 〈g0〉, then the result is shown. Otherwise, G\〈g0〉 is non-empty and finite with N elements. Chooseg1 ∈ G\〈g0〉. Again, if 〈g1〉 = G, then the result is shown. Otherwise, since g1 /∈ 〈g0〉 and both 〈g0〉 and 〈g1〉 are subgroupsof G, by hypothesis 〈g0〉 ⊆ g1. Further, G\〈g1〉 is non-empty and has at most N − 1 elements. Choose g2 ∈ G\〈g1〉, andcontinue the process. Since the number of elements in G\〈gk〉 is strictly decreasing, there is some j such that G\〈gj〉 isempty, implying that 〈gj〉 = G. We conclude that G is cyclic. Designate the generator of G as g.

Now we will show that G has pn elements where p is some prime and n ≥ 0. Let M = |G|, and assume that M containstwo distinct primes p1 and p2. By Theorem 2.4.9, 〈gM/p1〉 is a subgroup of 〈g〉 of order p1 and 〈gM/p2〉 is a subgroup of〈g〉 of order p2. However, because p1 and p2 are relatively prime, by Theorem 2.4.9 neither can be the subgroup of theother, contradicting our hypothesis. Therefore M cannot contain two distinct prime factors. As a result, M = pn.

Section 2.5

Problem 2.5.1.

Solution: Part (a) Note that R is an additive group and GL2(R) is a multiplicative group. If r, s ∈ R, then:

α(r + s) =

(1 r + s0 1

)=

(1 r0 1

)(1 s0 1

)= α(r)α(s),

so α is a homomorphism.

The function α is not surjective. The matrix

(0 11 0

)is contained in GL2(R) but is not in the image set of α.

The function is injective. If α(r) = α(s) for some r, s ∈ R, then:(1 r0 1

)=

(1 s0 1

).

Hence r = s.Part (b) If a, b ∈ G, then α(ab) = (ab, ab) = (a, a)(b, b) = α(a)α(b), so α is a homomorphism. The function is also

injective. If α(a) = α(b), then (a, a) = (b, b), so a = b. It is not, however, surjective. If |G| > 1, then given distinctelements a, b ∈ G, the tuple (a, b) is not in α’s image set. Note, however, that it is surjective if G is a singleton.

Problem 2.5.2.

Solution: Given u = (g1, g2), v = (h1, h2) ∈ G×G, we have:

π1(uv) = π1(g1h1, g2h2) = g1h1 = π1(g1, g2)π1(h1, h2) = π1(u)π1(v).

Thus π1 is a homomorphism. If g1 ∈ G, then π1(g1, g2) = g1 for any g2 ∈ G, so π1 is surjective.Given a, b ∈ G:

σ1(ab) = (ab, 1) = (a, 1)(b, 1) = σ1(a)σ1(b),

so σ1 is a homomorphism. If σ1(a) = σ1(b), then (a, 1) = (b, 1), so a = b, establishing that σ1 is injective.

Problem 2.5.3.

Page 31

Solution: Given a group G, define α : G→ G by α(g) = g−1 for all g ∈ G. If α is a homomorphism, then given u, v ∈ Gwe have:

α(uv) = (uv)−1 = v−1u−1 = α(v)α(u) = α(vu) = (vu)−1,

so (uv)−1 = (vu)−1. Therefore uv = vu, so G is abelian. Conversely, if G is abelian, then given u, v ∈ G:

α(uv) = u−1v−1 = v−1u−1 = α(v)α(u),

so α is a homomorphism.

Problem 2.5.4.

Solution: Let m ∈ Z and abelian group G be given. Define α : G→ G by α(a) = am for any a ∈ G. Given u, v ∈ G:

α(uv) = (uv)α = uαvα = α(u)α(v),

so α is a homomorphism.

Problem 2.5.5.

Solution: Let αa be the inner automorphism of G determined by a ∈ G. If αa = 1G then given any g0 ∈ G:

αa(g0) = ag0a−1 = g0.

Multiplying both sides by a on the right, we have ag0 = g0a, so a ∈ Z(G). Conversely, if a ∈ Z(G), then given g1 ∈ G:

αa(g1) = ag1a−1 = g1aa

−1 = g1 = 1G(g1).

Problem 2.5.6.

Solution: Let C6 = 〈a〉 and C4 = 〈b〉 for some a ∈ C6 and b ∈ C4. If α : C6 → C4 is a homomorphism, thenα(a)6 = α(a6) = α(1) = 1, so o(α(a)) ∈ {0, 2}. Therefore α(a) = 1 or α(a) = b2, which by Theorem 2.5.2 completelydetermines α. As a result, there are only two possible homomorphisms from C6 to C4.

Problem 2.5.7.

Solution: Let G = Z = 〈1〉. Observe that o(1) =∞. Let G1 = Z/nZ = 〈1〉. Observe that o(1) = n. Define α : G→ G1

by α(m) = m for all m ∈ G. For r, s ∈ G, we have:

α(r + s) = r + s = r + s = α(r) + α(s),

so α is a homomorphism (but it is not an isomorphism since it is not injective, e.g., α(0) = α(n)).

Problem 2.5.9.

Solution: Let α : G→ G1 be a homomorphism and K = {g ∈ G : α(g) = 1G1}. We have 1G ∈ K because α(1G) = 1G1

by Theorem 2.5.1. Given a, b ∈ K:α(ab) = α(a)α(b) = 1G1

1G1= 1G1

,

so ab ∈ K. Finally, given c ∈ K where c−1 is its inverse in G:

α(c−1) = α(c)−1 = 1−1G1= 1G1 ,

so c−1 ∈ K. By the Subgroup Test, K is a subgroup of G.

Page 32

Problem 2.5.10.

Solution: Given g ∈ G, it follows that λ(g−1) = (g−1)−1 = g, so λ is surjective. If λ(a) = λ(b) for any a, b ∈ G, thena−1 = b−1, so a = b; therefore λ is injective. We conclude that λ is bijective.

Now suppose G is abelian. By exercise 2.5.3, λ is a homomorphism. Since λ is also a bijection, λ is a isomorphism.

Problem 2.5.11.

Solution: Let G be a group and a ∈ G. Given u ∈ G, by closure a−1ua ∈ G, so αa(a−1ua) = a(a−1ua)a−1 =(aa−1)u(aa−1) = u, so α is surjective. If α(v) = α(w) for some v, w ∈ G, then ava−1 = awa−1, so by right and leftcancellation v = w and α is injective. Hence α is a bijection.

Problem 2.5.12.

Solution: Part (a) Yes. This is a straightforward proof.Part (b) No because α is not a bijection. Given an odd m, there is no α(n) = m because α(n) is always even.

Part (c) No because α is not a bijection. α(1) = 12

= 1 = 42

= α(4), so α is not injective. The map α is also notsurjective since α(n) ∈ {1, 4} for all n ∈ G.

Part (d) Yes. This is a straightforward proof.Part (e) Yes. First, α is a homomorphism. Given a, b ∈ Z/7Z:

α(a+ b) = 2(a+ b) = 2a+ 2b = α(a) + α(b).

If α(a) = α(b), then 2a = 2b. Since 2−1

= 4, we have 4 · 2a = a = 4 · 2b = b, so α is injective.

Finally, given c ∈ Z/7Z, we must find x such that 2x = c. Multiplying both sides by 2−1

, we have x = 4c, so α issurjective.

Part (f) No because α is not surjective. There is no g ∈ Z/8Z such that α(g) = 2g = 7 (note 2 has no multiplicativeinverse).

Part (g) Yes. This is a straightforward proof.Part (h) No because α is not injective. o(g) =∞ for any g 6= ±1.Part (i) Yes. Any element x ∈ 2Z may be expressed as x = 2x′ and any y ∈ 3Z may be expressed as y = 3y′, where

x′, y′ ∈ Z. Given 2a, 2b ∈ 2Z, we have:

α(2a+ 2b) = α(2(a+ b)) = 3(a+ b) = 3a+ 3b = α(2a) + α(2b),

so α is a homomorphism. If α(2a) = α(2b), then 3a = 3b, so a = b and therefore 2a = 2b. If 3c ∈ 3Z, then α(2c) = 3c.Part (j) Yes. This is a straightforward proof.

Problem 2.5.13.

Solution: Designate:

U =

(0 −11 0

).

We have G = 〈U〉 with o(U) = 4 because:

U0 =

(1 00 1

), U2 =

(−1 00 −1

), U3 =

(0 1−1 0

).

Because G is a a closed, non-empty, finite subset of GL2(Z), the Finite Subgroup Test implies that it is a subgroup ofGL2(Z).

Now denote I = {1,−1, i,−i}. We have I = 〈i〉 with o(i) = 4 because i0 = 1, i2 = −1, and i3 = −i.Having found that both G and I are cylic groups of order 4, finding an isomorphism is straightforward. Define σ : G→ I

as σ(Uk) = ik for any Uk ∈ G. We will show that σ is well-defined. Any element of G is of the form Uk for some k ∈ Z.If U j = Uk for some k ∈ Z, then σ(U j) = ij = σ(Uk) = ik. Since o(U) = 4, it follows that j ≡ k (mod n), so (since I isfinite of order 4 are well) ij = ik. Thus σ is well-defined.

Page 33

Now we will show that σ is a homomorphism. We may express any A,B ∈ G as A = Ua and B = U b for some a, b ∈ Z.Therefore:

σ(AB) = σ(UaU b) = σ(Ua+b) = [σ(U)]a+b = ia+b = iaib = σ(A)σ(B).

Next we will show that σ is a bijection. If σ(A) = σ(B), then ia = ib, so a ≡ b (mod 4) by Theorem 2.4.2, from whichit follows that A = Ua = U b = B (since o(U) = o(i) = 4). Therefore σ is injective. It is also surjective: Given c ∈ I,we may express c = iγ for some γ ∈ Z, so σ(Uγ) = [σ(U)]γ = iγ = c. Since σ is a bijective homomorphism, it is anisomorphism.

Problem 2.5.14.

Solution: Let G be an infinite cyclic group where G = 〈a〉. If g ∈ G, then g = aj for some j ∈ Z. Define α : G→ Z byα(g) = j where aj = g. Since o(a) =∞, if k 6= j, then ak 6= aj , so there is a unique j such that aj = g. Therefore α iswell-defined. Given g0, g1 ∈ G where g0 = aj and g1 = ak, we have:

α(g0g1) = α(ajak) = α(aj+k) = j + k = α(aj) + α(ak) = α(g0) + α(g1).

Hence α is a homomorphism.If α(g0) = α(g1), then j = k where aj = g0 and ak = g1, so g0 = ak = g1 and α is injective. If l ∈ Z, then gl ∈ G

and α(gl) = l, so α is surjective. Thus G is isomorphic with Z.

Problem 2.5.15.

Solution: Let G = 〈a〉 where o(a) = n. If g ∈ G, then g = aj for some j ∈ Z. Define α : G→ Z/nZ by α(g) = j whereaj = g. We will show that α is well-defined. If g = aj = ak, Theorem 2.4.2 implies that j ≡ k (modn). As a result, j = k,so there is a unique residue in Z/nZ to which α maps g.

If g0, g1 ∈ G where g0 = aj and g1 = ak for j, k ∈ Z, then:

α(g0g1) = α(ajak) = α(aj+k) = j + k = j + k = α(aj) + α(ak) = α(g0) + α(g1),

so α is a homomorphism. If α(g0) = α(g1), then j = k, from which it follows that j ≡ k (mod n). By Theorem 2.4.2,g0 = aj = ak = g1, so α is injective. Finally, if l ∈ Z/nZ, then α(al) = l; since al ∈ G, the map α is surjective. Weconclude that G ∼= Z/nZ.

Problem 2.5.16.

Solution: Define σ : C∗ → C∗ by σ(z) = z∗ for z ∈ C∗. If z0, z1 ∈ C∗, then:

σ(z0z1) = (z0z1)∗ = z∗0z∗1 = σ(z0)σ(z1),

so σ is a homomorphism. Further, if σ(z0) = σ(z1), then z∗0 = z∗1 , so (z∗0)∗ = z0 = (z∗1)∗ = z1, hence σ is injective.Finally, given z2 ∈ C∗, we have σ(z∗2) = z2, so σ is surjective. Thus σ is an automorphism.

Problem 2.5.17.

Solution: Let g and h be elements of group G. Define α : 〈gh〉 → 〈hg〉 by α(gh) = hg. Let a, b ∈ 〈gh〉 where a = (gh)j

and b = (gh)k for some j, k ∈ Z. Since α(a) = α((gh)j) = α(gh)j = (hg)j , the map is defined for all of 〈gh〉. Further,by exercise 2.4.23(b), 〈gh〉 and 〈hg〉 are the same order. Suppose (gh)j = (hg)k. If o(gh) = n then j ≡ k (mod n), so(hg)j = (hg)k because o(hg) = n, as well; on the other hand, if o(gh) =∞, then j = k, so (hg)j = (hg)k. It follows thatα is well-defined.

Now we will show that α is an isomorphism. We have:

α(ab) = α((gh)j+k) = (hg)j+k = (hg)j(hg)k = α((hg)j)α((hg)k) = α(a)α(b),

so α is a homomorphism. If α(a) = α(b), then (hg)j = (hg)k, and from the argument above we infer that a = gj = gk = b.Given c ∈ 〈hg〉, we have c = (hg)l for some l ∈ Z, so α((gh)l) = c and (gh)l ∈ 〈gh〉. Thus α is bijective, so 〈gh〉 ∼= 〈hg〉.

Page 34

Problem 2.5.18.

Solution:

Problem 2.5.19.

Solution: Let σ : G → G1 be an isomorphism. If σ(a) ∈ σ[Z(G)], then ab = ba for all b ∈ G. Because σ is surjective,given any c ∈ G1 there is some σ(b) = c. Therefore:

σ(ab) = σ(a)σ(b) = σ(a)c = σ(ba) = σ(b)σ(a) = cσ(a).

Since this is true for any c ∈ G1, it follows that σ(a) ∈ Z(G1), so σ[Z(G)] ⊆ Z(G1).Conversely, suppose d ∈ Z(G1). The inverse σ−1 exists and is an isomorphism by Theorem 2.5.3. Given any e ∈ G, let

f = σ(e). We have:

σ−1(df) = σ−1(d)σ−1(f) = σ−1(d)e = σ−1(fd) = σ−1(f)σ−1(d) = eσ−1(d),

so σ−1(d) ∈ Z(G). Therefore σ(σ−1(d)) = d ∈ σ[Z(G)], so Z(G1) ⊆ σ[Z(G)]. We conclude that Z(G1) = σ[Z(G)].

Problem 2.5.20.

Solution: Define nZ = {nk : k ∈ Z}. Thus any x ∈ nZ may be expressed as x = ny for some y ∈ Z. Defineα : nZ → mZ by α(na) = ma for any na ∈ nZ. Given na, nb ∈ nZ, this map is well-defined because if na = nb, thena = b so α(na) = ma = mb = α(nb). If α(na) = α(nb), then ma = mb, so a = b and na = nb. If mc ∈ mZ, thenα(nc) = mc and nc ∈ nZ. Thus α is a bijection. Finally:

α(na+ nb) = α(n(a+ b)) = m(a+ b) = ma+mb = α(na) + α(nb),

so α is a homomorphism. Thus nZ ∼= mZ.

Problem 2.5.21.

Solution: Suppose α : (Z/10Z)× → (Z/12Z)× be an isomorphism. Observe that 3 and 7 are multiplicative inverses ofeach other in (Z/10Z)×. Designate a = α(3) and b = α(7). Since α is injective, a 6= b. Then α(3 · 7) = a · b = α(1) = 1,

so a = b−1

. But every element of (Z/12Z)× is its own inverse, so a = b, a contradiction. We infer that (Z/10Z)× is notisomorphic to (Z/12Z)×.

Problem 2.5.22.

Solution: Suppose α : R→ R∗ is an isomorphism. Find c ∈ R such that α(c) = −1. We have:

α(c+ c) = [α(c)]2 = (−1)2 = 1 = α(0),

so c + c = 0 because α is injective. Therefore c = −c, so by the injectivity of α, it must be that c = 0. However, thiscontradicts the fact that α(0) = 1 by Theorem 2.5.1. We conclude that R is not isomorphic to R∗.

Alternative Proof. Suppose α : R → R∗ is an isomorphism. Observe that if a ∈ R, then o(a) = 1 if a = 0 ando(a) = ∞ otherwise. By contrast, the element −1 of R∗ is order 2 (since (−1)2 = 1). Any isomorphism betwen thetwo groups preserves order. Since there is no element of R of order 2, there is no x ∈ R such that α(x) = −1. By thecontrapositive of Theorem 2.5.4, α is not an isomorphism.

Problem 2.5.23.

Page 35

Solution: Suppose α : C0 → R∗ is an isomorphism. Let a = eiπ/2, so a2 = −1. Designate b = α(a) and c = α(−1).Since α is injective, b, c 6= 1 (since α(1) = 1). Therefore:

α(a2) = b2 = α(−1) = c,

andα((−1)2) = α(1) = c2 = 1.

In sum, b2 = c and c2 = 1, so b, c ∈ {−1,+1}. But because c 6= 1, it cannot be that b = ±1, a contradiction. Weconclude that C0 is not isomorphic to R∗.

Problem 2.5.25.

Solution: No, Z and Q are not isomorphic. Suppose α : Z → Q is an isomorphism. The set Z is cyclic with generator 1(see example 2.4.3). By Corollary 2 to Theorem 2.5.1, Q = 〈α(1)〉. However, Q is not cyclic. Given any q ∈ Q, we mayexpress q = a/b where a ∈ Z and b ∈ Z>0 and gcd(a, b) = 1. Obviously 0 cannot generate Q, so assume q 6= 0. There isno k ∈ Z such that kq = k(a/b) = a/(2b), so there is no generator of Q. Consequently, there is no isomorphism α betweenthe two groups.

Problem 2.5.26.

Solution: We will show that (Z/14Z)× and (Z/18Z)× are both cyclic and of order 6. A straightforward calculation shows

that (Z/14Z)× = 〈3〉 and (Z/18Z)× = 〈5〉 where o(3) = o(5) = 6. Thus any a ∈ (Z/14Z)× may be expressed as a = 3j

and any b ∈ (Z/18Z)× may be expressed as b = 5k

where j, k ∈ Z.

Define α : (Z/14Z)× → (Z/18Z)× where α(3j) = 5

j. This map is well-defined because if 3

j= 3

k, then j ≡ k (mod 6),

so 5j

= 5k

by Theorem 2.4.2. It is straightforward to show that α is an isomorphism, so we will omit the proof here.

Problem 2.5.27.

Solution: Suppose G = 〈a〉 and G1 = 〈b〉 where o(a) = o(b) = 6. By example 2.5.8, any isomorphism λ : G→ G1 mustbe such that G1 = 〈α(a)〉. Further, λ is completely determined by its effect on the generating set of G, so we only needto worry about how the generator a of G is mapped to G1. Theorem 2.4.8 implies that bj is a generator of G1 if and onlyif gcd(j, 6) = 1. There are two such j: 1 and 5. Therefore if σ0, σ1 : G → G1 are distinct isomorphisms, it must be thatσ0(a) = b and σ1(b) = b5 = b−1. We therefore define:

σ0(ak) = bk and σ1(ak) = b−k.

We will show these are isomorphisms. In the case of σ0, if c ∈ G1, then c = bk for some k, so σ0(ak) = bk.Suppose d, e ∈ G where d = aj and e = al for some j, l. If σ0(d) = σ0(e), then bj = bl, so j ≡ l (mod 6); it followsthat d = aj = al = e. Thus σ0 is a bijection. Further, σ0(de) = σ0(aj+l) = bj+l = bjbl = σ0(d)σ0(e), so σ0 is ahomomorphism. We conclude that σ0 is an isomorphism. A similar argument shows σ1 is an isomorphism.

There is no other isomorphism λ from G to G1. If g ∈ G1\{b, b−1}, then g = br where r ∈ {0, 2, 3, 4}. But r is notrelatively prime to 6, so br cannot generate G1. Therefore if λ(a) = br, then G1 6= 〈λ(a)〉; by example 2.5.8, λ cannot bean isomorphism.

Problem 2.5.28.

Solution: Note that G = Z>0 × C0 is a multiplicative group, with any element in C0 of the form eiφ. Intuitively, we canview this product G as a complex exponential reiφ with a phase φ given by the C0 part (since it is on the unit circle) anda modulus r given by Z>0. Since reiφ = r(cosφ+ i sinφ), we end up with the coordinate of this complex number in thecomplex plane. Since reiφ = 0 + i0 only if r = 0, we see the origin of the complex plane is not associated with any elementof G. It is easy to see that every other point in the complex plane is expressible by a unique reiφ.

Page 36

Let’s formalize this approach. Define α : G→ C∗ by φ(r, eiφ) = reiφ. Let (r0, eiφ0), (r1, e

iφ1) ∈ G. We have:

α[(r0, eiφ0), (r1, e

iφ1)] = α(r0r1, ei(φ0+φ1)) = r0r1e

i(φ0+φ1) = r0eiφ0r1e

iφ1 = α(r0, eiφ0)α(r1, e

iφ1).

Hence α is a homomorphism.Any element a ∈ C∗ is of the form a = reiφ where r 6= 0. Since r ∈ Z>0 and eiφ ∈ C0, it follows that α(r, eiφ) = a,

so α is surjective. Finally, if α(r0, eiφ0) = α(r1, e

iφ1), then r0eiφ0 = r1e

iφ1 . Dividing each complex number into its realand imaginary parts, we see that r0 cosφ0 = r1 cosφ1 and r0 sinφ0 = r1 sinφ1. Squaring both expressions and adding, weget r20(cos2 φ0 + sin2 φ0) = r21(cos2 φ1 + sin2 φ1), so r20 = r21, from which it follows that r0 = ±r1. Since r0 and r1 mustbe positive, we have r0 = r1. It follows that eiφ0 = eiφ1 , so φ0 = φ1 Accordingly, (r0, e

iφ0) = (r1, eiφ1), so α is injective.

We conclude that α is an isomorphism.

Problem 2.5.29.

Solution: Each τa,b ∈ G1 is a bijection. Given y ∈ R, τa,b((y−b)/a) = y (which is valid since a 6= 0). If τa,b(x0) = τa,b(x1),then ax0 + b = ax1 + b, so x0 = x1. Therefore each τa,b is a permutation of R, so G1 ⊆ SR.

The identity permutation ε is contained in G1 because ε = τ1,0. Given τa0,b0 , τa1,b1 , for any x ∈ R we haveτa0,b0τa1,b1(x) = (a0a1)x+(b0+a0b1). Since a0a1 6= 0 and b0+a0b1 ∈ R, it follows that τa0,b0τa1,b1 = τa0a1,b0+a0b1 ∈ G1,so G1 is closed. Finally, given τa,b ∈ G1, its inverse in SR is τ−1a,b (y) = (y − b)/a (which is easily checked). But

τ−1a,b = τ1/a,−b/a ∈ G1 (which is valid since a 6= 0). By the Subgroup Test, G1 is a subgroup of SR.Next we will show that G is a subgroup of GL2(R). The identity matrix I is obviously an element of G with a = 1 and

b = 0. Given A0, A1 ∈ G, we may express each as:

A0 =

(a0 b00 1

)and A1 =

(a1 b10 1

), (3)

where a0, a1 6= 0. It is easily shown that:

A0A1 =

(a0a1 b0 + a0b1

0 1

).

Since a0a1 6= 0, it follows that A0A1 ∈ G, so G is closed. (Notice the similarity to the closure of G1!)It is easily shown that the inverse of A0 in GL2(R) is:

A−10 =

(1a0− b0a0

0 1

).

Since 1/a0 6= 0 and is valid, A−10 ∈ G. By the Subgroup Test, G is a subgroup of GL2(R).Having established that G and G1 are subgroups, we will show that they are isomorphic. Define α : G → G1 by

G(A0) = τa0,b0 where A0 is defined as in equation (3). Since A0 is uniquely defined by the values of a0, b0, this map iswell-defined. Given A0, A1 ∈ G:

α(A0A1) = α(

(a0a1 b0 + a0b1

0 1

)) = τa0a1,b0+a0b1 = τa0,b0τa1b1 = α(A0)α(A1),

where we used the results from our argument that G1 is closed. Hence α is a homomorphism. Given τa,b ∈ G1, we haveα(A0) = τa,b where a0 = a and b0 = b. Finally, if α(A0) = α(A1), then τa0,b0(x) = a0x+ b0 = a1x+ b1 = τa1,b1(x) forany x ∈ R. If x = 0, we have b0 = b1. If x = 1, then a0 + b0 = a1 + b1 = a1 + b0, so a0 = a1. Therefore A0 = A1. Weconclude that α is an isomorphism, so G ∼= G1.

Problem 2.5.31.

Solution: Let G = 〈a〉 where o(a) = n.Part (a) Suppose n = 2, so |G| = 2. We infer that G = {1, a}. Let α : G → G be an isomorphism. Theorem 2.5.1

requires that α(1) = 1, so α(a) = a, hence α = 1G. Since this is the only possible automorphism, autG = {1G}.Part (b) Suppose n = 3, so |G| = 3 and G = {1, a, a2}. For any automorphism α, Theorem 2.5.4 implies that

o(α(a)) = 3. As a result, α(a) ∈ {a, a2}. There are therefore two possible automorphisms: α0, α1 : G → G where

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α0(ak) = ak (which equals 1G) and α1(ak) = (a2)k = a−k. Therefore autG = {1G, α1}.

Problem 2.5.32.

Solution: Suppose G is an infinite cyclic group, which we can denote G = 〈g〉 for some g ∈ G where o(g) = ∞. Letα : G → G be an automorphism of G. Designate α(g) = gj for some j ∈ Z. Since α is surjective, there is some gk suchthat α(gk) = g. Therefore:

α(gk) = [α(g)]k = (gj)k = gjk = g.

Since o(g) is infinite, by Theorem 2.4.3 jk = 1. Therefore j = ±1, so α(g) = g or α(g) = g−1. Therefore the twopossible automorphisms of G are α0, α1 where α0(gk) = gk (so α0 = 1G) and α1(gk) = g−k. Thus autG = {1G, α1}.

Problem 2.5.33.

Solution: Suppose for a given group G that Z(G) = {1}. It follows that if a ∈ G\{1} then ag 6= ga for some g ∈ G.Define θ : G → inn G by θ(a) = σa where σa : G → G such that σa(g) = aga−1. We know from example 2.5.17 that θis a homomorphism. We will show that θ is bijective. Let a, b ∈ G be given. If θ(a) = θ(b), then σa = σb. As a result,if g ∈ G then aga−1 = bgb−1. We then have ga−1 = a−1bgb−1, so ga−1b = a−1bg. Therefore a−1b ∈ Z(G) = {1}, soa−1b = 1. We infer that a = b, hence θ is injective.

If σc ∈ G for some c ∈ G, then θ(c) = σc, so θ is surjective. We conclude that θ is an isomorphism, so G ∼= innG.

Problem 2.5.34.

Solution: Let Gz be given for some z ∈ Z(G). Let a, b, c ∈ Gz. The set Gz is a group under the operation ∗:

• Gz is closed. We have a ∗ b = abz−1 ∈ G by the closure of G, so Gz is closed.

• The ∗ operation is associative. We have:

(a ∗ b) ∗ c = (abz−1) ∗ c = (abz−1)cz−1 = a(bcz−1)z−1 = a ∗ (b ∗ c),

where we used the fact that z ∈ Z(G).

• z is the unity of Gz We have:z ∗ a = zaz−1 = azz−1 = a = a ∗ z.

• Every element of Gz has an inverse Given a, its inverse in Gz is a−1z2 since (using the commutativity of z):

a ∗ (a−1z2) = aa−1z2z−1 = z,

and(a−1z2) ∗ a = a−1z2az−1 = a−1az2z−1 = z.

Next we will show that Gz is isomorphic with G. Define α : Gz → G where α(a) = az−1 for a ∈ Gz. Given a, b ∈ Gz,we have:

α(a ∗ b) = α(abz−1) = abz−2 = (az−1)(bz−1) = α(a)α(b),

so α is a homomorphism. Given g ∈ G, it follows that α(gz) = gzz−1 = g, so α is surjective. If α(a) = α(b), thenaz−1 = bz−1, so by right cancellation a = b, hence α is injective. We conclude that Gz ∼= G.

Problem 2.5.36.

Solution: Part (a) Define a ∼ b for a and b in group G if b = gag−1 for some g ∈ G. We will show that ∼ is anequivalence relation. The relation is reflexive because 1Ga1G = a, so a ∼ a for any a ∈ G. If a ∼ b, then b = gag−1 forsome g ∈ G, so g−1bg = a; hence a ∼ b, so ∼ is symmetric. Finally, if a ∼ b and b ∼ c, there are g0, g1 ∈ G such thatb = g0ag

−10 and c = g1bg

−11 , so c = (g1g0)a(g−10 g−11 ) = (g1g0)a(g1g0)−1. Therefore a ∼ c, so ∼ is transitive.

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Part (b) The following two results are true of any group G. [1] = {1} for if a ∼ 1 then 1 = gag−1 for some g ∈ G, soa = g−1g = 1. If c ∈ Z(G), then [c] = {c} because given c ∼ b, there is some g ∈ G such that b = gcg−1 = cgg−1 = c.

If G is abelian, then every [d] for d ∈ G is a singleton. For if e ∼ d, then there is some g ∈ G such that d = geg−1 =egg−1 = e.

Problem 2.5.37.

Solution: Suppose G = 〈X〉 for some non-empty subset X of G and σ : G→ G1 is an onto homomorphism with a groupG1. If σ(x) ∈ σ(X), then σ(x) ∈ G1, so σ(X) ⊆ G1. By Theorem 2.4.10, 〈X〉 is a subset of G1. Conversely, let g ∈ G1.Because σ is surjective, there is some a ∈ G such that σ(a) = g. Because G is generated by the set X, we may express:

a =

M∏j=1

xαj

j ,

where xj ∈ X and αj ∈ Z for some M ∈ N. Because σ is a homomorphism, it follows that:

σ(a) = σ

M∏j=1

xαj

j

=

M∏j=1

[σ(xj)]αj =

M∏j=1

yαj

j ,

where yj = σ(xj) ∈ σ(X). Accordingly, g is contained in 〈σ(X)〉, so G1 ⊆ 〈σ(X)〉 by Theorem 2.4.10. We conclude thatG1 = 〈σ(X)〉.

Section 2.6

Problem 2.6.1.

Solution: The easiest way to solve this problem is to capitalize on Theorem 2.6.1. If aj ∈ Hak, then Haj = Hak (andsimilarly for left cosets). Remember that the left/right cosets form a partition of G!

Part (a)H1 = 1H = HHa = aH = {a, a5, a9, a13, a17}Ha2 = a2H = {a2, a6, a10, a14, a18}Ha3 = a3H = {a3, a7, a11, a15, a19}

K1 = 1K = KK2 = {a, a3, a5, a7, a9, a11, a13, a15, a17, a19}

Part (b) All the permutations of S4 are listed in example 1.4.9. Since G = A4, we are only concerned about thosepermutations that are a product of an even number of transpositions. By Theorem 1.4.6, those would be permutationsthat are a product of an even number of disjoint transpositions (column 3 in example 1.4.9) and the 3-cycles (column 2,because Theorem 1.4.6 implies that a 3-cycle is equal to a product of 2 transpositions). Even though there are 12 elementsin A4, Theorem 2.6.1 reduces this universe substantially because if any coset B contains an element σ of A4, then thecoset generated by σ equals B and therefore σ may be ignored.

Observe that K = {ε, (1 2 3), (1 3 2) }.

Hε = εH = H(1 2 3)H = { (1 2 3), (1 3 4), (2 4 3), (1 4 2) } = H(1 2 3)(1 2 4)H = { (1 2 4), (1 4 3), (1 3 2), (2 3 4) } = H(1 2 4)Kε = εK = K(1 2 4)K = { (1 2 4), (1 4)(2 3), (1 3 4) }(2 3 4)K = { (2 3 4), (1 3)(2 4), (1 4 2) }(1 4 3)K = { (1 4 3), (1 2)(3 4), (2 4 3) }K(1 2 4) = { (1 2 4), (1 3)(2 4), (2 4 3) }K(2 3 4) = { (2 3 4), (1 2)(3 4), (1 3 4) }K(1 4 3) = { (1 4 3), (1 4)(2 3), (1 4 2) }

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Part (c) Note that H = 2Z = {2k : k ∈ Z}, which is an additive group. We have:

0H = H0 = H and 1H = H1 = {2k + 1 : k ∈ Z}.

If j ∈ Z then j = 2l or j = 2l + 1 for some l ∈ Z, so j ∈ 0H or j ∈ 1H. We infer that jH = 0H or jH = 1H (andsimilarly for right cosets). Therefore there are no other cosets of H.

Note that K = 3Z = {3k : k ∈ Z}, which is also an additive group. Using the same argument as with H, we have:

0K = K0 = K, 1K = K1 = {3k + 1 : k ∈ Z}, and 2K = K2 = {3k + 2 : k ∈ Z}.

Part (d) Note that H = {0, 3, 6, 9} and K = {0, 2, 4, 6, 8, 10} (overbars omitted; remember these are residue classes,not integers). We have:

0H = H0 = H1H = H1 = {1, 4, 7, 10}2H = H2 = {2, 5, 8, 11}

0K = K0 = K1K = K1 = {1, 3, 5, 7, 9, 11}

These are all the left and right cosets of H and K. For H, given j ∈ G, we may express j = 3l + r by the divisionalgorithm, so 0 ≤ r ≤ 2. The element 3l ∈ H so 3l+r = 3l + r = j ∈ rH by definition (remember rH = {3k+r : k ∈ G}).The same is true for K.

Part (e) H = 〈a2〉 = {1, a2}. The left and right cosets of H are:

1H = H1 = HaH = {a, a3} = HabH = {b, ba2} = bHbaH = {ba, ba3} = Hba

Note that the right cosets required judicious use of the condition aba = b (e.g., in Hb, we found a2b = a(aba)a−1 =aba−1 = aba3 = (aba)a2 = ba2).

K = 〈b〉 = {1, b}. The left and right cosets of K are:

1K = K1 = KaK = {a, ba3}a2K = {a2, ba2} = Ka2

a3K = {a3, ba}Ka = {a, ba}Ka3 = {a3, ba3}

Part (f) Let group G be given and H some subgroup of G with index 2. We will argue only with respect to the leftcosets; the argument for the right cosets of H is analogous. There are exactly two left cosets of H by hypothesis, and oneof them must be 1H. If a ∈ H, then a−1 ∈ H, so aa−1 = 1 ∈ aH, from which it follows that aH = 1H by Theorem2.6.1(4). As a result, the other left coset of H must be generated by an element not contained in H. Assume b /∈ H.Clearly b ∈ bH, and since b /∈ 1H, it follows that 1H 6= bH. By Theorem 2.6.1(6), 1H and aH are disjoint. Since1H ∪ aH = H, we infer that aH = G\H. Therefore the two left cosets of H are H itself and the set of all elements of Gthat are not in H.

Problem 2.6.2.

Solution: Let G be a group. We will show the only coset of subgroup G is G. Let a ∈ G be given. Observe that aG andGa are subsets of G by the closure of G as a group. If g ∈ G, designate x0 = ga−1 and x1 = a−1g, so x0, x1 ∈ G. Itfollows that x0a = ax1 = g. Therefore g ∈ Ha and g ∈ aH. We conclude that aG = Ga = G for any a ∈ G.

Let b ∈ G be given. We have b{1} = {1}b = {b}. Therefore the cosets of G are the set of singletons of the elementsin G (i.e., {{a} : a ∈ G}).

Problem 2.6.3.

Page 40

Solution: No, Ha = Hb for a, b ∈ G does not imply that aH = bH. Let G = D4 and H = 〈b〉 = {1, b}. We haveHa = {a, ba} = Hba. However, aH = {a, ba3}, but baH = {a3, ba}, so aH 6= baH.

Problem 2.6.4.

Solution: Let K ⊆ H ⊆ G be finite subgroups. By Lagrange’s Theorem:

|G||H| = |G : H| and |H|

|K| = |H : K| .

As a result:

|G : H| · |H : K| = |G||H|· |H||K|

=|G||K|

= |G : K| .

Problem 2.6.5.

Solution: Let H be a subgroup of G. Define a ≡ b for a, b ∈ G if b−1a ∈ H.Part (a) We will show that ≡ is an equivalence relation. Let a, b, c ∈ G. Since a−1a = 1 ∈ H, it follows that a ≡ a.

If a ≡ b, then b−1a ∈ G, so its inverse (b−1a)−1 = a−1b ∈ H, hence b ≡ a. If b ≡ c as well, then c−1b ∈ H. Combiningthe two elements, we have (c−1b)(b−1a) = c−1a ∈ H, so a ≡ c. As a result, ≡ is an equivalence relation.

Part (b) Let a ∈ G. If b ∈ [a], then b−1a ∈ H. It follows that a−1b ∈ H, so a(a−1b) = b ∈ aH. Therefore [a] ⊆ aH.Conversely, suppose c ∈ aH. It follows that there is some h ∈ H such that ah = c, so h = a−1c ∈ H. Therefore a ≡ cand (by symmetry of equivalences) c ∈ [c]. Thus aH ⊆ [a]. We conclude that [a] = aH.

Problem 2.6.6.

Solution: Showing H is a subgroup of G is trivial, so we’ll skip the proof.The element (a, b) ∈ G represents a point at coordinate (a, b) in the x-y plane. The right coset (identical to the left

coset) of H generated by (a, b) is:H + (a, b) = {(x+ a,mx+ b) : x ∈ R}.

The coset H + (a+ b) corresponds to the line y = m(x− a) + b. This line has the same slope m as the line in H buthas been translated to the left along the x-axis such it intersects the point (a, b). Another way of looking at H + (a, b) isthat it is the same line as in H but translated such that it passes through the points (0, b−ma) and (a, b).

Problem 2.6.7.

Solution: Suppose H is a subgroup of G and Ha = bH for some a, b ∈ G. It follows that a ∈ bH, so there is someu0 ∈ H such that bu0 = a. If h ∈ aH, there is some u1 ∈ H such that au1 = h, so (bu0)u1 = b(u0u1) = h. Sinceu0u1 ∈ H, we infer that b(u0u1) = h ∈ bH. By Theorem 2.6.1(5), since aH and bH are not disjoint, aH = bH. By thesame argument, Hb = Ha. By the transitivity of equivalences, aH = bH = Ha = Hb. (Note we’ve proven the strongerstatement that the left and right cosets of H generated by a and b are all equal.)

Problem 2.6.8.

Solution: Given subgroups H and K of G, suppose Ha ⊆ Kb for some a, b ∈ G. If h ∈ H, then ha ∈ Kb, so there issome k ∈ K such that ha = kb. But a ∈ Kb, so there is some k′ ∈ K such that a = k′b. Putting these together, wehave ha = hk′b = kb, so hk′ = k by right cancellation and h = (k′)−1k. Since (k′)−1k ∈ K, it follows that h ∈ K, henceH ⊆ K.

Problem 2.6.9.

Page 41

Solution: Part (a) Given g ∈ R∗, we have gH = Hg = {hg : h ∈ R>0}. If g > 0, then obviously gH ⊆ R>0 (andthe same for Hg). If x ∈ R>0, then (x/g) · g = x, and x/g ∈ R>0 (all valid because g 6= 0). Therefore R>0 ⊆ gH, soR>0 = gH = Hg. On the other hand, if g < 0, then gH ⊆ R<0. If x′ ∈ R<0, then (x′/g)g = x′ and x′/g ∈ R>0 (sincex′ and g are both negative), so R<0 ⊆ gH. Therefore gH = Hg = R<0.

Part (b) Let z ∈ C∗, so z = reiφ where r ∈ R∗ and φ ∈ R. Thus zH = Hz = {hz : h ∈ R>0}. Accordingly, forz′ ∈ zH we have z′ = (hr)eiφ where the complex angle is fixed for all elements of zH. The magnitude hr ranges over allof R∗: Given x ∈ R∗, we have (x/h) · h = x where x/h ∈ R∗. As a result, zH = Hz corresponds to a line in the complexplane that excludes the origin and makes an angle φ with the real axis.

Part (c) Let x ∈ R. We have xH = Hx = {n+x : n ∈ Z}. This corresponds to all points that are an integer distancefrom x. Alternatively, it is the set of points that are a distance x− bxc from each integer.

Part (d) Let z ∈ C, so z = a + ib for some a, b ∈ R. Accordingly, zH = Hz = {(x + a) + ib : x ∈ R}. This cosetrepresents that points on a vertical line in the complex plane with constant imaginary part ib.

Problem 2.6.10.

Solution: Part (a) Let G = 〈a〉 with o(a) = 30 and H = 〈a6〉. By the corollary to Theorem 2.4.5, o(a6) = o(a)/6 = 5.By Lagrange’s Theorem, |G : H| = |G| / |H| = 30/5 = 6, which is the index of H.

Part (b) Let G = 〈a〉 and o(a) = n. Suppose d | n. The corollary to Theorem 2.4.5 implies that o(ad) = n/d = k, so∣∣〈ad〉∣∣ = k. By Lagrange’s Theorem, the index of 〈ad〉 is |G| / |〈a〉| = n/k = d.

Problem 2.6.11.

Solution: Let H and K be subgroups of group G.Part (a) Let u ∈ Ha ∩ Ka. It follows that there are elements h0 ∈ H and k0 ∈ K such that h0a = k0a = u. By

right cancellation, h0 = k0, so h0 ∈ H ∩K. As a result, h0a = u ∈ (H ∩K)a, so Ha ∩Ka ⊆ (H ∩K)a. Conversely, ifv ∈ (H ∩K)a, then v = h1a = k1a for some h1 ∈ H and k1 ∈ K. Therefore v ∈ Ha and v ∈ Ka, so v ∈ Ha ∩Ka,hence (H ∩K)a ⊆ Ha ∩Ka. We conclude that the two sets are equal.

Part (b)

Problem 2.6.12.

Solution: Part (a) By Corollary 2.6.2, g12 = 1, so if n = o(g), then n | 12. But n cannot equal 4 or 6 or any divisorthereof (otherwise g4 = 1 or g6 = 1), so it must be that n = 12. By Theorem 2.4.2, G = 〈g〉.

Part (b) We use the same argument as in part (a). Since g40 = 1, it must be that o(g) | 40, but o(g) cannot divide 8,20, or any divisor thereof. Therefore o(g) = 40, so G = 〈g〉.

Part (c) We use the same argument: g60 = 1 so o(g) | 60, but o(g) cannot divide 12, 20, 30, or any divisor thereof.Therefore o(g) = 60, so G = 〈g〉.

Problem 2.6.13.

Solution: Since H contains a 3-cycle which we will designate σ, it follows that 〈σ〉 is a subgroup of H. It follows that|〈σ〉| = o(σ) = 3. Lagrange’s Theorem implies that 3 | |H|. But it is easily shown that K is a subgroup of H, so|K| = 4 | |H|. Since 3 and 4 are relatively prime, 12 | |H|. But |H| ≤ |A4| = 12, so |H| = 12. Hence H = A4.

Problem 2.6.14.

Solution: Lagrange’s Theorem implies that 45 | |G| and 75 | |G| where |G| < 400. The only possibility is |G| = 225.

Problem 2.6.15.

Page 42

Solution: Let H and K be subgroups of group G with |H| = p where p is prime. The set H∩K is a subgroup of H and K,so by Lagrange’s Theorem |H ∩K| | |H|. Since |H| is prime, |H ∩K| ∈ {1, p}. If |H ∩K| = 1, then H ∩K = {1} (sincethe unity is always a member of a group). Otherwise, |H ∩K| = p, so it equals all of H; accordingly, H ∩K = H ⊆ K.

Problem 2.6.16.

Solution: Let G be a group of order n and m ∈ Z where gcd(m,n) = 1.Part (a) Suppose gm = 1. By corollary 2.6.2, gn = 1. Since m and n are relatively prime, there are x, y ∈ Z such that

mx+ ny = 1. As a result:g1 = g = gmx+ny = (gm)x(gn)y = 1x · 1y = 1.

Part (b) Let g ∈ G be given. Again by corollary 2.6.2, gn = 1. Thus:

gny = (gn)y = 1 = g1−mx = g · g−mx.

Accordingly, g = (gx)m = am where a = gx. It follows that a is the mth root of g.

Problem 2.6.17.

Solution: Suppose G is a group where |G| = p2 such that p is prime. Let H be a proper subgroup of H. By Lagrange’sTheorem, |H| divides p2, so |H| ∈ {1, p} (it cannot equal p2 because H 6= G). If |H| = 1, then H = 〈1〉. If |H| = p, thenby corollary 2.6.3 H is cyclic.

Problem 2.6.18.

Solution: Suppose G is a non-cyclic group such that |G| = p3 where p is prime. Given g ∈ G, by corollary 2.6.2 gp3

= 1.Since 〈g〉 6= G, it cannot be that |〈G〉| = p3. By Lagrange’s Theorem, |〈g〉| divides p3, so |〈g〉| = o(g) ∈ {1, p, p2}. Since

gp2

= (gp)p = (g)p2

, we infer that gp2

= 1 for any of those three possible values of o(g).

Problem 2.6.19.

Solution: Let a and b be elements of group G where ak = bk for some k ∈ Z. Suppose o(a) = m and o(b) = n wheregcd(m,n) = 1. It follows that |〈a〉| = m and |〈b〉| = n. Further ak ∈ 〈b〉 by hypothesis. By corollary 2.6.4, 〈a〉∩〈b〉 = {1},so ak = bk = 1. Accordingly, m | k and n | k. Since m and n are relatively prime, mn | k by Theorem 1.2.4.

Problem 2.6.20.

Solution: Let n ≥ 3 be given. Observe that n+ (−1)(n− 1) = 1, so by Theorem 1.2.4 n and n− 1 are relatively prime,

hence n− 1 ∈ (Z/nZ)×. Further n− 12

= n2 − 2n+ 1. But n2 − 2n+ 1 ≡ 1 (mod n), so n− 12

= 1. Since n− 1 6= 1,it follows that o(n− 1) = 2. Corollary 2.6.1 implies that 2 | |(Z/nZ)×|, so |(Z/nZ)×| is even.

Problem 2.6.21.

Solution: Let n ≥ 1. Given a ∈ Z, by the division algorithm we have a = nq + r where 0 ≤ r ≤ n − 1. Clearlya(nZ) = (nZ)a by the commutativity of integer addition. We have:

a(nZ) = (nZ)a = {nm+ a : m ∈ Z} = {n(m+ q) + r : m ∈ Z} = {nm′ + r : m′ ∈ Z} = r(nZ) = (nZ)r.

In short, if a ≡ r (mod n) where 0 ≤ r ≤ n − 1, then a(nZ) = r(nZ) (and the same for right cosets). We can thereforelimit our search for distinct cosets to those generated by r ∈ {0, 1, . . . , n− 1}.

Now we will show that the left cosets r(nZ) (and hence the right cosets, as well) are distinct for 0 ≤ r ≤ n − 1.Suppose r0(nZ) = r1(nZ). Since r0 ∈ r0(nZ), it follows that r0 ∈ r1(nZ). As a result, there is some m ∈ Z such thatnm+ r1 = r0, or nm = r0 − r1. Thus n | r0 − r1. But |r0 − r1| ≤ n− 1, so r0 − r1 = 0. We conclude that r0 = r1, sothe left cosets r(nZ) are distinct (and the same for the right cosets). Therefore |Z : nZ| = n.

Page 43

Problem 2.6.22.

Solution: Let G be a group of order n. Define σ : G→ G by σ(g) = gm for all g ∈ G and where gcd(m,n) = 1. We willshow that σ is a bijection. Suppose σ(a) = σ(b) for some a, b ∈ G, so am = bm. Since m and n are relatively prime, thereare x, y ∈ Z such that 1 = mx+ ny. We have:

(am)x = amx = a1−ny = a · (an)−y = a · 1−y = a = (bm)x = bmx = b1−ny = b · (bn)−y = b · 1−y = b,

where we used corollary 2.6.2 for the fact that an = bn = 1. Accordingly, a = b, so σ is injective. Given g ∈ G, by exercise2.6.16(b) there is an mth root a ∈ G such that am = g. Therefore σ(a) = am = g, so σ is surjective.

Now suppose further that G is abelian. Given d, e ∈ G, we have:

σ(de) = (de)m = dmem = σ(d)σ(e),

hence σ is a homomorphism. Therefore if G is abelian then σ is an automorphism of G.

Problem 2.6.23.

Solution: Let G be a group of order pk where p is prime and k ≥ 1. Corollary 2.6.2 implies that if g ∈ G then gpk

= 1.Since |G| > 1, there is some g0 ∈ G where g0 6= 1, so n = o(g0) > 1. It follows that n | pk, from which it follows thatn ∈ {p, p2, . . . , pk}. Since n = pj where 1 ≤ j ≤ k, from Theorem 2.4.5 o(gj−10 ) = pj/pj−1 = j. Thus G has an element

gj−10 of order p.

Problem 2.6.24.

Solution: Suppose G is a group of order pq where p and q are prime. Let H be a proper subgroup of G, in which case|H| 6= pq. Lagrange’s Theorem implies that |H| divides pq, so |H| ∈ {1, p, q}. If |H| = 1, then H = 〈1〉; otherwise, |H| isprime, and by corollary 2.6.3, H is cyclic.

Problem 2.6.25.

Solution: Part (a) The proposition is clear if k = 0 since a0ba0 = b. We will prove that akbak = b for k ∈ N bymathematical induction. If k = 1, then aba = b by the condition of Dn for any n. If the inductive hypothesis holds forsome k ∈ N, then:

ak+1bak+1 = a(akbak)a = aba = b.

Therefore the proposition holds for all k ∈ N. If k < 0, then −k ∈ N. Since b−1 = b, we have:

(akbak)−1 = a−kb−1a−k = a−kba−k = b.

As a result, akbak = b−1 = b.Part (b) Let k ∈ Z. Observe that bak 6= 1 for any k; otherwise, ak = b−1 = b, which contradicts that b /∈ 〈a〉. We

infer that o(bak) > 1. Using the result in part (a), we have (bak)2 = b(akbak) = b2 = 1. Therefore o(bak) = 2 for allk ∈ Z.

Problem 2.6.26.

Solution: Let u ∈ Z(Dn) where n ≥ 3. To solve this problem, we capitalize on the fact that ug = gu for any g ∈ G.Judicious choice of g will show which elements cannot be in Z(G).

Suppose n is odd. If u = ak for some 0 ≤ k ≤ n − 1, then (from exercise 2.6.25(a)) ak(bak) = b = (bak)ak = ba2k.By left cancellation, a2k = 1, so n | 2k. Since n and 2 are relatively prime, n | k; since 0 ≤ k ≤ n − 1, it follows thatk = 0, hence u = a0 = 1. Obviously 1 ∈ Z(G), so no ak where 1 ≤ k ≤ n− 1 is contained in Z(G). If u = baj , then:

(baj)an−j = ban = b = an−j(baj) = an−2j(ajbaj) = an−2jb.

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By right cancellation, an−2j = 1, so a2j = an = 1. Again, this implies that j = 0, but b /∈ Z(G) since ab = (aba)an−1 =ban−1 6= ba if n ≥ 3. Therefore no baj ∈ Z(G). We conclude that Z(G) = {1} is n is odd.

Now suppose n is even where n = 2m for some m > 1. If u = ak for some 0 ≤ k ≤ n− 1, then (again) a2k = 1 andn | 2k. Thus 2m | 2k so m | k. Since m > 1 and k < 2m, we conclude that m = k. It suffices to show that given bal, wehave am(bal) =)bal)am. We have:

am(bal) = am(bal)a2m = (ambam)am+l = bam+l = (bal)am = bm+l,

so am ∈ Z(Dn). If u = baj for some 0 ≤ j ≤ n − 1, then again a2j = 1 = a2m. Therefore 2m | 2j, so m | j. We inferthat j ∈ {0,m}. But b, bam /∈ Z(Dn), for ba 6= ab = ban−1 (since n ≥ 3) and (bam)a = bam+1 6= abam = bam−1. As aresult, Z(Dn) = {1, am} if n is even.

Problem 2.6.27.

Solution: If D5 × C3 is isomorphic to D3 × C5, Theorem 2.5.4 implies that n = o(g) for some g ∈ D5 × C3 if and onlyif there is some element of D3 × C5 also of order n. Denote C3 = 〈c〉. Observe that in D5 × C3 the order of (a, 1) is 5.Given any u ∈ D3 × C5, we have u = (v, cj) where v ∈ D3 and 0 ≤ j ≤ 4. If o(u) = 5, then (v5, c5j). However, the onlyelement v ∈ D3 where v5 = 1 is v = 1 (since a5 = a2, (a2)5 = a, and (bak)5 = ((bak)2)2)bak = b3ak = bak by exercise2.6.25(b)). Similarly, the only cj ∈ C5 where (cj)5 = 1 is j = 1. Accordingly, the only element of order 5 in D3 × C5 is1, 1), but (a, 1) cannot be mapped to (1, 1) because any isomorphism can map only D5 × C3 to the unity of D3 × C5.Therefore the two groups are not isomorphic.

Problem 2.6.29.

Solution: Let G be a group and p be a prime.Part (a) Suppose H and K are subgroups of G of order p. The set H ∩K is a subgroup of both H and K and cannot

be empty since it contains 1. Let u ∈ H ∩K. It follows that 〈u〉 is a subgroup of H and K, so |〈u〉| divides |H| and |K|.Therefore |〈u〉| ∈ {1, p}. If |〈u〉| = 1, then 〈u〉 = {1}, so u = 1. If |〈u〉| = p, then 〈u〉 = H = K. We conclude that eitherH ∩K = {1} or H = K.

Part (b) Suppose there are N distinct subgroups {H1, H2, . . . ,HN} of G, each of order p. Given 1 ≤ i, j ≤ N , theresult in part (a) implies that either Hi = Hj or Hi ∩ Hj = {1}. Consequently, i 6= j implies that Hi ∩ Hj = {1}.Accordingly, each Hi has p − 1 non-unity elements that are not contained in any Hj where i 6= j. The union of all Hi

therefore contains each of these p− 1 non-unity elements plus unity, so:∣∣∣∣∣∣N⋃j=1

Hj

∣∣∣∣∣∣ = 1 +N(p− 1).

Part (c) Suppose G is a group of order 15. Choose g0 ∈ G\{1}. If o(g0) = 15 (in which case G is cyclic), we’re done.Otherwise, Lagrange’s Theorem implies that o(g0) ∈ {3, 5}. If o(g0) = 5, choose g1 ∈ G\〈g0〉. (Note we are assured g1 6= 1

since 1 ∈ 〈g0〉.) Again, o(g1) ∈ {3, 5}. If o(g1) = 5, choose g2 ∈ G\(〈g0〉 ∪ 〈g1〉). If o(g2) = 5, denote H =⋃2j=0〈gj〉.

Observe that 〈g0〉, 〈g1〉, and 〈g2〉 are distinct since g1 /∈ 〈g0〉, g2 /∈ 〈g1〉, and g2 /∈ 〈g0〉. Since each subgroup is of order 5(a prime), part (b) implies that |H| = 1 + 3(4) = 13. Therefore there are exactly two elements in G that do not belongto H. Choose g3 ∈ G\H. Since 〈g3〉 is a subgroup of G not equal to {1}, its order must be either 3 or 5. But its ordercannot be 5: If it were, then |H ∪ 〈g3〉| = 16, an impossibility. Therefore o(g3) = 3.

Problem 2.6.30.

Solution: Let G be a group with |G| ≥ 2 and no subgroups other than {1} and G. Let g ∈ G with g 6= 1 (possible since Gcontains at least two elements). Since the subgroup 〈g〉 6= {1}, by hypothesis 〈g〉 = G, showing that G is cyclic. Assumeo(g) =∞. Since 〈g2〉 = 〈g3〉 = G it follows that g2 ∈ 〈g3〉, so g2 = g3j where j ∈ Z. Theorem 2.4.3 implies that 2 = 3j,which is impossible. Therefore G is finite with o(g) = n. We will show that n is prime. Let d ≥ 1 divide n. By Theorem2.4.9, for every k | n there is a subgroup of G of order k. The contrapositive states that if G does not have a subgroup oforder k, then k - n. Since the only subgroups of G are of order 1 and n, we infer that only 1 and n divide n, hence n isprime.

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Problem 2.6.31.

Solution: Let K ⊆ H ⊆ G be groups. Suppose |G : H| = m and |H : K| = n. Denote Kh1,Kh2, . . . ,Khn be the ndistinct cosets of K in H. The set {Khj}nj=1 form a partition of H. Let g ∈ G be given and Lg =

⋃nk=1Khkg. Observe

that hkg is contained in G, so each Khkg is a coset of K in G generated by hkg. First, we will show that Hg = Lg. Ifu ∈ Lg, there is some j such that u ∈ Khjg, hence there is some k ∈ K ⊆ H such that u = khjg. Since khj ∈ H, itfollows that khj ∈ H, so u ∈ Hg, thus Lg ⊆ Hg. Conversely, if v ∈ Hg, there is some h ∈ H such that v = hg. SinceLg forms a partition of H, we have h ∈ Khk for some k, so v ∈ Khkg, so Hg ⊆ Lg. We conclude that Hg = Lg.

Now we will prove the proposition. Since |G : H| = m, there are m distinct cosets Hg1, Hg2, . . . ,Hgm of H in G, whichform a partition of G. As just shown, each Hgl equals Lgl , the union of n cosets of K in G. Since this forms a partitionof G, we infer that the set of {Lgl}mk=1 are the distinct cosets of K in H. There are |G : K| = mn = |G : H| · |H : K|such cosets of K in G.

Conversely, suppose |G : K| is finite. It follows that |H : K| is finite (the cosets {Kh}h∈K of K in H are a subset ofthe cosets {Kg}g∈K of K in G). Since each Hg for g ∈ G equals Lg, which is a subset of the cosets of K in G and thereare a finite number of distinct cosets of K in G, we infer that |G : H| is finite as well.

Problem 2.6.32.

Solution: Let H and K be subgroups of G with |G : H| = m and |G : K| = n.Part (a) By exercise 2.6.11(a), (H ∩K)g = Hg ∩Kg for any g ∈ G. Clearly if Hg0 = Hg1 and Kg0 = Kg1 for some

g0, g1 ∈ G, then Hg0 ∩Kg0 = Hg1 ∩Kg1. By the contrapositive, a necessary condition for Hg0 ∩Kg0 6= Hg1 ∩Kg1 isfor either Hg0 6= Hg1 or Kg0 6= Kg1. There are m choices of Hg and n choices of Kg, so there are at most mn distinctHg ∩Kg = (H ∩K)g. Therefore |G : H ∩K| = |G : L| ≤ mn.

Part (b) Suppose gcd(m,n) = 1. Denote L = H ∩ K, which is a subgroup of H and K. Since |G : L| is finiteby part (a), it must be that |H : L| and |K : L| are finite, as well. Applying excercise 2.6.31, |H : L| and |K : L|are finite. Let m = |H : L| and n = |K : L|. Again by exercise 2.6.31, |G : L| = |G : H| |H : L| = |G : H|m and|G : L| = |G : K| |K : L| = |G : K|n, so m | |G : H| and n | |G : K|. Since m and n are relatively prime, mn | |G : L|.But 1 ≤ |G : L| ≤ mn by part (a), so |G : L| = |G : H ∩K| = mn.

Problem 2.6.33.

Solution: Let H1, H2, . . . ,HN be subgroups of G, each of finite index. We will prove that proposition that⋂Nj=1Hj is of

finite index by mathematical induction on N . If N = 1, then H1 is of finite index by hypothesis. If the inductive hypothesisholds for some N ∈ N, then

⋂N+1j=1 Hj = HN+1 ∩H ′ where H ′ =

⋂Nj=1Hj . The inductive hypothesis implies that H ′ is

of finite index. By exercise 2.6.32(a), HN+1 ∩H ′ is of finite index. Thus the proposition holds for all n ∈ N.

Problem 2.6.35.

Solution: Part (a) Let a, b, c ∈ G. Obviously a = 1a1 and 1 ∈ H and 1 ∈ K, so a ≡ a. If a ≡ b, there is some h ∈ Hand k ∈ K such that a = hbk, from which it follows that b = h−1bk−1; thus b ≡ a. Finally, if a ≡ b and b ≡ c, there areh0, h1 ∈ H and k0k1 ∈ K such that a = h0bk0 and b = h1ck1, so a = (h0h1)c(k0k1). As a result, a ≡ c, and ≡ is anequivalence on G.

Part (b) If a ∈ G, then [a] = {b ∈ G : a ≡ b}. Define HaK = {b ∈ G : b = hak for some h ∈ H and k ∈ K}.Note that HaK is a double coset consisting of the union of the left cosets of K in G generated by the elements of theright cosets of H in G generated by a. If b ∈ [a], then b ≡ a, so b = h′ak′ for some h′ ∈ H and k′ ∈ K, so b ∈ HaK.Conversely, if c ∈ HaK, then c = h′′ak′′ for some h′′ ∈ H and k′′ ∈ K, so a ≡ c. Thus c ∈ [a]. We conclude that[a] = HaK.

Section 2.7

Problem 2.7.1.

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Solution: The group of motions are {ε, (1 3), (2 4), (1 2)(3 4), (1 4)(2 3), (1 3)(2 4), (1 2 3 4), (1 4 3 2) }. Since allthe cycles are disjoint, note that products of cycles are the same irrespective of the order of terms (e.g., (1 3)(2 4) = (24)(1 3)).

Problem 2.7.2.

Solution:Label the vertices as follows on a cube of equal length sides:

1

23

4

5

67

8

A reflection across the vertical plane parallel to the left and right faces of the cube is a symmetry (preserves all distancesbetween the vertices) but cannot be achieved through any motion because 1 and 4 cannot be transposed. This reflectionσ is given by σ = (1 4)(2 3)(5 8)(6 7), with the end result:

4

32

1

8

76

5

Problem 2.7.3.

Solution: Part (a) The group of motions G = {ε, (1 2 3)( 1 3 2) }], since vertex 4 must remain fixed in any motion.Part (b) There are symmetries not contained in G by reflections through each of vertices 1, 2, and 3. The symmetries

are H = {ε, (1 2), (1,3), (2 3), (1 2 3), (1 3 2) } = S3.

Problem 2.7.4.

Solution: Part (a) {ε, (1 3)(2 4), (1 2 3 4), (1 4 3 2) }.Part (b) {ε, (1 3), (2 4), (1 3)(2 4), (1 2 3 4), (1 4 3 2) }.

Problem 2.7.5.

Solution: {ε, (1 4)(2 3), (1 2)(3 4), (1 3)(2 4) } = K4.

Section 2.8

Problem 2.8.1.

Solution: We will use this result.

Proposition 3. If aj , bak ∈ Dn, then ajbak = ban+k−j .

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Proof. We have:aj · bak = aj(bak)an = (ajbaj)an+k−j = ban+k−j ,

where we used the result from exercise 2.7.25(a) that ambam = b for any m ∈ Z.

Part (a) Showing H is a subgroup of D12 by the Subgroup Test is just a brute force effort. Obviously 1 ∈ H. Wehave (a6)2 = 1, b · a6 = ba6, a6 · b = a6ba12 = (a6ba6)a6 = ba6, and (ba6)2 = 1, so H is closed. (Note we used exercise2.7.25’s result quite a bit.) Finally (a6)−1 = a6, b−1 = b, and (ba6)−1 = ba6, all of which belong to H.

The subgroup H is not normal to D12. We have aH = {a, a7, ba11, ba5}, but Ha = {a, a7, ba, ba7}.Part (b) It is easily shown H is a subgroup of G by the Subgroup Test by brute force. Just use the result in exercise

2.7.25 as needed.But H is not normal to D12. We have aH = {a, a5, a9, ba11, ba3, ba7} and Ha = {a, a5, a9, ba, ba5, ba9}.Part (c) We’ll skip showing that H is a subgroup of D12.If u ∈ D12, either u = aj or u = baj for some 0 ≤ j ≤ 11. Given v ∈ H, either v = a2m or v = ba2m for some

0 ≤ m ≤ 5. If uHu−1 ⊆ H for all u ∈ D12, then H CD12 by Theorem 2.8.3. We have two cases:

1. Suppose u = aj . If v = a2m then uvu−1 = aja2ma−j = a2m ∈ H. If v = ba2m, then uvu−1 = ajba2ma−j =ba12+2m−2j = ba12a2(m−j) = ba2(m−j) ∈ H. For the last part, we will show that a2(m−j) = a2r where 0 ≤ r ≤ 5.By the division algorithm, m − j = 6q + r where q ∈ Z and 0 ≤ r ≤ 5. As a result, 2(m − j) = 12q + 2r, so2(m− j) ≡ 2r (mod 12), hence a2(m−j) = a2r.

2. Suppose u = baj . If v = a2m, then:

uvu−1 = baja2m(baj)−1 = baja2ma−jb−1 = ba2mb−1 ∈ H,

by closure in H. If v = ba2m, then:

uvu−1 = bajba2ma−jb−1 = bajbaja12+2m−2j = bba12+2m−2j = a2(m−j) ∈ H.

Accordingly, H is normal.

Problem 2.8.2.

Solution: The normal subgroups of D4 are {1}, {1, a2}, {1, a2, b, ba2}, {1, a3, b, ba3}, 〈a〉, and D4. The subgroups {1}and D4 are trivially normal. By Lagrange’s Theorem, any other subgroup of D4 must be of order 2 or 4. We will evaluatetype each in turn.

• If H is a subgroup of D4 of order 4, it must be normal because |D4 : H| = 2 by Lagrange’s Theorem. Obviously〈a〉 = {1, a, a2, a3} is such a subgroup. There are no other subgroup of D4 of order 4. The other subgroups of order4 are {1, a2, b, ba2} and {1, a, b, ba3} are also normal. There are no other subgroups of D4 of order 4.

• If K is a subgroup of D4 of order 2, designate K = {1, k} where k ∈ D4\{1}. Exercise 2.8.7 implies that k ∈ Z(D4).The only non-unity element of Z(D4) is a2 (see exercise 2.6.26), hence K = {1, a2}.

Problem 2.8.3.

Solution: Since |A4| = 12, any non-trivial subgroup must be of order 2, 3, 4, or 6 by Lagrange’s Theorem. By exercise2.6.34, there are no subgroups of A4 of order 6. We will evaluate the other possible groups.

1. Groups of Order 4. The precise set K was shown to be normal in exercise 2.8.3. We will show there are no othersubgroups of order 4. Let H be a subgroup of A4 where |H| = 4 and H 6= K. The set H must contain a 3-cycle τ(since there are only three non-3-cycles in A4). But o(τ) = 3, so τ4 = τ 6= 1, contradicting Corollary 2.6.2. ThereforeK is the only subgroup of order 4.

2. Groups of Order 3. Suppose L is a subgroup of A4 of order 3. Let α and β be distinct non-identity elements ofL. It cannot be that both α and β are each products of disjoint transpositions because their product would be theproduct of disjoint transpositions not contained in L (see Proposition 4 below). Thus we may assume α is some3-cycle in A4. But o(α) = 3, so 〈α〉 = L with inverse (b d c). If we designate α = (a b c), then L = {ε, (a b c), (ac b) }. But (a b d) ∈ A4 with inverse (a d b), and (a b d)(a b c)(a d b) = (b d c) /∈ L, so L is not normal in A4.

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3. Groups of Order 2. Suppose J is a subgroup of A4 of order 2. Let σ be the non-identity element of J . It followsthat σ2 = 1 by Corollary 2.6.1. Therefore σ is the product of disjoint transpositions since no other element of A4 isof order 2. Given µ ∈ A4, Theorem 2.8.3 implies that µσµ−1 ∈ J . Designate σ = (a b)(c d), and let µ = (a b d).But:

(a b d)(a d)(c d)(a d b) = (b d c) /∈ J.

Therefore J 6 A4.

We conclude that K is the only normal subgroup of A4 not equal to A4 or {1}.

Proposition 4. If α, β, γ are the three distinct products of disjoint transpositions in A4, then αβ = γ.

Proof. Designate α = (a b)(c d), β = (a c)(b d), and γ = (a d)(b c) where a, b, c, d are distinct elements of {1, 2, 3, 4}.It follows that:

αβ = (a d)(b c) = γ.

Problem 2.8.4.

Solution: Since |H| = 4 and |K| = 2, Lagrange’s Theorem implies that |H : K| = 2, so K C H by Theorem 2.8.4.Similarly, since |D4| = 8, it follows that |D4 : H| = 2, so H CD4.

But K is not normal in D4. We have a ∈ D4, so aba−1 = aba−1a4 = (aba)a3 = ba3 /∈ K. By Theorem 2.8.3,K 6 D4.

Problem 2.8.6.

Solution: Suppose H is a subgroup of G and if a ∈ G then there is some b ∈ G such that aH = Hb. We will usethe following two results to show H C G. First, since a ∈ aH, it follows that a ∈ Hb, so a = h0b for some h0 ∈ H.Second, we will show that a−1H = Hb−1. Given a−1h′ ∈ H, since aH = Hb we have ah = h′b for some h′ ∈ H. Itfollows that h = a−1h′b, so hb−1 = a−1h′ ∈ Hb−1. Conversely, given h′b−1 ∈ H, we have h′b = ah for some h ∈ H, soa−1h′ = hb−1 ∈ a−1H. Accordingly, a−1H = Hb−1.

Now we may complete the proof. Let a ∈ G be given. If h ∈ H, then aha−1 = ah(h0b)−1 = a(hb−1)h−10 . Because

a−1H = Hb−1, there is some h1 ∈ H such that hb−1 = a−1h−11 . As a result, aha−1 = a(a−1h1)h−10 = h1h−10 ∈ H.

Theorem 2.8.3 implies that H CG.

Problem 2.8.7.

Solution: Suppose H CG and |H| = 2. Accordingly, H = {1, h} for some h 6= 1. If g ∈ G, Theorem 2.8.3 implies thatghg−1 ∈ H. It cannot be that ghg−1 = 1 because this would imply that h = 1. Therefore ghg−1 = h, the only otherelement of H. Accordingly, gh = hg for all g ∈ G, so h ∈ Z(G). Since 1 ∈ Z(G) as well, we conclude that H ⊆ Z(G).

Problem 2.8.8.

Solution: Suppose H be a subgroup of G and K CH. Let h ∈ H be given. If j ∈ H ∩K, then j ∈ H and j ∈ K. Byclosure of the binary operation in H, it must be that hjh−1 ∈ H. Because K is normal in H and h ∈ G, it follows thathjh−1 ∈ K. Consequently, hjh−1 ∈ H ∩K, so Theorem 2.8.3 implies that H ∩K CH.

It is not true that H ∩K CK. Let K = S3, which is trivially normal in S3, and let H = {ε, τ}. We have H ∩K = H.The subgroup H is not normal in S3: Observe that στσ−1 = (στσ)σ = τσ /∈ H. Obviously then H 6 K.

Problem 2.8.9.

Page 49

Solution: Given a group G, define D = {(g, g) : g ∈ G}. First we will show that D is a subgroup of G × G. Obviously(1, 1) ∈ D. If u = (g0, g0), v = (g1, g1) ∈ D, then uv = (g0g1, g0g1) where g0g1 ∈ G, so uv ∈ D. Finally,the inverse of uin G is (g−10 , g−1), which is clearly contained in D. By the Subgroup Test, D is a subgroup of G.

If G is abelian, then G × G is abelian (given a = (g0, g1), b = (g2, g3) ∈ G × G, we have ab = (g0g2, g1g3) =(g2g0, g3g1) = ba). By Theorem 2.8.2, D is normal in G×G.

Conversely, suppose D is normal in G × G. Let g0, g1 ∈ G be given. The inverse of (1, g0) is (1, g−10 ). We have(1, g0) · (g1, g1) · (1, g−10 ) = (g1, g0, g1g

−10 ) ∈ D. It follows that g0g1g

−10 = g1, so g0g1 = g1g0. Since this is true for any

g0, g1 ∈ G, we conclude that G is abelian.

Problem 2.8.10.

Solution: Let N CG and K CG. Let g ∈ G be given. If a ∈ N ∩K, then a ∈ N and a ∈ K. As a result, gag−1 ∈ Nbecause N is normal in G and gag−1 ∈ K because K is normal in G; therefore gag−1 ∈ N ∩ K. By Theorem 2.8.3,N ∩K CG.

Problem 2.8.11.

Solution: Suppose G is a group of order pq where p and q are distinct primes and G has a unique subgroup H of order pand a unique subgroup K of order q. Since H and K are the only subgroups of G of their order, by Corollary 2.8.2 H CGand K CG. By Corollary 2.6.3, H and K are both cyclic, and exercise 2.4.25 implies that H ×K is cyclic.

We will show that H ∩K = {1}. If u ∈ H ∩K, then 〈u〉 is a subgroup of both H and K. By Lagrange’s Theorem,|〈u〉| divides both p and q. Since p and q are distinct primes, |〈u〉| = 1, so u = 1.

With these facts established, Corollary 2.8.2 implies that G ∼= H ×K. Since H ×K is cyclic, H ×K = 〈(h0, k0)〉 forsome h0 ∈ H and k0 ∈ K. Let σ : G→ H ×K be an isomorphism. Since σ is surjective, there is some g0 ∈ G such thatσ(g0) = (h0, k0) By Theorem 2.4.4., o(g0) = o(h0, k0) = pq = |G|. Since G is finite, it follows that 〈g0〉 = G, so G iscyclic.

Problem 2.8.12.

Solution: Suppose K CG where K is cyclic, which we may denote K = 〈k〉 for some k ∈ K. Let H be a subgroup of K.By Theorem 2.4.7, H is cyclic, so we may denote it as H = 〈h〉 for some h ∈ H. Since h ∈ K, it follows that h = kj forsome j ∈ Z, so H = 〈aj〉.

We know there is an inner automorphism of K, so aut K is not empty. Let α : K → K be an automorphism of G.Designate a = α(k), hence o(a) = o(k) by Theorem 2.5.4. If h ∈ H, then h = (kj)l = kjl for some l ∈ Z. Further, sincea ∈ K we have a = ks for some s ∈ Z. We then have:

α(h) = α(kjl) = (α(k))jl = ajl = (ks)jl = (kj)sl.

Because sl ∈ Z, we infer that (kj)sl ∈ H. Therefore α(H) ⊆ H, so H is a characteristic subgroup in K. Corollary 2.8.3implies that H CG.

Problem 2.8.14.

Solution: Let G = H ×K where G is finite and H and K are groups with |H| = m and |K| = n. Clearly |G| = mn.Define H1 = {(h, 1) : h ∈ H} and K1 = {(1, k) : k ∈ K}. It is trivial to show that H1 and K1 are subgroups of G, so wewill omit the proof here. Obviously |H1| = m and |K1| = n, so |H1| · |K1| = |G|.

We will show that H1∼= H and K1

∼= K. Define σh : H1 → H by σh(h, 1) = h for h ∈ H. Let (h0, 1), (h1, 1) ∈ H1

be given. We have:σh((h0, 1) · (h1, 1)) = σh(h0h1, 1) = h0h1 = σh(h0, 1)σh(h1, 1),

so σh is a homomorphism. The map is obviously surjective. If σh(h0, 1) = σh(h1, 1), then h0 = h1, so (h0, 1) = (h1, 1),and σh is injective. It follows that σh is an isomorphism, so H1

∼= H. A similar argument shows K1∼= K.

If u = H1 ∩ K1, then u = (h2, 1) = (1, k2) for some h2 ∈ H and k2 ∈ K. We infer that h2 = k2 = 1, sou = (1, 1) = 1G. Therefore H1 ∩K1 = {1G}.

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Finally, we will show that H1 and K1 are normal in G. Let g = (a, b) ∈ G where a ∈ H and b ∈ G. We haveg−1 = (a−1, b−1). If (h, 1) ∈ H1, then:

g · (h, 1) · g−1 = (a, b)(h, 1)(a−1, b−1) = (aha−1, bb−1) = (aha−1, 1).

Since aha−1 ∈ H by closure of H, it must be that (aha−1, 1) ∈ H. Theorem 2.8.3 implies that H1 C G. A similarargument shows that K1 CG.

Problem 2.8.15.

Solution: Suppose K is a subgroup of group G of index 2. Therefore K CG.Part (a) Suppose a ∈ G\K and b ∈ G\K. First, note that b−1 /∈ K because b /∈ K. Second, it must be that

a ∈ Kb−1: If it were not, then a ∈ K (since |G : K| = 2, it follows that G = K ∪Kb−1), contradicting that a /∈ K. As aresult, Ka = Kb−1, and since K is normal in G, we have aK = Ka = Kb−1. There is some k ∈ K such that ak = b−1,so ab = k ∈ K.

Part (b) Remember that H ∩ K is a subgroup of both H and K, so |H ∩K| divides both |H| and |K|. Chooseh0 ∈ H\K, which must exist since H is not a subset of K. Then define α : H ∩K → H\(H ∩K) where α(h) = hh0for h ∈ H ∩ K. Given h, h′ ∈ H ∩ K, if α(h) = α(h′), then hh0 = h′h0, so h = h′ by right cancellation; therefore αis injective. Now we will show that α is surjective. Let a ∈ H ∩ (H ∩K), from which it follows that a /∈ K (otherwisea ∈ H ∩K, a contradiction). Since the right cosets of H ∩K partition H, there is some h−10 ∈ a−1(H ∩K)

COME BACK TOSince α is a bijection, |H ∩K| = |H\(H ∩K)|. Obviously H = (H ∩K) ∪ (H\(H ∩K)). Because the two sets are

disjoint, |H| = |H ∩K|+ |H\(H ∩K)| = 2 |H ∩K|. As a result, |H : H ∩K| = |H| / |H ∩K| = 2.

Problem 2.8.16.

Solution: Let G be a group. Give µ ∈ autG, if σa ∈ innG, then for g ∈ G:

µσaµ−1(g) = µσa(µ−1(g)) = µ(aµ−1(g)a−1) = µ(a)µ(µ−1(g))[µ(a)]−1 = bgb−1,

where b = µ(a). As a result, µσaµ−1 = σb ∈ innG. By Theorem 2.8.3, innGC autG.

Problem 2.8.17.

Solution: Let Dn be given.Part (a) Since |〈a〉| = n, it follows that |Dn : 〈a〉| = |Dn| / |〈a〉| = 2n/n = 2, so 〈a〉 is normal in Dn. Suppose H is a

proper subgroup of 〈a〉. Since H is cyclic by Theorem 2.4.7, H = 〈aj〉 for some j ∈ Z. If c ∈ Dn, then either c = ak orc = bak for some k ∈ Z. Let h ∈ H where h = (aj)l = ajl for some l ∈ Z. If c = ak, then akha−k = akajla−k = ajl ∈ H.If c = bak, then:

(bak)ajl(bak)−1 = bakajla−kb−1 = bajlb = bajlbajlan−jl = b2an−jl = (aj)−l ∈ H.

Theorem 2.8.3 implies that H CDn.Part (b) Let n be odd where n = 2m− 1 for some m ∈ N and K CDn. Assume there is some bar ∈ K where r ∈ Z

(in which case K is not a subset of 〈a〉). Since K is normal in Dn, we have:

a−1(bar)(a−1)−1 = a−1(bar)a = (a−1ba−1)ar+2 = bar+2 ∈ K.

Since K is closed, (bar)(bar+2) = (bar)(bar)a2 = a2 ∈ K. Because gcd(2, 2m − 1) = 1, Theorem 2.4.8 implies that〈a2〉 = 〈a〉, and by Theorem 2.4.10 〈a〉 ⊆ H. But |〈a〉| = n, so |H| > n = |Dn| /2. Since Lagrange’s Theorem requiresthat |H| divide |Dn| and Dn is finite, H = Dn. Therefore H is either a subset of 〈a〉 or equals Dn.

Problem 2.8.19.

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Solution: Let H and K be subgroups of G. We will prove the cycle of implications in turn.

1. Hk is a subgroup of G. Suppose HK is a subgroup of G. Theorem 2.8.2 implies that HK = KH, so HK ⊆ KH.

2. HK is a subset of KH. Suppose HK ⊆ KH. Let kh ∈ KH. It follows that u = h−1k−1 ∈ HK, so u ∈ KH, henceu = k′h′ where h′ ∈ H and k′ ∈ K. Therefore u−1 = kh = (h′)−1(k′)−1, which is contained in HK. AccordinglyKH is a subset of HK (from which it follows that KH = HK).

3. KH is a subset of HK. Suppose KH ⊆ HK. By the result in (2), HK ⊆ KH, so HK = KH. By Lemma 2.8.2,HK is a subgroup of G.

Problem 2.8.20.

Solution: Let G = HK where H and K are subgroups of G and hk = kh if h ∈ H and k ∈ K. If g ∈ G, then g = h0k0where h0 ∈ H and k0 ∈ K. Given h ∈ H, we have:

ghg−1 = (h0k0)h(h0k0)−1 = (h0k0)h(k−10 h−10 ) = h0(k0k−10 )(hh−10 ) = h0hh

−10 ∈ H.

Thus gHg ⊆ H, so Theorem 2.8.1 implies that H CG. A similar argument shows that K CG.

Problem 2.8.21.

Solution: Note the mistake in the hint: Nicholson intends to refer to Theorem 2.4.10 rather than 2.4.8. I spent a whilescratching my head, wondering what was the relevance of Theorem 2.4.8 to this problem!

Suppose H ⊆ G and K ⊆ G are subgroups of group G with HK = KH. By Lemma 2.8.2, HK is a subgroup of G.If a ∈ HK, then a = h0k0 where h0 ∈ H and k0 ∈ K. Obviously a ∈ 〈H ∪ K〉 by definition. Next we will show thatH ∪K is contained in HK. If h ∈ H, then h · 1 = h ∈ HK. If k ∈ K, then 1 · k = k ∈ HK. Therefore H ∪K ⊆ HK,so Theorem 2.4.10 implies that 〈H ∪K〉 ⊆ HK. We conclude that HK = 〈H ∪K〉.

Problem 2.8.22.

Solution: Since G is finite and |H| and |K| are relatively prime, Corollary 2.6.2 implies that H ∩K = {1}. By Corollary2.8.1, |HK| = |H| · |K|. Since HK is a subset of G and G is finite, HK = G. Applying exercise 2.8.20, H C G andK CG. By Theorem 2.8.6, HK ∼= H ×K.

Problem 2.8.23.

Solution: Part (a) Let n = 2m where m is an odd positive integer and C2 be a cyclic group of order 2. DesignateH = {1, am}, which is a subgroup of Dn. We will show H is normal in Dn. If d ∈ Dn, then d = ar or d = bar forsome r ∈ Z. If d = ar, then damd−1 = arama−r = ar ∈ H. If d = bar, then damd−1 = (bar)am(a−rb) = bamb =b(ambam)am = b2am = am ∈ H. Thus H CDn. Obviously H ∼= C2 and |H| = 2.

Next, designate K = {1, a2, . . . , an−2, b, ba2, . . . , ban−2}. It is easily shown that K is a subgroup of Dn of order 2m.Further, am /∈ K because m is odd and K contains only even powers of a; therefore H ∩ K = {1}. Finally, K C Dn

because |Dn : K| = |Dn| / |K| = 4m/(2m) = 2. The subgroup K is isomorphic with Dm. In K, we have o(a2) = m ando(b) = 2. Further a2ba2 = a(aba)a = aba = b.

To sum up, we have normal subgroups H and K of Dn where |H| · |K| = 2(2m) = n = |Dn| and H ∩ K = {1}.Corollary 2.8.2 implies that Dn

∼= H × K. Since H ∼= C2 and K ∼= Dm, by example 2.5.13 H × K ∼= C2 × Dm. Weconclude that Dn

∼= C2 ×Dm (see Corollary 2.5.1 for the transivity of isomorphisms).Part (b) In order for D12

∼= C3×D4, there must be a subgroup K of D12 that is isomorphic with D4. Therefore theremust be an element k ∈ K such that o(k) = 4, since o(a) = 4 in D4. The only such elements in D12 are a3 and a9, hencea3 ∈ K or a9 ∈ K. But then K = {1, a3, a6, a9}, so there is no element of K of order 2 (i.e., b ∈ D4). Consequently,there is some subgroup of D12 isomorphic with D4.

Problem 2.8.24.

Page 52

Solution: Part (a) Let H be characteristic in group G. Thus given g ∈ G, the inner automorphism σg : G → G is suchthat σg(H) ⊆ H (i.e., gHg−1 ⊆ H). By Theorem 2.8.3, H CG.

Part (b) Suppose H is characteristic in group G. If σ ∈ aut G, then σ(H) ⊆ H. Since σ−1 ∈ aut G, we also haveσ−1(H) ⊆ H, hence given h0 ∈ H there is some h1 ∈ H such that σ−1(h0) = h1. Accordingly, σ(σ−1(h0)) = h0 = σ(h1),so H ⊆ σ(H). We conclude that σH = H.

Part (c) Let G = C2 × C2 and H = C2 × {1} where C2 = {1, a}. It’s easy to show that H is a subgroup of G; we’llomit the proof. Since |G| = 4 and |H| = 2, it follows that |G : H| = 2, so H is normal in G. But H is not characteristicin G. Define α : G → G where α(x, y) = (y, x) for x, y ∈ G. We’ll omit the proof that α is an automorphism. Becauseα(H) = {(1, 1), (1, c)} 6= H, we infer that H is not characteristic in G.

Part (d) Let α ∈ autG. Given z ∈ Z(G) and g ∈ G, there is some g′ ∈ G such that α(g′) = g because α is surjective.We have:

α(zg′) = α(z)α(g′) = α(z)g = α(g′z) = α(g′)α(z) = gα(z).

Since this is true for every g ∈ G, we conclude that α(z) ∈ Z(G). Therefore α(Z(G)) ⊆ Z(G), so Z(G) is characteristicin G.

Part (e) Suppose H ⊆ KCG where H is characteristic in K. Since KCG, if αa : G→ G is in the inner automorphismof G determined by a ∈ G, then αa(K) = K since aKa−1 = K by Theorem 2.8.3. As a result, the restriction αa | Kis a surjective map K → K. Since it inherits the properties of injectivity and homomorphism from αa, it follows thatαa | K is an automorphism of K. Because H is characteristic in K, we have (αa | K)(H) ⊆ H for a ∈ G. It follows thatα(H) ⊆ H, so H CG.

Part (f) Let H and K be subgroups of group G where H is characteristic in K and K is characteristic in G. Letα ∈ aut G. By hypothesis, α(K) = K. The restriction α | K is an automorphism K → K (it is surjective on domain K,and it inherits injectivity and the homomorphic property from α), so (α | K)(H) = H. Consequently, α(H) = H for anyα ∈ autG. We conclude that H is characteristic in G.

Part (g) Let G = 〈g〉 for some g ∈ G where o(g) = n and H be a subgroup of G. By Theorem 2.4.9, H = 〈gd〉 forsome d | n. Let α ∈ aut G. We may express h ∈ H as h = (gd)j for some j ∈ Z and α(g) = gk for some k ∈ Z. As aresult:

α(h) = [α(g)]dj = (gk)dj = (gd)jk ∈ H.

Accordingly, α(H) ⊆ H, so H is characteristic in G.The proposition does not hold if G is a general abelian group that is not cyclic. Part (c) serves as a counterexamplePart (h) Let H and K be characteristic in group G and α ∈ aut G. If u ∈ H ∩K, then α(u) ∈ H and α(u) ∈ K by

hypothesis. As a result, α(u) ∈ H ∩K, so α(H ∩K) ⊆ H ∩K. We conclude that H ∩K is characteristic in G.Part (i) Suppose H is a subgroup of G. Define K = {g ∈ G : g ∈ σ(H) for all σ ∈ aut G}. Let σ ∈ aut G. We infer

that K ⊆ σH. First we will show that K is a subgroup of G. Since σ(1) = 1, it follows that 1 ∈ K (and K is non-empty).Given k0, k1 ∈ K, there are h0, h1 ∈ H such that σ(h0) = k0 and σ(h1) = k1. Therefore k0k1 = σ(h0)σ(h1) = σ(h0, h1),so h0h1 ∈ K. Finally, σ(h−10 ) = k−10 , so every element of K has an inverse. By the Subgroup Test, K is a subgroup of G.

Next we will show that K ⊂ H. Since 1G ∈ autG, if g ∈ G\H, then g /∈ 1G(H), so g /∈ K. As a result, K ⊆ H.Now we will show that K is characteristic in G. If σ ∈ autG, then given k1 ∈ K ⊆ H there is some k2 ∈ K such that

σ(k1) = k2 by the definition of K. Thus σ(K) ⊆ K for all σ ∈ autG, so K is characteristic in H.Finally, we will show that every characteristic subgroup of H is contained in K. Let J be characteristic in G and

contained in H. Consequently, µ(J) = J for all µ ∈ autG, so J ⊆ K by definition of K.

Problem 2.8.28.

Solution: Part (a) Proved in exercise 2.3.9.Part (b) Let c ∈ C(K) for some K CG. If h ∈ H, then:

gcg−1 · h = (gc)(g−1hg)g−1 = g(g−1hg)(cg−1) = h · gcg−1,

since g−1hg ∈ H. Therefore gcg−1 ∈ C(K), so g[C(K)]g−1 ⊆ C(K). By Theorem 2.8.3, C(K) CG.

Section 2.9

Problem 2.9.1.

Page 53

Solution: Part (a) First we will find Z(D6). If k ∈ Z, then a(bak) = (aba)ak−1 = bak−1 6= bak+1, so bak /∈ Z(D6).Further, a simple calculation shows that akb 6= bak for any 1 ≤ k ≤ 5 other than k = 3, so such ak /∈ Z(D6). On theother hand, a3b = (a3ba3)a3 = ba3, so a3 ∈ Z(D6). We conclude that Z(D6) = {1, a3}.

The right cosets of K are:K1 = K = {1, a3} = Ka3,Ka = {a, a4} = Ka4,Ka2 = {a2, a5} = Ka5,Kb = {b, ba3} = Kba3,Kba = {ba, ba4} = Kba4,Kba2 = {ba2, ba5} = Kba5.

Accordingly D6/K = {K,Ka,Ka2,Kb,Kba,Kba2}. By Theorem 2.8.1, K = Z(D6) is normal in D6, so D6/K formsa group by Theorem 2.9.1. We can fill out the Cayley table with the rule Ka ·Kb = Kab = Kba4 = Kba2. As a result:

K Ka Ka2 Kb Kba Kba2

K K Ka Ka2 Kb Kba Kba2

Ka Ka Ka2 K Kba2 Kb KbaKa2 Ka2 K Ka Kba Kba2 KbKb Kb Kba Kba2 K Ka Ka2

Kba Kba Kba2 Kb Ka2 K KaKba2 Kba2 Kb Kba Ka Ka2 K

The group D6/K is isomorphic with D3. Observe that (Ka)3 = Ka3 = K, so o(Ka) = 3, and o(Kb) = 2, andKa ·Kb ·Ka = Kaba = Kb.

Part (b) The center of the quarternian group Z(Q) = {1,−1} = K, for if α, β ∈ Q\{±1} are distinct, thenαβ = −βα, hence α, β /∈ Z(Q). The factor group Q/K = {{1,−1}, {i,−i}, {j,−j}, {k,−k}. Observe that K1 = K(−1),Ki = K(−i), Kj = K(−j), and Kk = K(−k). The Cayley table is:

K Ki Kj KkK K Ki Kj KkKi Ki K Kk KjKj Kj Kk K KiKk Kk Kj Ki K

The factor group Q/K is isomorphic with the Klein group K4 because o(Ki) = o(Kj) = o(Kk) = 2 and the productof any two of i, j, k is the third.

Part (c) It is easy to show that K is a subgroup of G and K CG; the proof is omitted here.We will show that given α ∈ A and β ∈ B, the right coset K(α, β) = K(1, β). Since α ∈ A, it follows that

(α, β) ∈ K(1, β); Theorem 2.6.4(1) implies that K(α, β) = K(1, β). Thus |G/K| = |B|. It follows that the Cayley tablefor G/K is the same as the Cayley table for B since K(1, b) ·K(1, b′) = K(1, bb′) for any b, b′ ∈ B.

The factor group G/K is isomorphic with B. Define σ : G/K → B by σ(K(1, β)) = β. This map is obviouslysurjective. If σ(K(1, β)) = σ(K(1, β′)) for some β, β′ ∈ B, then β = β′, from which it follows that K(1, β) = K(1, β′).Thus the map is injective. Finally:

σ(K(1, β) ·K(1, β′)) = σ(K(1, ββ′)) = ββ′ = σ(K(1, β)) · σ(K(1, β′)),

so σ is a homomorphism. We conclude that G/K ∼= B.Part (d)

Problem 2.9.2.

Solution: Let n be an exponent for group G and G/K be the factor group of G by subgroup K. If Kg ∈ G/K, then(Kg)n = Kgn = K1 = K, which is the unity of K/G. As a result, n is an exponent for the factor group G/K.

Problem 2.9.3.

Page 54

Solution: Part (a) Theorem 2.9.1(4) implies that since G = 〈a〉 is cyclic, G/K = 〈Ka〉. By Lagrange’s Theorem,|G/K| = |G : K| = |G| / |K| = 24/2 = 12, hence o(Ka) = 12. Since Ka2 = (Ka)2, Theorem 2.4.5 implies thato(Ka2) = 6. Similarly, o(Ka3) = 4 and o(Ka4) = 3. Since 5 and 12 are relatively prime, Ka5 = (Ka)5 generates G/K,so o(Ka5) = o(Ka) = 12.

Part (b) We have |G/H| = |G| / |H| = 24/4 = 6. As in part (a), G/H = 〈Ha〉, so o(Ha) = 6. It follows thato(Ha2) = 3 and o(Ha3) = 2. Since gcd(5, 6) = 1, we have o(Ha5) = o(Ha) = 6. Since Ha4 = (Ha2)2, it followsthat 〈Ha4〉 is a subgroup of 〈Ha2〉. But gcd(2, o(Ha2)) = gcd(2, 3) = 1, so by Theorem 2.4.8 Ha4 generates 〈Ha2〉.Accordingly, o(Ha4) = o(Ha2) = 3.

Problem 2.9.5.

Solution: Part (a) Observe that K = {1, a2, a4, a6, a8, a10}. By exercise 2.8.17(a), K CD12 because K is a subgroup of〈a〉.

• Ka2 = K because a2 ∈ K, so o(Ka2) = 1.

• Ka3 6= K and (Ka3)2 = Ka6 = K since a6 ∈ K. It follows that o(Ka3) = 2.

• Ka5 6= K and (Ka5)2 = Ka10 = K since a10 ∈ K. It follows that o(Ka5) = 2.

• Kba 6= K and (Kba)2 = Kb2 = K, so o(Kba) = 2.

Part (b) Observe K = {1, a3, a6, a9, b, ba3, ba6, ba9}. But K is not normal in D12 because aba−1 = (aba)a10 = ba10 /∈K. As a result, the operation KaKb = K(ab) is not well defined in D12/K. Accordingly, D12/K is not a group.

Problem 2.9.6.

Solution: The center of Q is Z(Q) = {1,−1} because if distinct α, β ∈ Q\{1,−1} then αβ = −αβ. Theorem 2.9.1(5)implies that |Q : Z(Q)| = |(|Q)/ |Z(Q)| = 8/2 = 4. Since every group of order 4 is either cyclic or isomorphic with theKlein group, it suffices to show that Q/Z(Q) is not cyclic. Assume Q/Z(Q) is cyclic and generated by Ka where a ∈ Q.As a result, (Ka)2 = Ka2 = K, so o(Ka) ≤ 2, contradicting that Ka generates Q/Z(Q).

Problem 2.9.7.

Solution: Remember the that operation in Q is addition. As an initial matter, Z C Q: Given q ∈ Q, if n ∈ Z thenq + n+ (−q) = n ∈ Z. Therefore Q/Z is a proper group.

First we will show that Q/Z is infinite. Let q0, q1 ∈ Q where 0 < q0 ≤ q1 < 1. If Zq0 = Zq1, then q1 + (−q0) =q1 − q0 ∈ Z by Theorem 2.6.1(3). Because q1 − q0 ∈ [0, 1), it must be that q1 = q0. Therefore each Zq0 is distinct. Sincethere are an infinite number of q0, it follows that Q/Z contains {Zq : q ∈ Q and 0 < q < 1}; consequently, Q/Z is infinite.

The fact that Q/Z is abelian follows from Zq0 + Zq1 = Z(q0 + q1) = Z(q1 + q0) = Zq1 + Zq0.Finally, we will show that every element of Q/Z has finite order. If q ∈ Q, then q = a/b where a, b ∈ Z and b 6= 0. As a

result, qb = ab/b = a ∈ Z, so b(Zq) = Z(bq) = Za = {n+ a : n ∈ Z} = Z. As a result, o(Zq) ≤ b and therefore is finite.

Problem 2.9.8.

Solution: We will rely on this proposition.

Proposition 5. If K is a subgroup of H and H is a subgroup of G and K CG, then K CH.

Proof. If h ∈ H, then hkh−1 ∈ K because K CG and h ∈ G. Theorem 2.8.1 implies that K CH.

Suppose K ⊆ H ⊆ G are finite groups and K C G. By Theorem 2.9.1 G/K is a group where |G/K| = |G| / |K|.Proposition 5 implies that H CK, so H/K is a group where |H/K| = |H| / |K|. Given Kh ∈ H/K, since h ∈ G we haveKh ∈ G/K, from which it follows that H/K is a subgroup of G/K. Lagrange’s Theorem implies that |G/K : H/K| =|G/K| / |H/K| = (|G| / |K|)/(|H| |K|) = |G| / |H| = |G : H|.

Page 55

Problem 2.9.9.

Solution: Suppose K C G and o(g) = n for all g ∈ G. It follows that (Kg)n = Kgn = K. By Theorem 2.4.2(1),o(Kg) | n.

Problem 2.9.10.

Solution: Suppose K C G and |G : K| = m. Corollary 2.6.2 implies that if g ∈ G then (Kg)m = Kgm = K. Sincegm ∈ Kg, it follows that gm ∈ K.

Problem 2.9.11.

Solution: Suppose KCG has index m, in which case |K/G| = |K : G| = m. Let g ∈ G where o(g) = n and gcd(m,n) = 1.By Corollary 2.6.2, (Kg)m = Kgm = K. Further, Theorem 1.2.4 implies 1 = mx+ ny for some x, y ∈ Z. Therefore:

Kg = Kgmx+ny = KgmxKgny = (Kgm)x(Kgn)y = K1xK1y = K.

Since g ∈ Kg, it must be that g ∈ K.

Problem 2.9.12.

Solution: Let K be a finite group where K C G. Suppose Kg ∈ G/K is of order n. Since (Kg)n = K, there is somek ∈ K such that kn = 1. By the definition of order, (Kg)j 6= K for 1 ≤ j < n; therefore 1 /∈ (Kg)j , so kj 6= 1. It followsthat o(k) = n.

Problem 2.9.13.

Solution: Let K CG, so G/K is a group.Part (a) Suppose K and G/K have the trivial center, so Z(K) = {1} and Z(G/K) = {K}. Let g0 ∈ Z(G), in

which case given any g1 ∈ G we have Kg0Kg1 = Kg0g1 = Kg1g0 = Kg1Kg0. It follows that Kg0g1 ∈ Z(G/K), soKg0g1 = Kg1g0 = K. Thus g0g1 ∈ K and g1g0 ∈ K. Since this is true for any g1, it must be that g0 ∈ Z(K), so g0 = 1.We conclude that Z(G) = {1}.

Part (b) Suppose every element of K and K/G has a finite order. If g ∈ G, then g ∈ Kg, and (Kg)n = Kgn = Kwhere n = o(Kg). By Theorem 2.6.1(2), gn ∈ K. By hypothesis, o(gn) = m, so (gn)m = gnm = 1. Therefore everyelement of G has a finite order.

Part (c) Suppose every element of K and G/K has order of a power of a fixed prime p. Note that this meansthe same prime is associated with the order of each element. (I misread the problem, thinking the prime was fixed onlyfor each element, and this caused me much wasted effort and consternation!) If g ∈ G, then o(Kg) = pj for some

j ∈ N. Thus (Kg)pj

= Kgpj

= K, so gpj ∈ K by Theorem 2.6.1(2). By hypothesis, o(gp

j

) = pk for some k ∈ N, so

(gpj

)pk

= gpj+k

= 1. As a result, o(g) | pj+k, so o(g) is some power of p (and the power l must be j ≤ l ≤ j + k).Part (d) Suppose K and G/K are finite generated. Denote K = 〈X〉 such that X = {xj}Mj=1 where M ≥ 1 and

xj ∈ K. Denote G/K = 〈Y 〉 such that Y = {Kgj}Nk=1 where N ≥ 1 and Kgj ∈ G/K. Given g ∈ G:

Kg =

R∏j=1

Kgαjaj = K

R∏j=1

gαjaj

,where Kgaj ∈ Y , R ≥ 1, and αj ∈ Z. Denote D =

∏Rj=1 g

αjaj . Since Kg = KD, by Theorem 2.6.13, Dg−1 ∈ K there is

some k ∈ K such that k = Dg−1, hence g = Dk−1. By hypothesis, k−1 =∏Sj=1 x

βk

bkwhere S ≥ 1, xbk ∈ X, and βk ∈ Z.

Therefore:

g =

R∏j=1

gαjaj

S∏j=1

xβk

bk

.Thus g is a product of powers of elements in the set W = {gj}Mj=1 ∪ {xk}Nk=1, from which it follows that G is finitelygenerated.

Page 56

Problem 2.9.14.

Solution: Let K C G with prime index p, so |G/K| = p. Corollary 2.6.3 implies that G/K is cyclic, and if Ka ∈ G/Kand Ka 6= K, then G/K = 〈Ka〉. There must be such a Ka because |G/K| ≥ 2. Theorem 2.4.2(5) implies that〈Ka〉 = {K,Ka,Ka2, . . . ,Kap−1} with each element distinct. Since the right cosets of K form a partition of G, it follows

that G =⋃p−1j=0 Ka

j is a disjoint union, as we sought to show.

Problem 2.9.15.

Solution: Let X be a non-empty subset of G where G = 〈X〉 and K CG. Designate Y = {Kx : x ∈ X}. If Kg ∈ G/K,

then g =∏Mj=1 x

αj

j where xj ∈ X, M ≥ 1, and αj ∈ Z. Since G/K forms a group:

Kg = K

M∏j=1

xαj

j

=

M∏j=1

K(xαj

j ) =

M∏j=1

(Kxj)αj ∈ 〈Y 〉,

so G/K ⊆ 〈Y 〉. Conversely, if w ∈ 〈Y 〉, then w =∏Nj=1(Kxj)

βj , so:

w =

N∏j=1

(Kxj)βj =

N∏j=1

K(xβj

j ) = K

N∏j=1

xβj

j

= Kg′ ∈ G/K

where g′ =∏Nj=1 x

βj

j , so g′ ∈ G. We conclude that G/K = 〈Y 〉.

Problem 2.9.16.

Solution: Le G be a group where H is a subset of G that is closed under the group operation and g2 ∈ H for all g ∈ G.First we will show that H is a subgroup of G. The unity belongs to H because 12 = 1 ∈ H. By hypothesis, if a, b ∈ Hthen ab ∈ H. Finally, since a ∈ H, then (a−1)2 = a−2 ∈ H, so a−1 = aa−2 ∈ H by closure. By the Subgroup Test, H isa subgroup of G.

Now we will show H C G. Let g ∈ G. Given h ∈ H, by hypothesis g2 ∈ H, so g2h−1 ∈ H. Similarly, (hg−1)2 ∈ H.As a result:

(g2h−1)(hg−1)2 = (g2h−1)(hg−1)(hg−1) = ghg−1 ∈ H,

from which it follows from Theorem 2.8.1 that H is normal in G.Finally, we will show G/H is abelian. Let g0, g1 ∈ G. We have g−20 ∈ H, so g−20 (g0g1) = g−10 g1 ∈ Hg0g1, and we infer

that Hg0g1 = Hg−10 g1. Similarly, g−21 ∈ H, so g−21 (g1g0) = g−11 g0 ∈ Hg1g0, and we infer that Hg1g0 = Hg−11 g0. Since(g−10 g1)(g−11 g0)−1 = (g−10 g1)(g−10 g1) = (g−10 g1)2 ∈ H, Theorem 2.6.1(3) implies that Hg−10 g1 = Hg−11 g0. By transitivity,Hg0g1 = Hg0Hg1 = Hg1g0 = Hg1Hg0. Therefore G/H is abelian.

Problem 2.9.17.

Solution: Part (a) Since o(1) = 1, it follows that 1 ∈ T (G). If a, b ∈ T (G), designate m = o(a) and n = o(b). Since G isabelian, (ab)mn = amnbmn = (am)n(bn)m = 1, so ab is of finite order. Finally, since am = 1, it follows that (a−1)m = 1,so a−1 is of finite order and belongs to T (G). The Subgroup Test implies that T (G) is a subgroup of G.

Part (b) If T (G)g ∈ T (G/T (G)), then [T (G)g]m = T (G)gm = T (G) where m = o(T (G)g). It follows thatgm ∈ T (G), so gm is of finite order n. Therefore (gm)n = gmn = 1, so g is also of finite order and belongs to T (G).Therefore T (G)g = T (G) by Theorem 2.6.1(4), from which it follows that T (G/T (G)) = {T (G)}. Thus T (G) is torsion-free.

Part (c) Although the problem doesn’t indicate such, we assume H CG because otherwise G/H doesn’t constitute agroup. Suppose G is a torsion group, so all of its elements are of finite order. Because H ⊆ G, obviously H is a torsiongroup. Given Hg ∈ G/H where m = o(g), we have (Hg)m = Hgm = H, so Hg is of finite order, hence G/H is a torsiongroup. Conversely, if H and G/H are both torsion groups, by exercise 2.9.13(b) G is a torsion group.

Problem 2.9.18.

Page 57

Solution: Suppose K ⊆ H ⊆ G are groups and K CG and |G : K| is finite. It follows that |G/K| = |G : K| is finite, soG/K is a finite group. By Proposition 5, K C H, so H/K is a group. If Kh ∈ H/K, then Kh ∈ G/K, so H/K is asubgroup of G/K and therefore is itself finite. Lagrange’s Theorem implies that |G/K : H/K| = |G/K| / |H/K|, whichis clearly finite. COME BACK TO FOR LAST PART; MAYBE FIND BIJECTION

Problem 2.9.19.

Solution: Part (a) The subgroup H = {1} is trivially normal in G. Since G is abelian, if a, b ∈ G, then HaHb = Hab =Hba = HbHa; therefore G/H is abelian. Theorem 2.9.3 implies that G′ ⊆ H = {1}, hence G′ = {1}.

If g ∈ G where g = aba−1b−1 for some a, b ∈ G, then g = aa−1bb−1 = 1. Thus 1 is the only commutator of G, soG′ = {1}.

Part (b) Obviously [1, 1] = [−1,−1] = 1, and [1,−1] = [−1, 1] = −1. If α, β ∈ Q\{−1, 1}, then [α, β] = αβα−1β−1 =(αβ)2 = −1. Finally, [α, 1] = [α,−1] = [1, α] = [−1, α] = −1. Therefore every commutator of Q is either 1 or −1, soG′ = {−1, 1}.

Part (c) Let H = 〈a2〉 = {1, a2, a4}. Obviously ajaka−j = ak ∈ K where j ∈ Z and ak ∈ H. Further, a2b = ba4 anda4b = ba2, so ba2b−1 = bab = a4 ∈ H and ba4b−1 = ba4b = a2 ∈ H. Thus H CG.

Since |D6/H| = 12/3 = 4, by Theorem 3 D6/H is abelian. By Theorem 2.9.3, G′ ⊆ H, so |G′| = 1 or |G′| = H. ButD6 is not abelian, so it cannot be that G′ = {1} (since G′ = {1} if and only if D6 is abelian as explained after the proofof Theorem 2.9.3). Thus G′ = H = {1, a2, a4}.

Theorem 3. Every group of order 4 is abelian.

Proof. Suppose G is a group of order 4. Designate the three non-unity elements of G as a, b, and c. Obviously if a = b−1

then ab = ba = 1. If a 6= b−1, then ab = c (since if ab = b then a = 1 by right cancellation, and similarly if ab = a). Bythe same argument, ba = c, so ab = ba. We conclude that G is abelian.

Part (d) The answer is G′ = An. I’m not entirely clear on the proof....

Problem 2.9.20.

Solution: Let σ : G → G be an automorphism of G. It suffices to show that all generators of G′ (i.e., the set of allcommutators in G) are mapped to elements of G′. Given [a, b] ∈ G′, we have:

σ([a, b]) = σ(aba−1b−1) = σ(a)σ(b)(σ(a))−1(σ(b))−1 = [σ(a), σ(b)] ∈ G′.

We infer that σ(G′) ⊆ G′, so G′ is a characteristic subgroup of G.

Problem 2.9.21.

Solution: Let G be a group. If u ∈ (G×H)′, then for some M ≥ 1 and αj , γj ∈ G and βj , δj ∈ H:

u =

M∏j=1

[(αj , βj), (γj , δj)] =

M∏j=1

(αjγjα−1j γ−1j , βjδjβ

−1j δ−1j ) = (

M∏j=1

[αj , γj ],

M∏k=1

[βk, δk]) ∈ G′ ×H ′,

so (G×H)′ ⊆ G′ ×H ′.Conversely, if v ∈ G′ ×H ′, then for some N ≥ 1 and αj , γj ∈ G and βj , δj ∈ H:

v = (

N∏j=1

[αj , γj ],

N∏k=1

[βk, δk]) = (N∏j=1

αjγjα−1j γ−1j ,

N∏k=1

βjδjβ−1j δ−1j ) =

N∏j=1

([αj , βj ], [γj , δj ]) ∈ (G×H)′,

so G′ ×H ′ ⊆ (G×H)′. Therefore the two derived groups are equal.

Problem 2.9.22.

Page 58

Solution: Let H be a subgroup of G. If h ∈ H ′, then h is the finite product of commutators in H, so h ∈ H by closure.But h ∈ G′ as well because the commutators of H belong to G′, so h′ ∈ H ∩G′. Thus H ′ ⊆ H ∩G′.

It is not necessarily the case that H ′ = H∩G′. From exercise 2.9.18(b), we see the derived subgroup of the quarterniongroup Q is Q′ = {−1, 1}. The set H = {−1, 1} is a subgroup of Q with H ′ = {1} 6= Q′ (since [1, 1] = [1,−1] = [−1, 1] =[−1,−1] = 1), but H ∩Q′ = {−1, 1} = Q′.

Problem 2.9.23.

Solution: Let K CG.Part (a) Suppose K ⊆ H where H is a subgroup of G. Since K C H by Proposition 5, H/K is a factor group.

Obviously H/K is a subset of G/K. Further, K ∈ H/K because 1 ∈ H. Let Ka,Kb ∈ H/K. Since a, b ∈ H becausea ∈ Ka and b ∈ Kb (which partition the elements of H), we have KaKb = Kab ∈ H/K because ab ∈ H. Finally,because a ∈ H, it must be that a−1 ∈ H, so Ha−1 ∈ H/K. By the Subgroup Test, H/K is a subgroup of G/K.

Part (b) Let χ be a subgroup of G/K, and define H = {h ∈ G : Kh ∈ χ}. Since χ is a subgroup of G/K, it mustcontain K, so K1 ∈ χ, hence 1 ∈ H. If a, b ∈ H, then Ka,Kb ∈ χ, so KaKb = Kab ∈ χ and ab ∈ H. Finally, if a ∈ H,then Ka ∈ χ, so (Ka)−1 = Ka−1 ∈ χ, so a−1 ∈ H. Thus H is a subgroup of G by the Subgroup Test. Further, sinceK ∈ χ, it follows that if k ∈ K then Kk ∈ χ by Theorem 2.6.1(2), so k ∈ H. Therefore K is contained in H.

Now we will show χ = H/K. If Kc ∈ χ, then c ∈ H, so Kc ∈ H/K; thus χ ⊆ H/K. Conversely, if Kd ∈ H/K, thend ∈ H, so Kd ∈ χ, hence H/K ⊆ χ. The two sets are therefore equal.

Problem 2.9.24.

Solution: Suppose K CG and K ∩G′ = {1}. If k ∈ K, given g ∈ G we have [k, g] = k(gk−1g−1) ∈ K because K CG.Because [k, g] ∈ G′ by definition, [k, g] ∈ K ∩ G′, so [k, g] = kgk−1g−1 = 1. Therefore kg = gk for all g ∈ G, sok ∈ Z(G). We conclude that K ⊆ Z(G).

Let Ka ∈ Z(G/K), in which case if Kb ∈ G/K then KaKb = Kab = KbKa = Kba. It follows that ab(ba)−1 =aba−1b−1 ∈ K. But aba−1b−1 = [a, b] ∈ G′ by definition, so aba−1b−1 = 1 by hypothesis; therefore ab = ba, so a ∈ Z(G).Therefore Ka ∈ Z(G)/K. Conversely, if Kc ∈ Z(G)/K, then given d ∈ G we have KcKd = Kcd = Kdc = KdKc, soKc ∈ Z(G/K). We conclude that Z(G/K) = Z(G)/K.

Problem 2.9.25.

Solution: Let K CG.Part (a) Let a, b ∈ G be given. We have:

[Ka,Kb] = KaKb(Ka)−1(Kb)−1 = KaKbKa−1Kb−1 = K(aba−1b−1) = K[a, b].

Part (b) Suppose K ⊆ G′. By Theorem 2.9.3, G′ C G, so G′/K is a group. Observe that (G/K)′ = 〈{[Ka,Kb] :a, b ∈ G}〉 and G′/K = 〈{K[a, b] : a, b ∈ G}〉. By Theorem 2.4.10, it suffices to show that the generators of (G/K)′

and G′/K are each contained in the other. If [Ka,Kb] ∈ (G/K)′, then [Ka,Kb] = K[a, b] by the result in part (a),so [Ka,Kb] ∈ G′/K. Theorem 2.4.10 implies that (G/K)′ ⊆ G′/K. On the other hand, if [Kc,Kd] ∈ G′/K, then[Kc,Kd] = K[c, d] ∈ (G/K)′. Theorem 2.4.10 again implies that G′/K ⊆ (G/K)′. It follows that (G/K)′ = G′/K.

Problem 2.9.26.

Solution: The argument basically follows that of Theorem 2.9.2. Let K ⊆ Z(G) be a subgroup of G such that G/K = 〈X〉where X = {Kx1, . . . ,Kxn} and xixj = xjxi for all i, j. Since K CG, we are assured that G/K is a group. If a, b ∈ G,then:

Ka =∏Mj=1Kyj = K

[∏Mj=1 yj

]and Kb =

∏Nk=1Kzj = K

[∏Nk=1 zk

],

where M,N ≥ 1 and Kyj ,Kzj ∈ X.

Since a ∈ Ka and b ∈ Kb, it follows that a = k0∏Mj=1 yj and b = k1

∏Nk=1 zk for some k0, k1 ∈ G. Because all yj , zk

commute among each other and k0 and k1 are in the center of G (and therefore commute with all elements), we have:

ab =

k0 M∏j=1

yj

(k1 N∏k=1

zk

)=

k1 M∏j=1

yj

(k0 N∏k=1

zk

)=

(k1

N∏k=1

zk

)k0 M∏j=1

yj

= ba.

Page 59

This is true for all a, b ∈ G, so G is abelian.

Problem 2.9.27.

Solution: Suppose K ⊆ H ⊆ G are groups with K characteristic in G and H/K characteristic in G/K. By exercise2.9.24(a), K CG, so H CG by Proposition 5. Thus G/K and H/K are groups. We will show that H is characteristic inG.

Let σ : G→ G be an automorphism. Define α : G/K → G/K by α(Kg) = Kσ(g) for Kg ∈ G/K. First we will showthat α is an automorphism. If Kg0,Kg1 ∈ G/K, then:

α(Kg0Kg1) = α(K(g0g1)) = Kσ(g0g1) = K[σ(g0)σ(g1)] = Kσ(g0)Kσ(g1) = α(Kg0)α(Kg1),

so α is a homomorphism. The map α is obviously surjective because σ is surjective. If α(Kg0) = α(Kg1), then Kσ(g0) =Kσ(g1). Theorem 2.6.1 implies that σ(g0)[σ(g1)]−1 = σ(g0g

−11 ) = k0 for some k0 ∈ K. Since σ is bijective, we can take

the inverse, and since σ(K) = K, it must be that σ−1(k0) = k1 for some k1 ∈ K. Therefore:

σ−1(σ(g0g−11 )) = g0g

−11 = σ−1(k0) = k1 ∈ K.

Theorem 2.6.1 implies that Kg0 = Kg1, so α is injective. As a result, α is an automorphism.Since H/K is characteristic in G/K, if Kh ∈ H/K then α(Kh) = Kσ(h) = Kh1 ∈ H/K where h1 ∈ H. Therefore

σ(h)h−11 ∈ K ⊆ H, so σ(h) ∈ H. As a result, σ(H) ⊆ H, from which it follows that that H is characteristic in G.

Problem 2.9.28.

Solution: Part (a) Suppose |G : Z(G)| is prime, in which case |G/Z(G)| is prime. By Corollary 2.6.3, G/Z(G) is cyclic.Theorem 2.9.2 then implies that G is abelian. However, this then requires that Z(G) = G, so G/Z(G) = G. As a result,|G/Z(G)| = 1, which contradicts that |G/Z(G)| is prime. We conclude that |G : Z(G)| cannot be prime.

Part (b) Z(D4) = {1, a2} by exercise 2.6.26. Obviously D4 is not abelian: For example, ba 6= ab = ba3). Throughbrute force, I can show that D4/Z(D4) is abelian, but I don’t have an efficient or elegant proof for this proposition.

Problem 2.9.29.

Solution: Let k, n ≥ 2 and k | n. Define K = 〈ak〉 where a ∈ Dn. Obviously K is a subgroup of Dn. Given aj ∈ K, forakx ∈ K we have ajakxa−j = akx ∈ K. Given j ∈ Z, since (baj)akx(baj)−1 = b(ajakxa−j)b = bakxb, it suffices to showthat bakxb ∈ K. We have:

bakxb = b(akxbakx)a−kx = ak(−x) ∈ K.

Thus K CDn, so Dn/K is a group.We will now identify the elements of Dn/K.

• If Kaj = Kal, then aj(al)−1 = aj−l ∈ K by Theorem 2.6.1(3), from which it follows that aj−l = akq for someq ∈ Z. By Theorem 2.4.2(2), j − l ≡ kq (mod n) for some q ∈ Z. We may express ks = n by hypothesis, so(j − l)− kq = nt = k(st) for some t ∈ Z. Accordingly, j − l = k(st+ q), so j ≡ l (mod k). We prove the converseby the reversing the steps, so Kaj = Kal if and only if j ≡ l (mod k). As a result, Kaj ∈ {K,Ka, . . . ,Kak−1} forany j ∈ Z.

• If Kbaj = Kbal, then baj(bal)−1 = baj−lb = b(aj−1baj−l)al−j = al−j ∈ K, so al−j = akq for some q ∈Z. Following the same argument above, Kbaj = Kbal if and only if j ≡ l (mod k). Accordingly, Kbaj ∈{Kb,Kba, . . . ,Kbak−1} for any j ∈ Z.

Because we have covered all elements of Dn, we conclude Dn/K = {K,Ka, . . . ,Kak−1,Kb,Kba, . . . ,Kbak−1}.It’s now a simple matter to show that Dn/K is isomorphic with Dk. Clearly |Dn/K| = 2k = |Dk|. We have

(Ka)k = Kak = K because k ≡ 0 (mod k) and (Kb)2 = Kb2 = K. Further, KaKbKa = K(aba) = Kb. ThereforeDn/K ∼= Dk.

Page 60

Section 2.10

Problem 2.10.1.

Solution: Let g0, g1 ∈ G denoted by:

g0 =

(a0 b00 c0

)and g1 =

(a1 b10 c1

).

Define α : G→ R∗ × R∗ by α(g0) = (a0, c0). We will show α is a homomorphism. We have:

g0g1 =

(a0a1 a0b1 + b0c1

0 c0c1

),

so α(g0g1) = (a0a1, c0c1) = (a0, c0) · (a1, c1) = α(g0)α(g1). Thus α is a homomorphism.Obviously if k ∈ K, then α(k) = (1, 1) (the unity of the codomain), so k ∈ ker α. Conversely, if g1 ∈ ker α, then

α(g1) = (1, 1) so g1 ∈ K. As a result, K = ker α. Theorem 2.10.1 then implies that K CG.

Problem 2.10.2.

Solution: Let g ∈ G.(a) ⇒ (b): If α is the trivial homomorphism, then α(g) = 1G1 , so α(G) = {1G1}. Therefore ker α = G.(b) ⇒ (c): If ker α = G, then α(g) = 1G1 , so α(G) = {1G1}.(c) ⇒ (a): If α(G) = {1G1

}, then α(g) = 1G1. Therefore α is the trivial homomorphism.

Problem 2.10.3.

Solution: Since |G : H| = 2, the group H is normal in G, and |G/H| = 2. Therefore G/H = {H,Hg} for some g /∈ H.Since H and Hg partition G, an element of G is contained in H if and only if it is not contained in Hg. We will show αis a homomorphism. Let g0, g1 ∈ G. We have the following cases:

• If g0, g1 ∈ H, then g0g1 ∈ H, so α(g0g1) = 1 = 1 · 1 = α(g0)α(g1).

• If g0 ∈ H and g1 /∈ H, then Hg0Hg1 = H(g0g1) = H1 ·Hg = Hg. Therefore g0g1 ∈ Hg, so g0g1 /∈ H. Thereforeα(g0g1) = −1 = 1 · −1 = α(g0)α(g1).

• If g0 /∈ H and g1 ∈ H, then by the same argument g0g1 /∈ H, so α(g0g1) = −1 = α(g0)α(g1).

• If g0, g1 /∈ H, then Hg0Hg1 = H(g0g1) = HgHg = Hg2. Therefore g0g1 ∈ Hg2. But g2g−1 = g /∈ H, so byTheorem 2.6.1 Hg2 6= Hg, so Hg2 = H. Therefore g0g1 ∈ H, hence α(g0g1) = 1 = −1 · −1 = α(g0)α(g1).

We conclude that α is a homomorphism.If h ∈ H, then α(h) = 1, so H ⊆ kerα. If u ∈ kerα, then α(u) = 1, so u ∈ H and kerα ⊆ H. As a result, H = kerα.

Problem 2.10.6.

Solution: Suppose α : G → G1 is a group homomorphism where we denote K = ker α. Let a ∈ G. If g ∈ Ka, theng = ka for some k ∈ K. Therefore α(g) = α(ka) = α(k)α(a) = α(a). Thus Ka ⊆ {g ∈ G : σ(g) = σ(a)}. Conversely,if g ∈ {g ∈ G : α(g) = α(a)}, then α(g)[α(a)]−1 = α(ga−1) = 1, so ga−1 ∈ ker α = K. Thus Kg = Ka and g ∈ Ka,hence {g ∈ G : σ(g) = σ(a)} ⊆ Ka. We conclude Ka = {g ∈ G : σ(g) = σ(a)}.

Problem 2.10.8.

Solution: Part (a) Let α : C6 → K4 be a group homomorphism. Denote C6 = 〈d〉 for some d ∈ C6 and K4 = {1, a, b, c}where a2 = b2 = c2 = 1 and the product of any two distinct non-unity elements is the third non-unity element. Since C6

Page 61

is abelian, every subgroup of C6 is normal. It cannot be that ker α = {1} because then α is injective, which is impossiblebecause |C6| > |K4|. If kerα = {1}, then α is the trivial homomorphism (α(g) = 1 for all g ∈ C6).

Otherwise, we may assume without loss of generality that α(d) = a. Since α is determine entirely by its effect on C6’sgenerator, it follows that α(dj) = aj for j ∈ Z. We will show that α is well-defined. If dj = dk for some j, k ∈ Z, thenj ≡ k (mod 6), hence j = 6q + k for some q ∈ Z. Therefore:

α(dj) = aj = a6q+k = (a2)3qak = ak = α(dk).

Thus α is well-defined. Since there are four choices for α(d), there are four possible homomorphisms from C6 to K4.Part (b) Let α : C3 → A4 be a group homomorphism. Denote C3 = 〈a〉. Observe that |C3| = 3 and |A4| = 12. As

in part (a), every subgroup of C3 is normal. Since |C3| is prime, the only subgroups of C3 are {1} and C3. If ker α = C3,then α is the trivial homomorphism. If kerα = {1}, then α is injective. Since o(a) = 3, it follows from Corollary 2.5.1 thato(α(a)) divides o(a), so o(α(a)) = 3 (since o(α(a)) = 1 implies that α(a) = 1, so ker α = C3). The elements of A4 withorder 3 are the eight 3-cycles of A4. Thus α(a) = σ where σ ∈ { (1 2 3), (1 2 4), (1 3 4), (2 3 4), (1 3 2), (1 4 2), (1 43), (2 4 3) }. Since there are nine choices of α(a) (which defines all of α), there are nine homomorphisms from C3 to A4.

Part (c) Let α : D3 → C4 be a group homomorphism. Denote D3 = {1, a, a2, b, ba, ba2} (with a3 = 1, b2 = 1, andaba = 1) and C4 = 〈c〉 where c ∈ C4. Observe that |D3| = 6 and |C4| = 4. The normal subgroups of D3 are {1},{1, a, a2}, and D3. It cannot be that kerα = {1} because then α would be injective, which is impossible since |D3| > |C4|.If ker α = D3, then α is the trivial homomorphism.

If kerα = {1, a, a2}, then α(a) = 1. Since o(b) = 2, Corollary 2.5.1 implies that o(α(b)) divides 2. Since α(b) 6= 1,it must be that o(α(b)) = 2, so α(b) = a2 (since o(a) = o(a3) = 4). We can show that α so defined is well-defined andhomomorphic through the corollary to the Isomorphism Theorem (2.10.4). Let ϕ : D3 → D3/H be the coset mappingwhere H = {1, a, a2}, so ϕ(g) = Hg for g ∈ D3. Theorem 2.9.1 assets that ϕ is a surjective homomorphism. It is easilyshown that D3/H = {H,Hb}. Define σ : D3/H → 〈c2〉 by σ(H) = 1 and σ(Hb) = c2. Obviously σ is an isomorphism.Since ajb = ba4−j for any j ∈ Z, it suffices to show that σϕ is well-defined for bjak for j, k ∈ Z. We have:

σϕ(bjbak) = σ(K(bjak)) = σ(KbjKak) = σ(KbjK) = σ(Kbj) = [σ(Kb)]j = c2j .

It follows that α = σϕ, which is well-defined and homomorphic because it is the composition of two well-defined, homo-morphic maps. We conclude that there are two homomorphisms from D3 to C4.

Part (d) Let α : A4 → C3 be a group homomorphisms. Denote C3 = 〈c〉 where c ∈ C3. The only normal subgroups ofA4 are {ε}, K = {ε, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3), } (i.e., the elements equal to a product of disjoint transpositions),and A4. As before, if ker α = A4 then α is the trivial homomorphism, and ker α 6= {1} because α cannot be injective. Weare left with ker α = K. But if τ ∈ A4\K, then τ is a 3-cycle of order 3, so o(α(τ)) divides o(τ) = 3. Since the onlyelement of C3 of order 3 is c2, it follows that α(τ) = c2. Since α is a homomorphism, α(τ jσk) = c2j . Thus there are twohomomorphisms from A4 to C3.

Problem 2.10.11.

Solution: Let G be a group and θ : G→ G×G where θ(g) = (g, g) for all g ∈ G.Part (a) Given a, b ∈ G, we have θ(ab) = (ab, ab) = (a, a) · (b, b) = θ(a)θ(b), so θ is a homomorphism. Observe that

θ(a) = (1, 1) if and only if a = 1, so ker α = {1}, hence θ is injective by Theorem 2.10.3.Part (b)

(1) ⇒ (2) Suppose G is abelian. Given (g0, g1) ∈ G×G, we have for g ∈ G:

(g0, g1)(g, g)(g0, g1)−1 = (g0gg−10 , g1gg

−11 ) = (gg0g

−10 , gg1g

−11 ) = (g, g) ∈ θ(G).

Therefore θ(G) CG×G.(2) ⇒ (3) Suppose θ(G) C G × G. Therefore given (g0, g1) ∈ G × G, if g ∈ G then (g0, g1)(g, g)(g0, g1)−1 =

(g0gg−10 , g1gg

−11 ) ∈ θ(G). By definition of θ, it follows that g0gg

−10 = g1gg

−11 . Rearranging, we have:

g0gg−11 = gg−11 g0, (4)

for all g, g0, g1 ∈ G. We will use this result below.Define ϕ : G×G→ G by ϕ(g0g1) = g−10 g1. Given (g0, g1), (g2g3) ∈ G×G, we have:

ϕ((g0, g1) · (g2, g3)) = ϕ(g0g2, g1g3) = (g0g2)−1(g1g3) = (g−12 g−10 g1)g3 = (g−10 g1)(g−12 g3) = ϕ(g0, g1)ϕ(g2, g3),

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where we used equation (4). As a result, ϕ is a homomorphism. Further, if (g, g) ∈ θ(G), then ϕ(g, g) = g−1g = 1, soθ(G) ⊆ ker ϕ. Conversely, if (g0, g1) ∈ ker α, then ϕ(g0, g1) = g−10 g1 = 1, so g0 = g1, hence ker ϕ ⊆ θ(G). We concludethat ker ϕ = θ(G).

(3) ⇒ (1) Suppose there is a homomorphism ϕ : G × G → G such that ker ϕ = θ(G). Let g ∈ G by given. If(g′, g′) ∈ G, we have:

ϕ(g′g, gg′) = ϕ(g′, 1)ϕ(g, g)ϕ(1, g′) = ϕ(g′, 1)ϕ(1, g′) = ϕ(g′, g′) = 1,

since (g, g), (g′, g′) ∈ ker ϕ. By the definition of θ, it must be that g′g = gg′. Since this is true for every g and g′, weconclude that G is abelian.

Problem 2.10.12.

Solution: Suppose G is simple, in which case {1} and G are the only normal subgroups of G. By definition of simplicity,G 6= {1}. Let σ : G → G1 be a non-trivial homomorphism where G1 is some group. Since ker σ is a normal subgroup ofG that cannot equal G (since σ is non-trivial), it must be that ker σ = {1}. Theorem 2.10.3 implies that σ is injective.

Conversely, suppose every nontrivial homomorphism G→ G1 is injective. If H is a normal subgroup of G not equal toG, then by Theorem 2.10.2 H = ker ϕ where ϕ is the coset mapping. By hypothesis, kerϕ = {1}, so H = {1}. Since Gis a always a normal subgroup of G, we conclude that G is simple.

Problem 2.10.13.

Solution: Let G be a simple group and G1 be an arbitrary group. Suppose there is a non-trivial homomorphism α : G→ G1.Exercise 2.10.12 implies that ker α = {1}. Further, α(G) is a subgroup of G1 by Theorem 2.10.1. By the IsomorphismTheorem, α(G) ∼= G/ker α = G/{1}. But G/{1} is isomorphic to G, so α(G) ∼= G by Corollary 2.5.1 (proving that ∼= isan equivalence). Thus G1 has a subgroup isomorphic to G.

Conversely, suppose G1 has a subgroup H isomoprhic to G. There must exist an isomorphism σ : G → H, whereker σ = {1} because σ is injective. Since H ⊆ G, the map τ : G→ G1 where τ(g) = σ(g) for g ∈ G is a homomorphism.Moreover, τ is non-trivial because τ(g) 6= 1 if g ∈ G\{1}.

Problem 2.10.14.

Solution: Suppose n is odd. Let α : Dn → A4 be a homomorphism. By exercise 2.8.17, every normal subgroup of Dn iseither a subgroup of 〈a〉 or equal to Dn. Since o(α(a)) divides o(a) (which must be odd), it follows that o(α(a)) cannotbe 2. Since every element of A4 is of order 1, 2, or 3, it must be that o(α(a)) ∈ {1, 3}. There are nine such choices (ε andthe 3-cycles in A4). Because o(b) = 2, it must be that o(α(b)) ∈ {1, 2}. There are four such options (ε and the productsof disjoint transpositions in A4). Accordingly, there are at most thirty-six possible homomorphisms Dn → A4.

Problem 2.10.15.

Solution: Let G be a group where |G| ≥ 2 and autG is cyclic. Accordingly, autG = 〈α〉 where α is some automorphism ofG. By Theorem 2.10.5, innG ∼= G/Z(G). Since innG is a subgroup of autG, it must be that innG is cyclic, so G/Z(G)is cyclic. Theorem 2.9.2 implies that G is abelian.

Next we will show that autG is finite. Since |G| ≥ 2, the inverse mapping automorphism σ : G→ G where σ(g) = g−1

for all g ∈ G is not equal to the identity mapping. Thus σ = αj for some j ∈ Z where j 6= 0. Since σσ = 1G, it followsthat (αj)2 = α2j = 1G, so autG is finite. Observe that {1G, σ} is a subgroup of autG of order 2. By Lagrange’s Theorem,autG is even.

Problem 2.10.16.

Solution: Suppose autG is simple. By exercise 2.8.16, innG is normal in autG, so innG equals either {1G} or autG. IfinnG = {1G}, then given a ∈ G we have σa = 1G, so given g ∈ G we have σa(g) = aga−1 = g. Therefore ag = ga, so Gis abelian.

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On the other hand, if innG = autG, the innG is simple. Let H/Z(G) be a normal subgroup of G/Z(G), in which caseH is a subgroup of G containing Z(G). It follows that given Z(G)g ∈ G/Z(G), if h ∈ H, then Z(G)h ∈ H/Z(G) and:

[Z(G)g][Z(G)h][Z(G)g]−1 = Z(G)(ghg)−1 = Z(G)h′ ∈ H/Z(G),

where h′ ∈ H. Therefore ghg−1(h′)−1 ∈ H, so ghg−1 ∈ H. Consequently, H is normal in G if H/Z(G) is normal inG/Z(G).

Define K = {σh ∈ inn G : h ∈ H}. We will show that K is a subgroup of inn G. Since 1 ∈ H, it follows thatσ1 = 1G ∈ K. If u, v ∈ H, then σuσv(g) = σu(vgv−1) = (uv)g(uv)−1 = σuv(g) for g ∈ G where uv ∈ H, so σuv ∈ K.Finally, the inverse of σu is σu−1(g) = u−1gu; since u−1 ∈ H, it must be that σu−1 ∈ K.

Let σa ∈ innG. If σh ∈ K, then for g ∈ G:

σaσh[σa]−1(g) = (aha−1)g(aha−1) = h0hh−10 ,

where h0 = aha−1 is contained in H because H is normal in G. Since σaσh[σa]−1 belongs to K, it follows that K isnormal in aut G. But since inn G is simple, K equals either {1G} or inn G. If K = {1G}, then σw ∈ K implies σw = 1G,so w ∈ Z(G) by exercise 2.5.5. Since H contains Z(G), it follows that H = Z(G). If K = innG, then H = G. Thus theonly normal subgroups of G/Z(G) are Z(G)/Z(G) = {Z(G)} and G/Z(G). We conclude that G/Z(G) is simple.

Problem 2.10.17.

Solution: Part (a) If [a, b] is a commutator of G, then α([a, b]) = α(aba−1b−1) = α(a)α(b)[α(a)]−1[α(b)]−1 =

[α(a), α(b)]. If g ∈ G′, then g =∏Mj=1[aj , bj ] where M ≥ 1 and aj , bj ∈ G. Therefore:

α(g) =

M∏j=1

α([aj , bj ]) =

M∏j=1

[α(aj), α(bj)],

which belongs to G′1. Therefore α(G′) ⊆ G′1.Part (b) Define α : G/G′ → G1/G

′1 by α(G′g) = G′1α(g) for all G′g ∈ G/G′. First we will show that α is well-

defined. Suppose G′g0 = G′g1, in which case g0g−11 ∈ G′. By the result in part (a), α(g0g

−11 ) = α(g0)[α(g1)]−1 ∈ G′1, so

G′1α(g0) = G′1α(g1). Consequently, α(G′g0) = α(G′g1). In addition, α is a homomorphism because:

α(G′g0G′g1) = α(G′(g0g1)) = G′1α(g0g1) = G′1α(g0)G′1α(g1) = α(G′g0)α(G′g1).

Finally, we will show that α as defined above is the only homomorphism such that αϕ = ϕ1α. If σ : G/G′ → G1/G′1

is a homomorphism that also meets this condition, then given g ∈ G:

σϕ(g) = σ(G′g) = ϕ1α(g) = G′1α(g).

Since we defined α identically, it follows that σ = α. Accordingly, there is a unique homomorphism α such that αϕ = ϕ1α.

Problem 2.10.18.

Solution: Suppose G = H ×K and K1 = {(1, k) : k ∈ K}. Let (h, k) ∈ G be given. If u′ ∈ K, then u′ = (1, u) foru ∈ K, so:

(h, k)(1, u)(h, k)−1 = (1, kuk−1) = (1, k′),

where k′ = kuk−1 ∈ K. It follows that (1, k′) ∈ K1, so K1 is normal in G.Define α : K1 → K by α(1, k) = k. If k0, k1 ∈ K, then α((1, k0)(1, k1)) = α(1, k0k1) = k0k1 = α(1, k0)α(1, k1).

Therefore α is a homomorphism. The map α is also surjective because given k ∈ K we have α(1, k) = k. Further, ifα(1, k) = 1 then k = 1, so ker α = {(1, 1)}. By The Isomorphism Theorem, α(K1) = K ∼= K1/{(1, 1)} ∼= K1.

Since K1 C G, the set G/K1 is a factor group. Define σ : G → H by σ(h, k) = h for (h, k) ∈ G. Given(h0, k0), (h1, k1) ∈ G, we have σ((h0, k0)(h1, k1)) = σ(h0h1, k0k1) = h0h1 = σ(h0, k0)σ(h1, k1), so σ is a homo-morphism. Clearly σ is surjective. Further, σ(h, k) = 1 if and only if (h, k) = (1, k); thus ker σ = K1. By the IsomorphismTheorem, σ(G) = H ∼= G/ker (σ) = G/K1.

Problem 2.10.19.

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Solution: Part (a) If g ∈ G and k ∈ K, then det(g) 6= 0 and det(k) = 1. Therefore:

det(gkg−1) = det(g)det(k)[det(g)]−1 = det(g)[det(g)]−1det(k) = det(k) = 1.

As a result, gkg−1 ∈ K, so K CG.Define α : G→ R∗ by α(g) = det(g) for all g ∈ G. Because no element of G has a determinant equal to 0, the domain

of α is correct. Let x ∈ R∗ be given. There exists some g ∈ G such that det(g) = x: We have det(I) = 1, and by theproperties of the determinant, det(x1/nI) = xdet(I) = x. As a result, α(x1/nI) = x, so α(G) = R∗. If α(g) = 1, thendet(g) = 1, so ker α ⊆ K; conversely, if k ∈ K then α(k) = 1. It follows that ker α = K. By the Isomorphism Theorem,α(G) = R∗ ∼= G/ker α = G/K.

Part (b) By the same argument as in part (a), K1 CG. Define β : G→ R>0 by β(g) = [det(g)]2 for g ∈ G. As shownin part (a), for any x ∈ R>0, there is some g ∈ G such that det(G) =

√x. As a result, β(G) = R>0. If β(g) = 1, then

det(g) = ±1, so g ∈ K1; conversely, if k ∈ K1, then β(g) = (±1)2 = 1. It follows that ker β = K1. The IsomorphismTheorem implies that β(G) = R>0

∼= G/ker β = G/K1.

Problem 2.10.20.

Solution: First some notation and helper results. Let g0, g1 ∈ G and k ∈ K, each denoted by:

g0 =

(a0 b00 c0

), g1 =

(a1 b10 c1

), k =

(1 β0 1

),

where a0, c0, a1, c1 6= 0. A simple calculation shows that:

g−10 =

(a−10 −b0c−10 /a00 c−10

),

which is, of course, in G.With these results in hand, we have:

g0kg−10 =

(a0a−10 −b0c−10 + a0c

−10 b1 + b0c

−10

0 c0c−10

)=

(1 a0b1c

−10

0 1

),

which belongs to K. As a result, K CG.Define α : G → R∗ × R∗ by α(g0) = (a0, c0). There is no g0 such that a0 = 0 or c0 = 0, so the domain of α is

correct. Since g0g1 =

(a0a1 a0b1 + b0c1

0 c0c1

), it follows that α(g0g1) = (a0a1, c0c1) = (a0, c0)(a1, c1) = α(g0)α(g1), so

α is a homomorphism. Given (a0, b0) ∈ R∗ ×R∗, we have α(g0) = (a0, b0) for g0 with those entries, so α(G) = R∗ ×R∗.Obviously ker α = K. By the Isomorphism Theorem, α(G) = R∗ × R ∼= G/ker α = G/K.

Problem 2.10.21.

Solution: Define σ : C∗ → R>0 by σ(z) = |z| for z ∈ C∗. Since |z| > 0 for all z ∈ C∗, the domain is valid. If z0, z1 ∈ C∗,then σ(z0z1) = |z0z1| = |z0| |z1| = σ(z0)σ(z1), hence σ is a homomorphism. If x ∈ R>0, then σ(x) = |x| = x, so σis surjective. If z ∈ ker σ, then |z| = 1, so z ∈ C0. Conversely, if z ∈ C0, then σ(z) = 1, so z ∈ ker σ. We infer thatker α = C0. The Isomorphism Theorem implies that σ(C∗) = R>0

∼= C∗/ker σ = C∗/C0.

Problem 2.10.22.

Solution: Define τ : R∗ → R>0 by τ(x) = |x| for x ∈ R∗. (Notice that τ is the restriction of σ in exercise 2.10.21to R∗). For the same reasons as in exercise 2.10.21, τ is a surjective homomorphism. If x ∈ ker τ , then |x| = 1, sox ∈ {−1, 1}; if y ∈ {−1, 1}, then τ(y) = 1, so y ∈ ker τ . Thus ker τ = {−1, 1}. The Isomorphism Theorem implies thatR>0

∼= R∗/{−1, 1}.

Problem 2.10.23.

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Solution: Define α : G → R∗ where α(τa,b) = a. We will show that α is well-defined. If τa,b = τc,d, then τa,b(x) =ax+ b = τc,d(x) = cx+ d for all x. By evaluating the functions at x = 0, we find that b = d; by evaluating the functionsat x = 1, we find that a = c. Therefore α(τa,b) = α(τc,d).

We will skip the straightforward proof that α a is homomorphism.If x ∈ R∗, then α(τx,0) = x, so α is surjective. If τa,b ∈ ker α, then α(τa,b) = a = 1, hence τa,b ∈ K. Conversely, if

τa,b ∈ K, then a = 1, so α(τa,b) = 1 ∈ ker α. Thus ker α = K. By the Isomorphism Theorem, G/K ∼= R∗.

Problem 2.10.24.

Solution: Remember that the binary operation on M2(Z is matrix addition, with the zero matrix as the unity. Let n ≥ 2.Denote mj ∈M2(Z) and sk ∈M2(nZ) for any j, k as:

mj =

(aj bjcj dj

), and sk =

(ntk nuknvk nwk

),

where all the entries are integers. It is easy to show that M2(nZ) is a normal subgroup of M2(Z), so we omit the proof.Define α : M2(Z)→M2(Z/nZ) by:

α(m0) =

(a0 b0c0 d0

).

If m0 = m1, then a0 = a1, implying that a0 = a1, and so on for all pairs of entries. Therefore α(m0) = α(m1), establishingthat α is well-defined. It is easy to show that α is a homomorphism by the basic principles of modular addition.

Let s0 ∈M2(nZ). It follows that nt0 ≡ 0 (mod n), so t0 = 0. Therefore:

α(s0) =

(nt0 nu0nv0 nw0

)=

(0 00 0

),

so s0 ∈ ker α. Conversely, if s1 ∈ ker α, then s1 ∈M2(nZ) with t1 = u1 = v1 = w1 = 0. Consequently, kerα = M2(nZ).By the Isomorphism Theorem, M2(Z)/M2(nZ) ∼= M2(Z/nZ).

Problem 2.10.25.

Solution: This problem is easy to solve with α : G×G×G→ G×G where α(a, b, c) = (ab−1, bc−1) for (a, b, c) ∈ G×G×G.

Problem 2.10.26.

Solution:

Proposition 6. If α : B → C and β : A→ B are homomorphisms, then αβ is a homomorphism.

Proof. If a, b ∈ A, then:αβ(ab) = α(β(a)β(b)) = [αβ(a)][αβ(b)],

so αβ is a homomorphism.

Let G be a group, K be a subgroup where K C G, and H be a group where G/K ∼= H. Let α : G/K → H be anisomorphism. Define ϕ : G → G/K as the coset map, which is surjective by Theorem 2.9.1. It follows that σ = αϕ is ahomomorphism by Proposition 6 and surjective.

Let k ∈ K. It follows that αϕ(k) = α(Kk) = α(K) = 1 because K is the unity of G/K. It follows that K ⊆ ker σ.On the other hand, if g ∈ ker σ, then σ(g) = 1, so αϕ(g) = α(Kg) = 1. Because α is injective, Kg = K. Thereforeg ∈ K, so ker σ ⊆ K. We conclude that ker σ = K. Thus σ is a surjective homomorphism from G→ H were K = ker σ.

Problem 2.10.27.

Page 66

Solution: Given α(g) ∈ α(G) and α(k) ∈ α(K), we have:

α(g)α(k)[α(g)]−1 = α(gkg−1) = α(k′) ∈ α(K),

where k′ = gkg−1 ∈ K because K CG. Thus α(K) C α(G), and α(G)/α(K) is a factor group.Define σ : G → α(G)/α(K) by σ(g) = α(K)α(g) for g ∈ G. Obviously σ is surjective. First we will show that

σ is well-defined. In all cases, let g0, g1 ∈ G. If g0 = g1, then g0g−11 = 1, so g0g

−11 ∈ ker α ⊆ K. Therefore

α(g0g−11 ) = α(g0)[α(g1)]−1 ∈ α(K), so α(K)α(g0) = α(K)α(g1). It follows that σ is well-defined.

Next we will show that σ is a homomorphism. We have:

σ(g0g1) = α(K)α(g0g1) = α(K)[α(g0)α(g1)] = α(K)α(g0)α(K)α(g1) = σ(g0)σ(g1).

Finally, we will show that ker σ = K. If u ∈ ker σ, then σ(u) = α(K)α(u) = α(K), so α(u) ∈ α(K). Thusα(u) = α(k′) for some k′ ∈ K, and α(u(k′)−1) = 1, so u(k′)−1 ∈ ker α ⊆ K. Thus ker σ ⊆ K. Conversely, if v ∈ K,then σ(v) = α(K)α(v) = α(K)k′′ = α(K) where k′′ = α(v) ∈ α(K). Thus K ⊆ ker σ, so ker σ = K.

By the Isomorphism Theorem, G/ker σ = G/K ∼= α(G)/α(K).

Problem 2.10.29.

Solution: Part (a) Obviously I ∈ G. Given g0, g1 ∈ G, we have:

g0g1 =

1 a0 b00 1 c00 0 1

1 a1 b10 1 c10 0 1

1 a1 b10 1 c10 0 1

=

1 a0 + a1 b0 + b1 + a0c10 1 c0 + c10 0 1

, (5)

which is contained in G. A simple calculation shows that:

g−10 =

1 −a0 a0c0 − b00 1 −c00 0 1

,

which is also contained in G. The Subgroup Test implies that G is a subgroup of M3(R)∗.From equation (5), we see that g0 ∈ Z(G) if and only if g0g1 = g1g0 for any g1 ∈ G, which requires that b0+b1+a0c1 =

b0b1 + a1c0, from which it follows that a0c1 = a1c0. Since this must hold for all a1 and c1, if we let a1 = 0 and c1 = 1,then a0 = 0. Similarly, if a1 = 1 and c1 = 0, then c0 = 0. Accordingly, Z(G) = {g0 ∈ G : a0 = c0 = 1}. Of course,Z(G) CG.

Define α : Z(G) → R by α(g) = b0. It is easy to show that α is a homomorphism, so we will omit the proof. Givenx ∈ R, we have α(g0) = x with the b0 entry of g0 equal to x, so α is surjective. Further, kerα = {I}. By the IsomorphismTheorem, α(Z(G)) = R ∼= Z(G)/{I} ∼= Z(G).

Part (b) Define σ : G→ R× R by σ(g0) = (a0, b0) for g0 ∈ G. Given g0, g1 ∈ G, from equation (5) we see that:

σ(g0g1) = (a0 + a1, c0 + c1) = (a0, c0) + (a1, c1) = σ(g0) + σ(g1),

so σ is a homomorphism. If (x, y) ∈ R×R, then σ(g0) = (x, y) if a0 = x and c0 = y, so σ is surjective. Finally, it is easilyshown that ker σ = Z(G). By the Isomorphism Theorem, R× R ∼= G/Z(G).

Problem 2.10.30.

Solution: Let m,n ∈ Z where m | n. Let mk = n where k ∈ Z. Define α : Z/nZ → Z/mZ by α(x) = x (wherethe residue class as an argument to α is modulo n and its image is a residue class in modulo m). From here on letx0, x1 ∈ Z/nZ. We will show that α is well-defined. If x0 = x1, then x0 − x1 = nq = m(qk) for some k ∈ Z, hencex0 ≡ x1 (modm). Thus α(x0) = α(x1).

Next we will show that α is a homomorphism:

α(x0 x1) = α(x0x1) = x0x1 = x0 x1 = α(x0)α(x1).

Given y ∈ Z/mZ, by the division algorithm y = mu+ r = n(ku) + r where u ∈ Z and 0 ≤ r ≤ m− 1. Consequently,y ≡ r (mod n) and y ≡ r (modm). Therefore α(r) = r = y, and α is surjective.

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If a ∈ ker α, then α(a) = a = 0 = m, so a ∈ 〈m〉. Conversely, if b ∈ 〈m〉, then b = cm (where m is modulo n) forsome c ∈ Z. Therefore α(b) = α(cm) = cα(m) = cm = c0 = 0. Accordingly, b ∈ ker α. We infer that ker α = 〈m〉.

The Isomorphism Theorem implies that Z/nZ/〈m〉 ∼= Z/mZ.

Problem 2.10.32.

Solution: Let G be a group and σ : Z → G be a homomorphism where σ(G) is infinite. Let u = σ(1). If x ∈ Z, thenx = x(1), so σ(x) = σ(x(1)) = xσ(1) = xu. It follows that σ(Z) = {ju : j ∈ Z} = 〈u〉. Since σ(Z) is infinte, itfollows that o(u) is infinite. By Theorem 2.4.3(1), ker σ = {0G}. As a result, Z/ kerσ = Z/{0} ∼= Z. By the IsomorphismTheorem, σ(G) ∼= Z/ kerσ ∼= Z.

Problem 2.10.33.

Solution: In all parts, let α : A→ X be a homomorphism where A is the set below and X is a given set. Remember thatthe kernel of an homomorphism is a normal subgroup of the homomorphism’s domain.

Part (a) Since Z/4Z is abelian, every subgroup (consisting of {0}, {0, 2},Z/4Z) is normal. Therefore:

• If ker α = {0}, then α(Z/4Z) ∼= Z/4Z/{0} ∼= Z/4Z.

• If ker α = Z/4Z, then α(Z/4Z) ∼= (Z/4Z)/(Z/4Z) ∼= {0}.

• If ker α = {0, 2}, then α(Z/4Z) ∼= Z/4Z/{0, 2} ∼= Z/2Z.

Part (b) The only normal subgroups of G are {1} and G, so:

• If ker α = {1}, then α(G) ∼= G/{1} ∼= G.

• If ker α = G, then α(G) ∼= G/G ∼= {1}.

Part (c) By exercise 2.5.3, the only normal subgroups of A4 are {ε}, {ε, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3) }, and A4.As a result:

• If ker α = {ε}, then α(A4) ∼= A4/{ε} ∼= A4.

• If ker α = A4, then α(A4) = A4/A4∼= {ε}.

• If kerα = {ε, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3) } = H, then α(A4) = A4/H = {K,K(1 2 3), K(1 3 2)} by example2.9.3. Thus α(A4) ∼= 〈(1 2 3)〉 ∼= C3.

Problem 2.10.34.

Solution: Let G be a group where |G| ≥ 3. The identity map 1G : G → G is an automorphism of G. If G is abelian,then the inverse map α : G → G where α(g) = g−1 is another automorphism of G. If G is not abelian, then Z(G) is aproper subset of G. By Theorem 2.10.5, G/Z(G) ∼= inn G where G/Z(G) 6= {1}. Therefore |G/Z(G)| ≥ 2, so there issome σa ∈ innG where σa 6= 1G. Thus there are at least two automorphisms of G.

Problem 2.10.35.

Solution: Suppose X is a subgroup of G. We will skip the straightforward proof the α(X) is a subgroup of G.Now suppose X C G. Choose g ∈ G, then let g′ = α(g) ∈ G. If a ∈ α−1(X), then α(a) = x for some x ∈ X. We

then have:α(gag−1) = g′α(a)g′−1 = g′xg′−1.

Since X is normal in G, it follows that g′xg′−1 ∈ X, so α(gag−1) ∈ X. By definition, so g′ag′−1 ∈ α−1(X), soα−1(X) C (G).

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Define σ : G → G1/X by σ(g) = Xα(g) for all g ∈ G. Let g0, g1 ∈ G. If g0 = g1, then g0g−11 = 1 ∈ X,

so σ(g0) = Xg0 = Xg1 = σ(g1); therefore σ is well-defined. Further, it follows that σ is a homomorphism by themultiplicative property of the factor group G1/X.

Given Xc ∈ G1/X, there is some g3 ∈ G that σ(g3) = c because α is surjective. Finally, if x ∈ α−1(X), thenσ(x) = Xα(x) = X because α(x) ∈ X; therefore α−1(X) ⊆ ker α. Conversely, if y ∈ ker α, then σ(y) = Xα(y) = X.Therefore α(y) ∈ X, so y ∈ α−1(X), from which it follows that α−1(X) = ker α.

By the Ismorphism Theorem, G/α−1(X) ∼= G1/X.

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