MATH 413-01 Abstract AlgebraHomework Solutions: #2 – #7

1 Homework #2

1. [Bachwich, A.] In D4, find all elements x such that

(a) x3 = F|Solution: S1 = {F|}

(b) x3 = R90

Solution: S2 = {R270}(c) x3 = R0

Solution: S3 = {R0}(d) x2 = R0

Solution: S4 = {R0, R180, F|, F−, F\, F/}(e) x2 = F−

Solution: S5 = ∅

2. [Leonard, R.] Prove that every Cayley table for a group is a Latin square. A Latin square of order nis an n× n array on n symbols such that each symbol appears exactly once in each row and column.Hint: use Theorem 2.2.

Let G be a group. By contradiction, assume that an element q appears twice in one column of theCayley table of G. This implies ∃ a, b, c ∈ G such that b ◦ a = q and c ◦ a = q. We know fromTheorem 2.2 (Cancellation) that b ◦ a = c ◦ a implies b = c. Therefore, by contradiction, q cannotappear more than once in each column. Similarly q cannot appear more than once in each row.

We must also prove an element q will appear at least once in each row. Each row has n elements,where n equals the order of the group. Because the operation is closed, there are n possible outcomesbetween any two elements in G. Thus there are n entries in each row of the Cayley table and npossible elements to fill it with. We know from the previous argument that no element can appearmore than once in each row, and we know that each entry in the Cayley table must be filled becauseevery operation a ◦ b ∈ G has a solution. By this, we conclude that each element appears at leastonce in each row because it takes n elements to fill a row of size n without duplicating an element.Similarly, each element has to appear once in each column.

Therefore, for every row and column, each element will appear exactly once. This makes the Cayleytable for G a Latin square.

1

3. [Berger, E.] Suppose the table below is a Cayley table for an unknown group. Fill in the blankentries. Do not assume e is the identity. Also, Problem 2 will be helpful.

e a b c d

e e − − − −a − b − − eb − c d e −c − d − a bd − − − − −

In the previous problem, we proved that any Cayley table is a Latin square. Thus, we can use therules of Latin squares to begin filling out the table. By these rules, we know d ◦ a = e, e ◦ a = a,b ◦ e = b, b ◦ d = a, c ◦ b = e, and c ◦ e = c.

By the definition of a group, we know there exists an identity element e ∈ G such that a◦e = a = e◦afor all a ∈ G. Coincidently, since e ◦ e = e, e ◦ a = a, b ◦ e = b, and c ◦ e = c and we know the identityis unique, we know that e is our identity element. As such, a ◦ e = a, d ◦ e = d, e ◦ b = b, e ◦ c = c,and e ◦ d = d.

We can solve the remaining entries, d ◦ d = c, a ◦ c = d, d ◦ c = b, d ◦ b = a and a ◦ b = c, using theproperties of Latin squares.

e a b c d

e e a b c da a b c d eb b c d e ac c d e a bd d e a b c

4. [Li, Y.] Suppose that in the definition of a group G, the condition that there exists an element ewith the property a ◦ e = e ◦ a = a for all a ∈ G is replaced by a ◦ e = a for all a ∈ G. Show thate ◦ a = a for all a ∈ G.

Assume that e ◦ a = x. If we operate by the same element, a, on both sides of the equation,

a ◦ e ◦ a = a ◦ x.

According to the definition of group, the operation is associative. Therefore

(a ◦ e) ◦ a = a ◦ x.

Since a ◦ e = a, we substitute a ◦ e with a. We then get a ◦ a = a ◦ x. By cancellation, x = a.Therefore, e ◦ a = a.

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5. [Dyke, M.] Suppose that in the definition of a group G, the condition that for each element a ∈ Gthere exists an element b ∈ G with the property a ◦ b = b ◦ a = e is replaced by a ◦ b = e. Show thatb ◦ a = e. (Note that under this assumption inverses only work on the right hand side.)

Consider the identity element property of groups:

e ◦ a = a ◦ e.

Use the given right inverse property to replace the e on the left side of the equation

(a ◦ b) ◦ a = a ◦ e.

Using the associativity property of groups, the equation can be rearranged as

a ◦ (b ◦ a) = a ◦ e.

Recall Theorem 2.2, which states that a ◦ b = a ◦ c implies b = c. Applying Theorem 2.2 toa ◦ (b ◦ a) = a ◦ e allows the cancellation of a on the left side of both halves of the equation

a ◦ (b ◦ a) = a ◦ e,

which yieldsb ◦ a = e,

thus proving the desired property b ◦ a = e.

3

2 Homework #3

1. [Hoffman, L.] Invent your own Color-five group. Provide a Cayley table and clearly state theidentity and each element’s inverse.

red

blue

clear

orange

oxblood

Figure 1: Color-five group.

R = Red, R−1 = Oxblood

B = Blue, B−1 = Orange

C = Clear, C−1 = Clear

O = Orange, O−1 = Blue

X = Oxblood, X−1 = Red

The identity is Clear.

C O X R B

C C O X R BO O X R B CX X R B C OR R B C O XB B C O X R

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2. [Spurlock, J.] Determine the following quantities:

(a) 51 (mod 13)

51 ≡ 12 (mod 13)

(b) 342 (mod 85)

342 ≡ 2 (mod 85)

(c) 82 · 73 (mod 7)

82 · 73 ≡ 5 · 3 = 15 ≡ 1 (mod 7)

(d) 51 + 68 (mod 7)

51 + 68 ≡ 2 + 5 = 7 ≡ 0 (mod 7)

(e) 35 · 24 (mod 11)

35 · 24 ≡ 2 · 2 = 4 (mod 11)

(f) 47 + 68 (mod 11)

47 + 68 ≡ 3 + 2 = 5 (mod 11)

3. [Brubaker, N.] For n = 5, 8, and 12, find all positive integers less than n that are relatively prime ton.

(a) {a ∈ Z : 0 < a < 5 , gcd(a, 5) = 1} = {1, 2, 3, 4}(b) {a ∈ Z : 0 < a < 8 , gcd(a, 8) = 1} = {1, 3, 5, 7}(c) {a ∈ Z : 0 < a < 12 , gcd(a, 12) = 1} = {1, 5, 7, 11}

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4. [Carroll, S.] Complete the following Cayley tables.

(a) (Z4,+)

+ 0 1 2 3

0 0 1 2 31 1 2 3 02 2 3 0 13 3 0 1 2

(b) (Z6,+)

+ 0 1 2 3 4 5

0 0 1 2 3 4 51 1 2 3 4 5 02 2 3 4 5 0 13 3 4 5 0 1 24 4 5 0 1 2 35 5 0 1 2 3 4

5. [Kranstz, S.] Prove the following statements concerning inverses in a group G.

(a) For all a ∈ G, prove (a−1)−1 = a.

Let x = (a−1)−1.Then a−1 ◦ x = e(a ◦ a−1) ◦ x = a ◦ e

e ◦ x = ax = a.

Therefore (a−1)−1 does, in fact, equal a.

(b) For all a, b ∈ G, prove (a ◦ b)−1 = b−1 ◦ a−1.We know

(a ◦ b) ◦ (a ◦ b)−1 = ea−1 ◦ (a ◦ b) ◦ (a ◦ b)−1 = a−1 ◦ e(a−1 ◦ a) ◦ b ◦ (a ◦ b)−1 = a−1

e ◦ b ◦ (a ◦ b)−1 = a−1

b ◦ (a ◦ b)−1 = a−1

(b−1 ◦ b) ◦ (a ◦ b)−1 = b−1 ◦ a−1e ◦ (a ◦ b)−1 = b−1 ◦ a−1

(a ◦ b)−1 = b−1 ◦ a−1.

Therefore (a ◦ b)−1 = b−1 ◦ a−1.

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6. [Rotert, A.] Let n and a be positive integers. Show that the equation ax ≡ 1 (mod n) has a solutionif and only if a and n are relatively prime.

Proof (Forward Implication):

Assume ax ≡ 1 (mod n). Then ax = qn+ 1 is an equivalent statement, where q ∈ Z+.

We can rewrite ax = qn + 1 as ax − qn = 1. If we consider the set S = {ax + by : (ax + by) >0 and x, y ∈ Z}, then ax − qn must be in S, where n = b and −q = y, and using the DivisionAlgorithm as we did in our gcd proof, it can be show that 1 ∈ S.

To show that 1 is also the greatest element, consider d′ such that d|a, d|n, d′|a, d′|n and d′ > d. Leta = d′h and b = d′u ∃h, u ∈ Z. Then d = as + bt = (d′h)s + (d′u)t = d′(hs + ut). This implies thatd′|d, implying that d > d′, a contradiction.

Thus, if ax ≡ 1 (mod n), then gcd(a, n) = 1.

Backward Implication:

Assume gcd(a, n) = 1. Consider the set S = {ax + by : (ax + ny) > 0 and x, y ∈ Z}. This meanswe have an equation of the form ax + ny = 1, because gcd(a, n) = 1. This can be rewritten asax = 1− qn. If we assume that q ≤ 0, then this becomes ax = qn+ 1.

This is equivalent to stating ax ≡ 1 (mod n), meaning if a and n are relatively prime, then ax ≡1 (mod n).

Therefore ax ≡ 1 (mod n) if and only if gcd(a, n) = 1.

7. [Wulff, A.] Suppose a and b are integers that divide the integer c. If a and b are relatively prime,show that ab divides c. Show by example that if a and b are not relatively prime, then ab need notdivide c. Hint: familiarize yourself with Theorem 0.3.

Presume a and b are relatively prime, and thus share no common factors, aside from 1. This can betaken from the Fundamental Theorem of Arithmetic: for a,b ∈ Z, they are either prime or a productof primes. Let a = f1 · f2 · f3 · · · fr and b = g1 · g2 · g3 · · · gs, where r and s need not be equal, and nofr = gs. Let A be the set of factors of a, B be the factors of b, and C the collection of factors of c. Ifeither a|c or b|c, then c is a composite of prime factors and A ⊆ C or B ⊆ C, respectively. It followsthat if a|c and b|c, then A ∪ B ⊆ C, and thus C contains all f1, f2, f3, . . . , fr and g1, g2, g3, . . . , gs.If A ∪ B = C, then C = {f1, f2, . . . , fr, g1, g2, . . . , gs}, and ab = c. In this case ab|c. If A ∪ B ⊂ C,then there exists other prime numbers h1, h2, . . . , ht ∈ C, and c = f1 · · · fr · g1 · · · gs · h1 · · ·ht. Letu = h1 · · ·ht, and then there exists an integer u such that c = abu, and ab|c.If a and b are not relatively prime, then ∃fr ∈ A, and ∃gs ∈ B such that, fr = gs. For exampleconsider a = 6 = 2 · 3, b = 8 = 23, and c = 24 = 23 · 3. In this case A = {2, 3}, and B = {2, 2, 2}.Note that A ∩B 6= ∅. So, while a|c, and b|c, ab = 48, which does not divide 24.

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3 Homework #4

1. For each group in the following list: (i) find the order of the group and the order of each element inthe group. (ii) Find all subgroups of the following groups. State the order of each subgroup.

(a) [Dyke, M.] Klein-four group

Klein-fourg |g| H |H|a 1 {a} 1b 2 {a, b} 2c 2 {a, c} 2d 2 {a, d} 2

{a, b, c, d} 4

(b) [Feather, R.] U(10)

U(10)g |g| H |H|1 1 {1} 13 4 {1, 9} 27 4 U(10) 49 2

(c) [Franck, D.] (Z5,+)

Z5

g |g| H |H|0 1 {0} 11 5 Z5 52 53 54 5

(d) [Kittler, S.] Hexaflexagon group

Hexaflexagong |g| H |H|n 1 {n} 1f 2 {n, f} 2↑ 3 {n, ↑, ↓} 3↓ 3 {n, ↑f} 2↑f 2 {n, ↓f} 2↓f 2 {n, f, ↑, ↓, ↑f , ↓f} 6

8

(e) [Ogden, A.] D4

D4

g |g| H |H|R0 1 {R0} 1R90 4 {R180, R0} 2R180 2 {F�, R0} 2R270 4 {F�, R0} 2F| 2 {F|, R0} 2

F− 2 {F−, R0} 2F� 2 {R0, R180, F�, F�} 4F� 2 {R0, R180, F−, F|} 4

{R90, R180, R270, R0} 4D4 8

(f) [Smith, C.] Quaternions

Quaternionsg |g| H |H|1 1 {1} 1−1 2 {1,−1} 2i 4 {1,−1, i,−i} 4−i 4 {1,−1, j,−j} 4j 4 {1,−1, k,−k} 4−j 4 {1,−1, i,−i, j,−j, k,−k} 8k 4−k 4

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4 Homework #5

1. [Berry, K.] List the six elements of GL(2,Z2). Show that the group is not abelian.

The group GL(2,Z2) is the set of 2 × 2 matrices with nonzero determinants with coefficients from

the group Z2 = {0, 1}. Let A =

[a bc d

]. The determinant of A is detA = ad − bc. If ad − bc 6= 0,

then A is nonsingular.

The six elements are

GL(2,Z2) =

{[1 10 1

],

[1 11 0

],

[0 11 1

],

[1 01 1

],

[1 00 1

],

[0 11 0

]}.

The group GL(2,Z2) is not abelian, as it is not commutative. Taking two elements from the groupand multiplying them together in a different order will not have the same result. For example:[

1 10 1

]·[1 11 0

]=

[0 11 0

],

and [1 11 0

]·[1 10 1

]=

[1 01 1

].

Thus GL(2,Z2) is not abelian.

2. [Cochran-Bjerke, L.] Find a cyclic subgroup of order 4 in U(40).

U(40) is all the numbers that are relatively prime to 40. Let’s start with 3 because 1 is the identity.

3 · 3 ≡ 9 (mod 40) = 9

3 · 3 · 3 ≡ 27 (mod 40) = 27

3 · 3 · 3 · 3 ≡ 81 (mod 40) = 1

Since we get back to the identity 1 we can say:

〈3〉 = {1, 3, 9, 27},

and 3 creates a cyclic subgroup of order 4 of U(40).

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3. [Deschamp, B.] Consider G = GL(2,R).

(a) Find C

([1 11 0

]).

Note that we need to know when[a bc d

]·[1 11 0

]=

[1 11 0

]·[a bc d

].

Or [a+ b ac+ d c

]=

[a+ c b+ da b

].

Equating entries yields

a+ b = a+ c a = b+ dc+ d = a c = b.

Combining these results yields b = a− d, and so c = a− d. This information tells us

C

([1 11 0

])=

{[a a− d

a− d d

]: a, d ∈ R and a, d 6= 0

}.

(b) Find C

([0 11 0

]).

Note that we need to know when[e fg h

]·[0 11 0

]=

[0 11 0

]·[e fg h

].

Or [f eh g

]=

[g he f

].

Equating entries yields

f = g e = hh = e g = f.

Combining these results yields f = g and e = h. This information tells us

C

([0 11 0

])=

{[e ff e

]: e, f ∈ R and e2 − f2 6= 0

}.

(c) Find Z(G). Hint: Use (a) and (b) to limit your search by using Problem 6.

We know the center is equal to the intersection of the centralizers, and so any matrix in thecenter must also match the forms of the matrices found in (a) and (b). From (b) we learn thata = d in (a). Also, it must be that a − d = f , but since a is used elsewhere in (a), the onlyoption is if f = 0. Then it must be that the only matrix that satisfies (a) and (b) has the form[

a 00 a

]= aI.

11

Consider an arbitrary matrix M ∈ GL(2,R). Note that

M · (aI) = a(M · I) = a(I ·M) = (aI) ·M.

Thus every matrix in GL(2,R) commutes with aI, and so

Z(G) = {aI : a ∈ R and a 6= 0} .

4. [Gates, S.] Assume that H,K ≤ G, where G is a group. Prove that

H ∩K = {g ∈ G : g ∈ H and g ∈ K}

is a subgroup of G.

By the two-step subgroup test, we must show that

(a)H ∩K is non-empty

(b)∀a, b ∈ H,K, a ◦ b ∈ H ∩K

(c)∀a ∈ H,K, a−1 ∈ H ∩K

(a) By the definition of H ∩K, any element is in H ∩K if and only if that element is in both Hand K. Because H and K are groups, they must, by definition, contain the identity: e ∈ H,K.Therefore e ∈ H ∩K and H ∩K is nonempty.

(b) Because H and K are groups, they are both closed: a ◦ b ∈ H and a ◦ b ∈ K. It follows thata ◦ b ∈ H ∩K.

(c) Because H and K are groups, they must, by definition, contain the inverse of all their elements:∀a ∈ H,K, a−1 ∈ H,K and a−1 ∈ H ∩K.

Because all parts have been proven, H ∩ K satisfies the Two-step subgroup test, and H ∩ K is agroup. Because H and K can only consist of elements of G, H ∩K can only consist of elements ofG, so H ∩K ≤ G : H ≤ G and K ≤ G.

5. [Cosand, K.] Prove that for each a in a group G the centralizer of a, C(a), is a subgroup of G.

Proof: Use the Two-step method. By the definition of a centralizer we know C(a) is a subset of G,o rC(a) ⊆ G. We know the centralizer is nonempty because it contains the identity, e ∈ C(a). Weneed to show: for every b, c ∈ C(a) that (a) b ◦ c ∈ C(a) and (b) b−1 ∈ C(a).

(a) Assume b, c ∈ C(a). We know b◦a = a◦b and c◦a = a◦c. We can now write (b◦c)◦a = b◦(c◦a),by associativity. Then, b ◦ (c ◦ a) = b ◦ (a ◦ c), since c ∈ C(a). Next, b ◦ (a ◦ c) = (b ◦ a) ◦ c,by associativity. Finally, (b ◦ a) ◦ c = (a ◦ b) ◦ c, since b ∈ C(a). We can now conclude thata ◦ (b ◦ c) = (b ◦ c) ◦ a, which means b ◦ c ∈ C(a).

(b) Take b ∈ C(a). We know b ◦ a = a ◦ b. If we perform a left operation with b−1, then b−1 ◦ b ◦ a =b−1 ◦ a ◦ b. This can be first simplified to e ◦ a = b−1 ◦ a ◦ b, and then a = b−1 ◦ a ◦ b. Next, if weperform a right operation with b−1, then a ◦ b−1 = b−1 ◦ a ◦ b ◦ b−1. This can first be simplifiedto a ◦ b−1 = b−1 ◦ a ◦ e, and then a ◦ b−1 = b−1 ◦ a. Therefore, we can conclude that b−1 ∈ C(a).Thus, since b ◦ c ∈ C(a) and b−1 ∈ C(a), by the Two-step test, C(a) ≤ G.

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6. [Gruenig, S.] Let G be a group. Show that

Z(G) =⋂a∈G

C(a).

In order for Z(G) =⋂

a∈GC(a), we must show that (1) Z(G) is a subset of⋂

a∈GC(a), and (2)⋂a∈GC(a) is a subset of Z(G).

(1) Consider b ∈ Z(G). Since b is an element of Z(G), we know g ◦ b = b ◦ g for all g ∈ G. Looking atthe definition of the centralizer, C(a) = {g ∈ G : g◦a = a◦g}, we can see that since b is commutativewith all g ∈ G it belongs to the centralizer of every element in G. In other words, if b is an elementof Z(G), it must also be an element of

⋂a∈GC(a). Therefore, Z(G) ⊆

⋂a∈GC(a).

(2) Now consider c ∈⋂

a∈GC(a). Since c ∈ C(a), we know that c◦a = a◦c. Furthermore, since c is in⋂a∈GC(a), we know c◦a = a◦c is true for all a ∈ G. And since Z(G) = {g ∈ G : g◦a = a◦g,∀a ∈ G},

c must also be an element of Z(G), and⋂

a∈GC(a) ⊆ Z(G).

We’ve now shown that (1) Z(G) ⊆⋂

a∈GC(a) and that (2)⋂

a∈GC(a) ⊆ Z(G). The only way thisis possible is if the two sets are equal. Therefore Z(G) =

⋂a∈GC(a).

7. [Norstedt, Z.] Prove that if an abelian group has more than three elements of order 2, then it has atleast seven elements of order 2. Find an example that shows this is not true for non-abelian groups.

Let G be an abelian group and a, b, c, d ∈ G such that |a|,|b|,|c|,|d|= 2 and a 6= b 6= c 6= d.

Consider a ◦ b = f , a ◦ c = g, and a ◦ d = h. Because b 6= c 6= d, then f 6= g 6= h. (Note that if f = g,then a ◦ b = a ◦ c, and then a = c, which cannot happen.) Next consider f2 = (a ◦ b) ◦ (a ◦ b). SinceG is abelian, f2 = (a ◦ a) ◦ (b ◦ b) = a2 ◦ b2. This is equivalent to e ◦ e = e, due to the fact that |a|= |b| = 2. Thus, f2 = e, and |f | = 2. This can be similarly shown for g2 and h2; only the symbolschange.

Therefore, if an abelian group has more than three elements with order 2, it has at least sevenelements with order 2.

For a non-abelian counter-example, consider D4.

There are exactly five elements in D4, {R180, F|, F−, F�, F�}, that have order 2. This is more thanthree and less than seven.

13

5 Homework #6

1. [Henson, T.] Let

α =

[1 2 3 4 5 6 7 82 3 4 5 1 7 8 6

]and β =

[1 2 3 4 5 6 7 81 3 8 7 6 5 2 4

].

(a) Write α and β as the product of disjoint cycles.

α = (12345)(678), β = (1)(23847)(56)

(b) Compute α ◦ β.

α ◦ β = (12345)(678)(1)(23847)(56) = (12485736)

or α ◦ β =

[1 2 3 4 5 6 7 82 4 6 8 7 1 3 5

](c) Compute β ◦ α.

β ◦ α = (1)(23847)(56)(12345)(678) = (13746285)

or β ◦ α =

[1 2 3 4 5 6 7 83 8 7 6 1 2 4 5

]2. [Hjelmfelt, C.] Write the Cayley table for S3.

ε (1)(23) (2)(13) (3)(12) (123) (132)

ε ε (1)(23) (2)(13) (3)(12) (123) (132)(1)(23) (1)(23) ε (123) (132) (2)(13) (3)(12)(2)(13) (2)(13) (132) ε (123) (3)(12) (1)(23)(3)(12) (3)(12) (123) (132) ε (1)(23) (2)(13)(123) (123) (3)(12) (1)(23) (2)(13) (132) ε(132) (132) (2)(13) (3)(12) (1)(23) ε (123)

14

3. [Quasney, R.] Suppose that a group is generated by two elements a and b (meaning if S = {a, b},then G = 〈S〉). Given that a3 = b2 = e and b ◦ a2 = a ◦ b, construct the Cayley table for the group.You will need to show your work for these computations. Associativity will play a prominent role inthese computations, and your work should reflect its use.

e a b a2 a ◦ b b ◦ ae e a b a2 a ◦ b b ◦ aa a a2 a ◦ b e b ◦ a bb b b ◦ a e a ◦ b a2 aa2 a2 e b ◦ a a b a ◦ ba ◦ b a ◦ b b a b ◦ a e a2

b ◦ a b ◦ a a ◦ b a2 b a e

First, we determine out elements can be e, a, b, a2, a ◦ b, b ◦ a, b ◦ a2, a2 ◦ b. We can remove b ◦ a2 sincewe know it equals a ◦ b. Also, we can remove a2 ◦ b because:

a ◦ b = b ◦ a2

(a2 ◦ a) ◦ b = a2 ◦ (b ◦ a2)a3 ◦ b = a2 ◦ b ◦ a2

e ◦ b = a2 ◦ b ◦ a2

b = a2 ◦ b ◦ a2

b ◦ a = a2 ◦ b ◦ (a2 ◦ a)

b ◦ a = a2 ◦ b ◦ a3

b ◦ a = a2 ◦ b ◦ eb ◦ a = a2 ◦ b

Non-obvious calculations:

(a)a ◦ (a ◦ b) = (a ◦ a) ◦ b

= a2 ◦ b= b ◦ a

(b)a ◦ (b ◦ a) = a ◦ (a2 ◦ b)

= (a ◦ a2) ◦ b= a3 ◦ b= e ◦ b= b

15

(c)b ◦ (a ◦ b) = b ◦ (b ◦ a2)

= (b ◦ b) ◦ a2

= b2 ◦ a2

= e ◦ a2

= a2

(d)b ◦ (b ◦ a) = (b ◦ b) ◦ a

= b2 ◦ a= e ◦ a= a

(e)a2 ◦ a2 = a ◦ a ◦ a ◦ a

= (a ◦ a ◦ a) ◦ a= a3 ◦ a= e ◦ a= a

(f)a2 ◦ (a ◦ b) = (a2 ◦ a) ◦ b

= a3 ◦ b= e ◦ b= b

(g)a2 ◦ (b ◦ a) = (a2 ◦ b) ◦ a

= (b ◦ a) ◦ a= b ◦ (a ◦ a)

= b ◦ a2

= a ◦ b

(h)(a ◦ b) ◦ a = (b ◦ a2) ◦ a

= b ◦ (a2 ◦ a)

= b ◦ a3

= b ◦ e= b

(i)(a ◦ b) ◦ b = a ◦ (b ◦ b)

= a ◦ b2

= a ◦ e= a

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(j)(a ◦ b) ◦ (a ◦ b) = (a ◦ b) ◦ (b ◦ a2)

= a ◦ (b ◦ b) ◦ a2

= a ◦ b2 ◦ a2

= a ◦ e ◦ a2

= a ◦ a2

= a3

= e

(k)(a ◦ b) ◦ (b ◦ a) = a ◦ (b ◦ b) ◦ a

= a ◦ b2 ◦ a= a ◦ e ◦ a= a ◦ a= a2

(l)(b ◦ a) ◦ a = b ◦ (a ◦ a)

= b ◦ a2

= a ◦ b

(m)(b ◦ a) ◦ b = (a2 ◦ b) ◦ b

= a2(b ◦ b)= a2 ◦ b2

= a2 ◦ e= a2

(n)(b ◦ a) ◦ a2 = b ◦ (a ◦ a2)

= b ◦ a3

= b ◦ e= b

(o)(b ◦ a) ◦ (a ◦ b) = b ◦ (a ◦ a) ◦ b

= (b ◦ a2) ◦ b= (a ◦ b) ◦ b= a ◦ (b ◦ b)= a ◦ b2

= a ◦ e= a

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(p)(b ◦ a) ◦ (b ◦ a) = (a2 ◦ b) ◦ (b ◦ a)

= a2 ◦ (b ◦ b) ◦ a= a2 ◦ b2 ◦ a= a2 ◦ e ◦ a= a2 ◦ a= a3

= e

4. [Cosand, K.] Find the order of the following permutations.

(a) (14)

(14) ◦ (14) = (1)(4)

|(14)| = 2

(b) (147)

(147) ◦ (147) = (174)

(147) ◦ (147) ◦ (147) = (1)(4)(7)

|(147)| = 3

(c) (14762)

(14762) ◦ (14762) = (17246)

(14762) ◦ (14762) ◦ (14762) = (16427)

(14762) ◦ (14762) ◦ (14762) ◦ (14762) = (12674)

(14762) ◦ (14762) ◦ (14762) ◦ (14762) ◦ (14762) = (1)(4)(7)(6)(2)

|(14762)| = 5

(d) Use (a), (b), and (c) to develop and prove a lemma concerning the order of α = (a1a2 · · · ak).Hint: Use the fact that α(ai) = ai+1, for 1 ≤ i < k, and α(ak) = a1.

Lemma: The order of a cycle α = (a1a2 · · · ak) is k, or the length of the cycle.

Proof: Consider the element ai. Then, α = (a1a2 · · · ai · · · ak). To permute from ai to ak wecan write αk−i(ai) = ak, and we know we will never cycle back to ai. To permute from akback to a1 we know this takes one permutation so α1(ak) = a1, and we do not ever get toai. Finally, to permute from a1 to ai, we know the number of permutations necessary is i − 1so αi−1(a1) = ai. We can apply all of these ideas to each ai. Putting all of this together,αk(ai) = αi−1(α(αk−i(a1))) = αi−1(α(ak)) = αi−1(a1) = ai. Therefore, the order of a cycleα = (a1a2 · · · ak) is k.

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5. [Keene, L.] Prove the following statements concerning order in groups.

(a) Prove that in a group an element and its inverse have the same order.

Proof: To prove that in a given group, an element and its inverse have the same order, let G bea group with the element a ∈ G. Let

|a| = n

and so an = e.

By applying a−1 to each side of the equation n times (as follows)

n times︷ ︸︸ ︷a ◦ a ◦ a · · · a = e

n−1 times︷ ︸︸ ︷a ◦ a ◦ a · · · a ◦ a ◦a−1 = e ◦ a−1

n−1 times︷ ︸︸ ︷a ◦ a ◦ a · · · a ◦ a ◦e = a−1

n−1 times︷ ︸︸ ︷a ◦ a ◦ a · · · a ◦a−1 = a−1 ◦ a−1

n−2 times︷ ︸︸ ︷a ◦ a ◦ a · · · a ◦e = a−1 ◦ a−1

...

a1 ◦ a−11 =

n times︷ ︸︸ ︷a−1 ◦ a−1 · · · a−1 · · · a−1

e =

n times︷ ︸︸ ︷a−1 ◦ a−1 · · · a−1 · · · a−1

e = (a−1)n

|a−1| = n.

Note that n is the smallest integer for which this is true. Thus we can determine that

|a| = |a−1|.

By this logic, we have determined that, in a given group, an element and its inverse have thesame order.

(b) Prove that in any group G, for a, b ∈ G, |a ◦ b| = |b ◦ a|.Proof: To prove that in any group G for a, b ∈ G, that |a ◦ b| = |b ◦ a|. Let

|a ◦ b| = n

n times︷ ︸︸ ︷(a ◦ b) ◦ (a ◦ b) ◦ · · · ◦ (a ◦ b) ◦ (a ◦ b) = e

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By the definition of a group, we know that operations are associative. Because of this fact, wecan rearrange the equation to the following.

n times︷ ︸︸ ︷(a ◦ b) ◦ (a ◦ b) ◦ · · · ◦ (a ◦ b) ◦ (a ◦ b) = e

a ◦n−1 times︷ ︸︸ ︷

(b ◦ a) ◦ (b · · · a) ◦ (b ◦ a) ◦b = e

a−1 ◦ a ◦ (b ◦ a)n−1 ◦ b = a−1 ◦ ee ◦ (b ◦ a)n−1 ◦ b = a−1 ◦ e(b ◦ a)n−1 ◦ b ◦ a = a−1 ◦ a

(b ◦ a)n−1 ◦ (b ◦ a) = e

Then(a ◦ b)n = e,

and so|a ◦ b| = n.

By this logic, we have determined that

|a ◦ b| = n = |b ◦ a|.

We now know that the order of a ◦ b is equal to the order of b ◦ a.

(c) If a, b, and c are elements of a group, give an example to show that it need not be the case that|a ◦ b ◦ c| = |c ◦ b ◦ a|. Hint: look in D4.

In D4, let a, b, and c be assigned as follows:

a = R90, b = F\, c = F|.

|a ◦ b ◦ c| = R90 ◦ F\ ◦ F| = R0, and |R0| = 1.

|c ◦ b ◦ a| = F| ◦ F\ ◦R90 = R180, and |R180| = 2.

We have therefore proven that |a ◦ b ◦ c| does not necessarily equal |c ◦ a ◦ b| in a given group G.

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6 Homework #7

1. [Bachwich, A.] Determine whether the following permutations are even or odd. Show your work.

(a) (135)

Solution: (135) = (31)(35). Since (135) can be expressed as the product of two transpositions,it is even.

(b) (1356)

Solution: (1356) = (53)(51)(56). Since (1356) can be expressed as the product of three trans-positions, it is odd.

(c) (13567)

Solution: (13567) = (65)(63)(61)(67). Since (1356) can be expressed as the product of fourtranspositions, it is even.

(d) (12)(134)(152)

Solution: (12)(134)(152) = (234)(51) = (15)(32)(34). Since (12)(134)(152) can be expressed asthe product of three transpositions, it is odd.

(e) (1243)(3521)

Solution: (1243)(3521) = (1)(2)(354) = (12)(21)(53)(54). Since (1243)(3521) can be expressedas the product of four transpositions, it is even.

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2. [Carroll, S.] In S4, find a cyclic subgroup of order 4 and a noncyclic subgroup of order 4.

Cyclic subgroup: Let α = (1234). Then α2 = (13)(24), α3 = (1432) and α4 = (1)(2)(3)(4), since asingle element in this set can generate all the other elements in the set, it is cyclic. To prove it’s asubgroup make a Cayley Table and check for closure.

Cayley Table for Cyclic Set:

(1)(2)(3)(4) (1234) (13)(24) (1432)

(1)(2)(3)(4) (1)(2)(3)(4) (1234) (13)(24) (1432)(1234) (1234) (13)(24) (1432) (1)(2)(3)(4)

(13)(24) (13)(24) (1432) (1)(2)(3)(4) (1234)(1432) (1432) (1)(2)(3)(4) (1234) (13)(24)

We have closure, so this is a cyclic subgroup of order 4.

Noncyclic subgroup: (1)(2)(3)(4), (12)(34), (13)(24), and (14)(23).

If α = (12)(34) then α2 = (1)(2)(3)(4)

If β = (13)(24) then β2 = (1)(2)(3)(4).

If γ = (14)(23) then γ2 = (1)(2)(3)(4).

Therefore no element can operate on itself to generate the other elements in the set. To prove it’s asubgroup, make a Cayley Table and check for closure.

Cayley Table for Noncyclic Set:

(1)(2)(3)(4) (12)(34) (13)(24) (14)(23)

(1)(2)(3)(4) (1)(2)(3)(4) (12)(34) (13)(24) (14)(23)(12)(34) (12)(34) (1)(2)(3)(4) (14)(23) (13)(24)(13)(24) (13)(24) (14)(23) (1)(2)(3)(4) (12)(34)(14)(23) (14)(23) (13)(24) (12)(34) (1)(2)(3)(4)

We have closure, so this is a noncyclic subgroup of order 4. Note that since each element is its owninverse all other elements in the subgroup are required to complete this subgroup of order 4, any oneelement operating on itself will only yield the identity and itself therefore it is noncyclic.

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3. [Cosand, K.] Let n be a positive integer. If n is odd, is an n-cycle an odd or an even permutation?If n is even, is an n-cycle an odd or an even permutation? Prove your result.

Proof: Let α = (a1a2 · · · an−1an). We know α can be written as the product of transpositions, α =(a1an)(a1an−1) · · · (a1a3)(a1a2). Since we have n symbols, we know we must have n−1 transpositionsbecause a1 pairs exactly once with each of the other symbols. Therefore, if n is odd, then n − 1 iseven, and we can conclude the n-cycle is an even permutation. Similarly, if n is even, then n − 1 isodd, and we can conclude the n-cycle is an odd permutation.

4. [Cosand, K.] Develop and prove a lemma that computes (a1a2 · · · ak)−1.

Lemma: (a1a2 · · · ak)−1 = (akak−1 · · · a2a1).Proof: Let α1 = (a1a2 · · · ak), and α2 = (akak−1 · · · a2a1). We need to show α1 ◦ α2 = ε.

1© Take α1 and α2 and decompose them into transpositions.

2© Take α1 ◦ α2 and determine if they equal ε.

Showing 1© :α1 = (a1a2 · · · ak) = (a1ak)(a1ak−1) · · · (a1a3)(a1a2)

andα2 = (akak−1 · · · a2a1) = (a1akak−1 · · · a3a2) = (a1a2)(a1a3) · · · (a1ak−1)(a1ak).

Now, to show 2©:

α1 ◦ α2 = (a1ak)(a1ak−1) · · · (a1a3)(a1a2)(a1a2)(a1a3) · · · (a1ak−1)(a1ak).

We know (aiaj)(aiaj) = ε. Therefore, in the middle of the composition, we can see the last transpo-sition for α1 is (a1a2) and the first transposition for α2 is (a1a2). Thus, we know (a1a2)(a1a2) = ε.The next two transpositions next to each other would be (a1a3) and (a1a3), which also reduces toε. By continuing this process until we reduce each pair of transpositions to ε, we eventually seeα1 ◦ α2 = ε. We know inverses are unique and we found one, so this must be the inverse. Therefore,we can conclude (a1a2 · · · ak)−1 = (akak−1 · · · a2a1).

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5. [Gates, S.] Prove that An is a subgroup of Sn.

By the two-step subgroup test, we must show that

An ⊆ Sn 1©An is non-empty 2©

∀α, β ∈ An, α ◦ β ∈ An 3©

∀α ∈ An, α−1 ∈ An 4©

By the definition of An, any element is in An if and only if that element is in Sn and is even, so weknow An ⊆ Sn, proving 1©.

The identity, ε, was proven to be even. This means ε ∈ An and An is non-empty, so 2© has beenproven.

Consider α = (a1a2 · · · ak) and β = (b1b2 · · · bj). Because α, β ∈ An, α and β are even, meaning theycan be decomposed to an even number of transpositions:

α = (a1ak)(a1ak−1) · · · (a1a2),

β = (b1bj)(b1bj−1) · · · (b1b2).

Following this, we have:

α ◦ β = (a1ak)(a1ak−1) · · · (a1a2)(b1bj)(b1bj−1) · · · (b1b2).

The product has the number of transpositions from α plus the number of transpositions from β.Because both of these numbers are even, the sum of these two numbers is even. Therefore α ◦ β iseven, and α ◦ β ∈ An, proving 3©.

First, write α as the product of disjoint cycles

α = α1α2 · · ·αk,

where each αi is a cycle. Second, consider, from homework 7 problem 4, that the inverse of a cycleuses the same number of transpositions as the original cycle. This means α−1 must have the samenumber of transpositions as α. Thus, if α had can be written as an even number of transpositions,then α−1 can also be written as an even number of transpositions. Therefore, α−1 ∈ An, and 4© isproven.

Because 1©, 2©, 3©, and 4© have been proven, An satisfies the Two-step subgroup test, and An ≤ Sn.

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