Israel N Herstein - Solutions

8/20/2019 Israel N Herstein - Solutions

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Jonathan Bergknoff

Herstein Solutions

Chapters 1 and 2

Throughout, G is a group and p is a prime integer unless otherwise stated. “ A

≤ B” denotes that A is a

subgroup of B while “A B ” denotes that A is a normal subgroup of B .

H 1.3.14* (Fermat’s Little Theorem) – Prove that if a ∈ Z then a p ≡ a mod p.Proceed by induction on (positive) integer a. The theorem holds for a = 1 because 1 p = 1 ≡ 1 mod p.Suppose that k p ≡ k mod p and then compute, by the binomial theorem,

(k + 1) p = p

i=0

pi

ki 1 p− i = k p + 1 + p(. . .) ≡ k + 1 mod p.

In the last step, we used the induction hypothesis. This proves the result for positive integer a. The expansion

(k + 1) p = k p + 1 + p(. . .) is justied by lemma 2 (below) which states that p | pn for n ∈ {1, . . . , p −1}.The case a = 0 is trivial: 0 p = 0 ≡ 0 mod p. Let a ∈ Z+ as before. As was just proven, there exists c ∈ Zsuch that a p −a = cp. Now, if p > 2 (odd), we have ( −a) p −(−a) = −a p + a = (−c) p so (−a) p ≡ −a mod p.On the other hand, if p = 2, we just need to see that “modulo 2” picks out even/odd parity. Regardless of

being positive or negative, an even integer squared is even and an odd integer squared is odd. Therefore

a2 ≡ a mod 2 and the theorem is proven in its entirety.Lemma 1 – Let n ∈ Z+ and r ∈ {0, . . . , n }. The binomial coefficient nr = n !r !(n − r )! is an integer.

Proof: Proceed by induction on n . For n = 1,10 = 1 and

11 = 1 are both integers, as claimed. Suppose

that n − 1r is an integer for all r ∈ {0, . . . , n −2}. Nown −1

r+

n −1r −1

= (n −1)!

r !(n −r −1)! +

(n −1)!(r −1)!(n −r )!

= ( n −1)!(n −r ) + rr !(n −r )!

=nr

.

Hence, when the above computation goes through, nr is a sum of integers and thus is, itself, an integer.

However, the computation fails from the outset for r = 0 and r = n (we’re interested in things like ( r −1)!and ( n −r −1)!), so those cases must be considered independently. We have

n0

= n!0!n!

= 1nn

= n!n!0!

= 1

and the claim is proven.

Lemma 2 – Let n ∈ {1, . . . , p −1}. Then p | pn .Proof: Intuitively, this lemma is true because the numerator has a factor of p and the denominator has no

factors that cancel it (relying crucially on the primality of p). By the fundamental theorem of arithmetic ( Z

is a UFD), we can write the denominator as n!( p −n)! = q a ii with q i the unique ascending list of prime


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divisors of n!( p−n)! and ai their respective powers. As every factor of n(n −1) · · ·2·1·( p−n)( p−n−1) · · ·2·1is smaller than p (this fails if n = 0 or n = p), p divides none of them and hence, as a prime, does not divide

their product. Then none of the q i is p. The factor of p in the numerator is then preserved upon taking the

quotient, and p | pn . Note that it makes no sense to talk about p dividing pn unless pn ∈ Z (lemma 1).

H 1.3.15 – Let m,n,a,b ∈ Z with (m, n ) = 1 . Prove there exists x with x ≡ a mod m and x ≡ b mod n .As m and n are coprime, there exist integers c, d such that cm + dn = 1. We see that ( ac)m + ( ad)n = a,

so that adn = a − acm ≡ a mod m. Similarly, ( bc)m + ( bd)n = b, so bcm = b − bdn ≡ b mod n . Setx = adn + bcm. Then we have

x ≡ adn mod m ≡ a mod m x ≡ bcm mod n ≡ b mod n.

H 2.3.3 – Let (ab)2 = a2b2 for all a, b ∈ G. Prove that G is abelian.The statement is that, for all a, b ∈ G, we have abab = a2b2 . Multiply both sides of the equation on the leftby a− 1 and on the right by b− 1 . Then we have ba = ab and hence G is abelian.

H 2.3.4* – Let G be such that, for three consecutive integers i, (ab)i = a i bi for all a, b ∈ G. Prove that Gis abelian.

Let N be the smallest of the three consecutive integers. Then we have that ( ab)N

= aN

bN

, (ab)N +1

=aN +1 bN +1 and ( ab)N +2 = aN +2 bN +2 for all a, b ∈ G. Inverting the rst equation and right multiplying it tothe second equation implies

ab = aN +1 bN +1 b− N a− N = aN +1 ba− N hence baN = aN b.

Inverting the second equation and right multiplying it to the third equation gives

ab = aN +2 bN +2 b− (N +1) a− (N +1) = aN +2 ba− (N +1) hence baN +1 = aN +1 b.

Therefore aN +1 b = baN +1 = baN a = aN ba and left multiplying by a− N yields ab = ba for arbitrary a, b ∈ G.Hence G is abelian.

H 2.3.8 – Let G be nite. Prove the existence of an N ∈ Z such that aN = e for all a ∈ G.Let a ∈ G. As G is nite and closed under multiplication, the set {a0 , a 1 , a 2 , . . .} is nite. Hence there existm, n ∈ {0, 1, . . .} such that am = an (without loss of generality, take m > n ). By the division algorithm, we


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can write m = kn + r with k, r ∈ Z and 0 ≤ r < n . Now we have the statement akn + r = an or a(k− 1) n + r = e.Enumerate G as {a1 , . . . , a n }. For each ai , there exists an N i = ( ki − 1)n i + r i , computed by the abovemethod, such that aN ii = e. Let N = N i . Then aN i = ( a

N ii )N/N i = eN/N i = e for each a i ∈ G.

H 2.3.9a – Let G have three elements. Prove that G is abelian.

Let G = {e,a,b}. In order to show that G is abelian, we only need to show that ab = ba because the identitycommutes with everything. Suppose ab = ba. We have only three choices: ab = e, ab = a or ab = b.(1) If ab = e, then a = b− 1 so ba = e = ab, which is a contradiction. Hence ab = e.(2) If ab = a, then b = e so ba = a = ab, which is a contradiction. Hence ab = a.(3) If ab = b, then a = e so ba = b = ab, which is a contradiction. Hence ab = b.Therefore ab = ba necessarily, and hence G is abelian.

H 2.3.10 – Let G be such that every element is its own inverse. Prove that G is abelian.

Let a, b ∈ G. Then ab = a− 1b− 1 = ( ba)− 1 = ba, so G is abelian.

H 2.3.11 – Let G have even order. Prove there exists a non-identity element a ∈ G with a2 = e.

If an element a of a group doesn’t satisfy a2

= e, then there exists a unique inverse element a− 1

= a in thegroup. Elements of this type can be counted in pairs {a, a − 1}. There are therefore an even number 2 k of elements with a2 = e. The identity satises e2 = e, so there are |G| −2k −1 ≥ 0 non-identity elements asatisfying a2 = e. If |G| is even, then |G| −2k −1 is non-zero and hence there exists a non-identity elementa ∈ G with a2 = e.

H 2.3.12 – Let G be a non-empty set closed under an associative product with an e ∈ G such that ae = a for all a ∈ G as well as the property that, for each a ∈ G, there exists y(a) ∈ G with ay(a) = e. Prove that G is a group under this product.

In order for this set to be a group, right inverses must also be left inverses and it must hold that ea = a for all

a ∈ G. Multiply the equation ay(a) = e by y(y(a)), the right inverse of y(a), on the right: by associativity,y(y(a)) = [ ay(a)]y(y(a)) = a[y(a)y(y(a))] = a.

Hence y(y(a)) = a is the right inverse of y(a). Now we see that, in addition to ay(a) = e, we have y(a)a = e


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so that y(a) is the inverse (both left and right) of a. Now we can trivially show the property ea = a. Multiply

the equation ae = a on the left by y(a) and on the right by a (notice that y(a)aa = [y(a)a]a = a):

a = y(a)[ae]a = [y(a)a]ea = ea.

Therefore G is a group.

H 2.5.1 – Let H and K be subgroups of the group G. Prove that H ∩K ≤ G.H ∩ K is closed: let a, b ∈ H ∩ K . Then a, b ∈ H, K so ab ∈ H, K because both are subgroups. Henceab ∈ H ∩K . H ∩K is closed under inverses: let a ∈ H ∩K . Then a− 1∈ H, K , and hence a− 1∈ H ∩K . Bylemma 2.3, H ∩K ≤ G.

H 2.5.2 – Let G have a subgroup H . Dene a left coset of H by aH = {ah | h ∈ H }. Show there is a bijection between left and right cosets of H in G.

Dene f : {aH | a ∈ G} → {Ha | a ∈ G} by f (aH ) = H a − 1 . This map is well-dened: suppose a1 , a 2 ∈ Gare such that a1H = a2H . Then there exists h ∈ H such that a1 = a2h and hence f (a1H ) = Ha − 11 =Hh − 1a− 12 = Ha

− 12 = f (a2H ). The map is trivially surjective: for any a ∈ G, H a is the image of a− 1H . The

map is injective: suppose a1H, a 2H are such that f (a1H ) = f (a2H ). Then H a − 11 = Ha− 12 which implies the

existence of h ∈ H such that a− 11 = ha − 12 . Inverting, we nd that a1 = a2h− 1 , i.e. that a1H = a2H whichproves injectivity.

H 2.5.3 – Let G have no proper subgroups. Prove that |G| is prime.Let g ∈ G with g = e. g is a subgroup of G. Because g = e, g is not the trivial subgroup {e}. Hence,because G has no proper subgroups, it must be that g = G which gives |g| = |G|. Suppose |G| = mn withm, n > 1 (so m,n < |G|). Then ( gm )n = e which implies that |gm | = n < |G|. The subgroup gm is proper,with 1 < n < |G| elements, which is a contradiction. Therefore |G| must be prime.

H 2.5.6* – Let H, K ≤ G have nite indices in G. Give an upper bound for the index of H ∩K .Missing.

H 2.5.7 – With a, b ∈ R, let τ ab : R → R be given by τ ab = ax + b. Let G = {τ ab | a = 0}. Prove that G is a group under composition. What is τ ab ◦τ cd ?


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(τ ab ◦τ cd )(x) = a(cx + d) + b = acx + ad + b = τ ac,ad + b(x). As R is a eld and a, c = 0, τ ac,ad + b ∈ G so thatG is closed under composition. The operation is associative:

(τ ab ◦τ cd ) ◦τ ef = τ ac,ad + b ◦τ ef = τ ace,acf + ad + b = τ ab ◦τ ce,cf + d = τ ab ◦(τ cd ◦τ ef ).

The set is closed under inverses: τ − 1ab = τ 1a ,− ba ∈ G. Finally, there is an identity element e = τ

1,0 . ThereforeG is a group. Note that G is non-abelian because τ ab ◦τ cd = τ ac,ad + b = τ ac,cb + d = τ cd ◦τ ab .

H 2.5.8 – Taking the group of 2.5.7, let H = {τ ab ∈ G | a ∈ Q}. Prove that H ≤ G and list the right cosets of H in G.

Because Q is a eld, we have that, for σab , σcd ∈ H , σab ◦σcd = σac,ad + b ∈ H and σ− 1ab = σ 1a ,− ba ∈ H . Bylemma 2.3, H ≤ G.

Now let τ ab , τ cd ∈ G be such that Hτ ab = Hτ cd . Then τ ab ◦τ − 1cd = τ

ac ,b−

adc

∈ H . Therefore a

c ∈ Q is thecondition on this situation. In particular, τ ab and τ ac are in the same right coset regardless of b and c, whichtells us that the right cosets are indexed by just one real parameter.

{Hτ ab } = {{τ cd ∈ G | c = qa for some q ∈ Q\{0}} | a ∈ R\{0}}.

H 2.5.9 – (a) In the context of 2.5.8, prove that every left coset of H in G is also a right coset of H in G.

(b) Give an example of a group G and a subgroup H of G such that the above is not true.

(a) Let τ ab , τ cd ∈ G be such that τ ab H = τ cd H . Then τ − 1ab ◦τ cd = τ ca , d − ba ∈ H . Again, this is the statementthat ca ∈ Q, so two elements of G are equivalent left-modulo H if the ratio of their rst parameters is rational.Therefore consider the cosets τ ab H and Hτ ab . If τ cd ∈ τ ab H then ca ∈ Q. Hence ac ∈ Q which gives thatτ cd ∈ Hτ ab . Therefore τ ab H ⊂ Hτ ab . The reverse inclusion is identical, so τ ab H = Hτ ab .(b) Consider G = S 3 = {e, (12) , (13) , (23) , (123), (213)} and the subgroup H = {e, (12)}. The left cosets are:

eH = {e, (12)} (213)H = {(213), (213)(12) = (23) } (123)H = {(123), (123)(12) = (13) }.

On the other hand, the right cosets are:

He = {e, (12)} H (213) = {(213), (12)(213) = (13) } H (123) = {(123), (12)(123) = (23) }.For this choice of G and H , there exist left cosets that are not right cosets and vice versa.

H 2.5.11 – Let G have subgroups of orders n and m. Prove that G has a subgroup of order lcm(m, n ).


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Missing.

H 2.5.12 – Let a ∈ G. Prove that N (a) = {g ∈ G | ga = ag} is a subgroup of G. ( N (a) is the “normalizer of a in G”)

Let g, h ∈ N (a). Then ( gh)a = gha = gah = agh = a(gh), so gh ∈ N (a). Additionally, multiply theequation ga = ag on the left and right by g− 1 to nd that g− 1gag − 1 = g− 1agg − 1 , so ag− 1 = g− 1a. Hence

g− 1∈ N (a). By lemma 2.3, N (a) ≤ G.

H 2.5.13 – Prove that Z (G) = {g ∈ G | gx = xg for all x ∈ G} ≤ G. ( Z is the “center of G”)The proof is identical to that of 2.5.12. Alternately, notice that Z (G) = a∈G N (a) ≤ G.

H 2.5.14 – Let G = g be cyclic and let H ≤ G. Prove that H is cyclic.If H is trivial, the result holds. Therefore let H be non-trivial. Then there exists a smallest positive n0 such

that gn 0 ∈ H (remember H is closed under inverses, so it can’t have just negative powers of g). The claim is

that H = gn 0 . It is clear, because H is a group containing gn 0 , that gn 0 ≤ H . Let gk ∈ H be arbitrary.By the division algorithm, we can write k = qn0 + r with q ∈ Z and r ∈ {0, 1, . . . , n 0 −1}. Now, by closure,gk g− qn 0 = gr ∈ H . By our assumption that n0 is the smallest positive exponent in H , we must conclude

that r = 0. Therefore n 0 | k and we see that every element of H is a power of gn 0 , whence H ≤ gn 0 . Thisproves the claim, so H = g

n 0 is cyclic and the result is shown.

H 2.5.15 – Let G be cyclic with |G| = n. How many generators does G have? Let g be a generator of G, i.e. G = g . The claim is that gm is also a generator if and only if m isrelatively prime to n. If m is relatively prime to n, then there exist a, b ∈ Z such that am + bn = 1. Nowg = gam + bn = ( gm )a (gn )b = ( gm )a which tells us that g ∈ gm so G = g ≤ gm ≤ G. Hence m, nrelatively prime implies that gm generates G. On the other hand, if G = gm , then g ∈ gm , so that forsome a ∈ Z we have (gm )a = g. This is impossible if am + bn > 1 for all a, b ∈ Z, as would be the case if m and n were not relatively prime. Hence G = gm implies that m is coprime to n . This proves the claim,and therefore the number of generators of G is φ(n), with φ the Euler totient function.

H 2.5.16 – Let a ∈ G. If am = e, prove that |a| divides m.By the division algorithm, we may write m = q |a| + r with q ∈ Z and r ∈ {0, 1, . . . , |a| − 1}. Now


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e = am = aq| a | + r = ( a | a | )qa r = a r . As r < |a| and |a| is the minimal exponent taking a to the identity, wemust conclude that r = 0 and hence |a| | m.

H 2.5.17 – Let a, b

∈ G be such that a5 = e and aba− 1 = b2 . What is

|b

|?

We have that a2ba− 2 = a(aba− 1)a− 1 = ab2a− 1 . It follows by induction that an ba− n = b2n

: suppose this is

true for n and compute

an +1 ba− (n +1) = a(an ba− n )a− 1 = ab2n

a− 1 = b2n +1

.

The last equality follows from raising the condition aba− 1 = b2 to powers: ( aba− 1)k = abk a− 1 = b2k .

Therefore a5ba− 5 = b32 , but, because a5 = a− 5 = e, the left hand side is simply b. We have, nally, b = b32 ,

or b31 = e. By 2.5.16, the order of b divides 31, so |b| = 1 or |b| = 31.

H 2.5.18* – Let G be nite, abelian and such that xn = e has at most n solutions for every n ∈ Z+ . Prove that G is cyclic.

Missing.

H 2.6.1* – Let H ≤ G be such that the product (Ha )(Hb) is again a right coset of H for a, b ∈ G. Prove that H G.

Consider the product ( Ha )(Ha − 1) for arbitrary a ∈ G. As sets, it is true that ( Ha )(Ha − 1) = H (aHa − 1):(h1a)(h2a− 1) ∈ (Ha )(Ha − 1) is h1(ah 2a− 1) ∈ H (aHa − 1) so (Ha )(Ha − 1) ⊂ H (aHa − 1); conversely,h1(ah 2a− 1) ∈ H (aHa − 1) is (h1a)(h2a− 1) ∈ (Ha )(Ha − 1) so H (aHa − 1) ⊂ (Ha )(Ha − 1). Then, by the con-dition of the problem, ( Ha )(Ha − 1) = H (aHa − 1) is a right coset of H . The set aH a − 1 contains e = aea − 1 ,

so in fact H (aHa − 1) = He = H which implies that aH a − 1⊂ H , i.e. that H G.

H 2.6.2 – Let H ≤ G have index 2. Prove that H G.Let g ∈ G\H . The right cosets of H in G may be enumerated as {H,Hg }. Because distinct cosets aredisjoint, H g is exactly the set G

\H of elements in G not belonging to H . It is trivial that H = He = eH is

a left coset in addition to being a right coset. Furthermore, gH is again G\H (The only alternative wouldbe gH = H , but that isn’t the case: ge = g ∈ H while ge ∈ gH .), so that gH = Hg. By lemma 2.10, H Gbecause every one of its right cosets in G is also a left coset in G.

H 2.6.3 – Let H ≤ G and N G. Prove that NH ≤ G.


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Let n1 , n 2 ∈ N and h1 , h2 ∈ H . Compute ( n1h1)(n2h2) = n1h1n2h− 11 h1h2 = n1(h1n2h− 11 )h1h2 . Because N is normal, there exists n3 ∈ N such that h1n2h− 11 = n3 . Then we have ( n1h1)(n2h2) = ( n1n3)(h1h2) ∈ N H so N H is closed under the group product. Furthermore, N H is closed under inverses: Let n ∈ N and h ∈ H .We have ( nh )− 1 = h− 1n− 1 = h− 1n− 1hh − 1 = n h− 1 ∈ N H , again using the existence of n ∈ N such that

n = h− 1

n− 1

h. Therefore, by lemma 2.3, N H is a group.

H 2.6.4 – Let M, N G. Prove that M ∩N G.Let x ∈ M ∩N (so x ∈ M , x ∈ N ) and g ∈ G. Because M and N are normal, we see that gxg− 1∈ M andgxg− 1∈ N so that gxg

− 1∈ M ∩N . Therefore g(M ∩N )g− 1⊂ M ∩N and hence M ∩N G.

H 2.6.5 – Let H ≤ G and N G. Prove that H ∩N H .Let x ∈ H ∩N and h ∈ H . We have hxh − 1 ∈ H because it is a product of three elements of H . We alsohave that hxh − 1∈ N because N is normal and x ∈ N . Therefore hxh − 1∈ H ∩N so H ∩N H .

H 2.6.6 – Let G be abelian. Prove that every subgroup of G is normal.

Let H ≤ G and let h ∈ H , g ∈ G. Then ghg− 1 = gg− 1h = h ∈ H , so H G.

H 2.6.7* – If every subgroup of a group G is normal, is G necessarily abelian?

No. Consider the order 8 quaternion group Q8 = {±1, ±i ± j ± k} with (−1)2 = 1, −1 commuting witheverything, and i2 = j 2 = k2 = ijk = −1. First observe that Q8 is non-abelian: ij = ( ij )(−k2) = ( −ijk )k =k while j i = j − 1i− 1 = ( ij )− 1 = k− 1 = −k.We can enumerate all the subgroups of Q8 by considering subgroups generated by combinations of elements.

First of all, we have the trivial subgroups 1 = {1} and i , j ,k = Q8 of orders 1 and 8 respectively. Both aretrivially normal. Next, the subgroup −1 = {−1, 1} is of order 2. Finally, the subgroups i , j , k of order4 round out the list. We can see that to be the case by noting that −1 is redundant as a generator if we includeany of

{i,j ,k

} because each already squares to

−1. Furthermore, any subgroup generated by 2 or more of

{i,j ,k } is all of Q8 : for instance, i, j = {1 = i4 , −1 = i2 , i, −i = i3 , j, − j = j 3 , k = ij, −k = ij 3} = Q8 .It is easy to see that −1 is normal because its elements commute with everything: i −1 = {i, −i} = −1 i,and so forth (in fact, it’s the center of Q8 . See 2.5.13). The rest of the subgroups are of order 4, so of index

2. By 2.6.2, i , j , k are all normal. Therefore all subgroups of Q8 are normal, but, as displayed above,Q8 is non-abelian.


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H 2.6.8 – Let H ≤ G. For g ∈ G, prove that gHg− 1 ≤ G.Let h1 , h2 ∈ H . We have gh1g− 1gh2g− 1 = gh1h2g− 1 ∈ gHg− 1 because h1h2 ∈ H . Furthermore,(gh1g− 1)− 1 = gh− 11 g− 1∈ gHg

− 1 because h− 1∈ H . By lemma 2.3, gHg− 1 ≤ G.

H 2.6.9 – Let G be nite and let H ≤ G be the only subgroup of G of order |H |. Prove that H G.The map f : H → gHg− 1 given by f (h) = ghg− 1 is a bijection. Injectivity: f (h1) = f (h2) impliesgh1g− 1 = gh2g− 1 so h1 = h2 . Surjectivity: any ghg− 1 ∈ gHg

− 1 is the image of h under f . Therefore

|gHg− 1| = |H |. Now conjugation brings H into another subgroup, according to 2.6.8, and that subgrouphas the same cardinality as H , by assumption. If H is the only subgroup of order |H |, then conjugation xesH , i.e. gHg− 1 = H for all g ∈ G. This is the condition that H G. The restriction to nite G is likely justto allow Herstein’s use of the word “order”.

H 2.6.11 – Let M, N G. Prove that NM G.

By 2.6.3, N M is a subgroup. Let g ∈ G and compute gNMg− 1 = gN g− 1gM g− 1 = N M . Hence N M G.

H 2.6.12* – Let M, N G be such that M ∩N = {e}. Prove that mn = nm for all m ∈ M and n ∈ N .Consider the commutator [ n, m ] = nmn − 1m− 1 . As M G and n ∈ G, there exists m ∈ M such thatm = nmn − 1 . Hence nmn − 1m− 1 = m m− 1 ∈ M . On the other hand, because N

G and m ∈ G,there exists n ∈ N such that n = mn

− 1m− 1 . Then nmn − 1m− 1 = nn ∈ N . Now we must have thatnmn − 1m− 1∈ M ∩N which is assumed to be trivial. Hence nmn − 1m− 1 = e or, multiplying on the right bym and then n , nm = mn .

H 2.6.13 – Let T G be cyclic. Prove that every subgroup of T is normal in G.

Say T = t . By 2.5.14, a subgroup S of T is again cyclic, so let S = tm . Because T is normal, thereexists a k such that gtk g− 1 = tk ∈ T (k depends on k and g). Then consider an element ( t

m )k of S .

g(tm )k g− 1 = ( gtk g− 1)m = ( tk )m = ( tm )k

∈

S . Therefore gSg− 1 = S and S G.

H 2.6.14* – Give an example of groups E ≤ F ≤ G with E F and F G but E not normal in G.Missing.


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H 2.6.15 – Let N G. For a ∈ G, prove that |Na | in G/N divides |a| in G.We have ( Na ) | a | = N (a | a | ) = N e = N , the identity in G/N . By 2.5.16, |Na | divides |a|.

H 2.6.16 – Let G be nite and let N G be such that [G : N ] and |N | are coprime. Prove that any element x ∈ G satisfying x|N | = e must be in N .First notice that x [G :N ]∈ N because the order of G/N is [G : N ] so N = ( Nx )

[G :N ] = N x [G :N ]. There exist

a, b ∈ Z such that a[G : N ] + b|N | = 1. Let x ∈ G be such that x | N | = e. Nowx = xa [G :N ]+ b| N | = ( x [G :N ])a (x | N | )b = ( x[G :N ])a ∈ N.

H 2.7.1 – Are the following maps homomorphisms? If yes, what are their kernels? (a) G = R\{0} under multiplication, φ : G → G given by φ(x) = x2 . (b) G as in (a) , φ : G → G given by φ(x) = 2 x . (c) G = Runder addition, φ : G → G given by φ(x) = x + 1 . (d) G as in (c), φ : G → G given by φ(x) = 13 x. (e) Gany abelian group, φ : G → G given by φ(x) = x5 .(a) Yes: φ(xy) = ( xy)2 = x2y2 = φ(x)φ(y). The identity in G is 1, so the kernel is the set of those x ∈ Gsuch that φ(x) = x2 = 1. Hence ker φ = {−1, 1}.(b) No: φ(2 ·3) = 2 6 = 64 while φ(2)φ(3) = 2 2 ·23 = 32.

(c) No: φ(x + y) = x + y + 1 while φ(x) + φ(y) = ( x + 1) + ( y + 1) = x + y + 2.

(d) Yes: φ(x + y) = 13( x + y) = 13 x + 13 y = φ(x) + φ(y). The identity in G is 0, so the kernel is the set of

those x ∈ G such that φ(x) = 13 x = 0. Hence ker φ = {0}.(e) Yes: φ(xy) = ( xy)5 = x5y5 = φ(x)φ(y). ker φ = {x ∈ G | x5 = e}.

H 2.7.2 – Let φ : G → G be given by φ(x) = gxg− 1 for xed g ∈ G. Prove φ is an isomorphism.φ is a homomorphism: φ(xy) = gxyg− 1 = gx(g− 1g)yg− 1 = φ(x)φ(y). The kernel of φ is trivial: if x

∈ G is

such that gxg− 1 = e, then x = e. Hence ker φ = {e} and therefore φ is an isomorphism.

H 2.7.3 – Let G be nite and let n ∈ Z be relatively prime to |G|. Prove that every x ∈ G may be written as g = xn with x ∈ G.There exist a, b ∈ Z such that an + b|G| = 1. For g ∈ G, we have g = gan + b| G | = ( ga )n .


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Herstein assumes, additionally, that G is abelian and hints to consider the map φ : G → G given byφ(y) = yn . This map is a homomorphism when G is abelian, but is not in general. The map, however, is

always surjective, as shown above ( g = φ(ga )). Therefore φ, the n-th power map, is a bijection for any nite

G and choice of n coprime to |G|. In that context, it makes sense to talk about a well-dened, unique nthroot of a group element.

H 2.7.4 – Let U ⊂ G. Let U ≤ G be the smallest subgroup of G containing U . (a) Prove that such a U exists. (b) If gug− 1∈ U for all g ∈ G and u ∈ U , prove that U G.(a) Let A be the collection of all subgroups V of G which contain U . G ∈ A, so A is not empty. U = V ∈A V is a subgroup of G which contains U and is a subset of every element of A. Hence any subgroup of G containingU also contains U . In this sense, U ts the criterion.

(b) By lemma 3 (below), U is exactly the set of all nite products of elements of U and their inverses.Note that if gug− 1 = u ∈ U then taking the inverse gives gu− 1g− 1 = u −1∈ U . Now an arbitrary elementof U may be written as uk 11 · · ·uk nn (with n ≥ 0 and ki = ±1) so conjugation gives

guk 11 · · ·uk nn g− 1 = guk 11 g− 1g · · ·g− 1guk nn g− 1 = u k 11 · · ·u k nn ∈ U ,where u i = gui g− 1 . Therefore U G.

Lemma 3 – Dene V = {uk 11 uk 22 · · ·uk nn ∈ G | n ∈ {0, 1, · · ·}, u i ∈ U, ki = ±1}, the set of all nite (or empty, in which case the result is e) products of elements from U or their inverses. Then V =

U .

Proof: As U is a subgroup containing U , it also contains all inverses of elements of U . Furthermore,it is closed under multiplication, so V ⊂ U immediately. Notice that V is both trivially closed undermultiplication and non-empty (because when n = 0 we see that e ∈ V ). Furthermore, the inverse of uk 11 · · ·uk nn is u− k nn · · ·u− k 11 which is again in V . By lemma 2.3, V ≤ G. Now U ⊂ V , so, by (a), U ≤ V .Therefore V = U and the claim is proven.

H 2.7.5 – Let U = {xyx − 1y− 1 | x, y ∈ G}. Dene G = U (the commutator subgroup of G). (a) Prove that G G. (b) Prove that G/G is abelian. (c) If G/N is abelian, prove that G ≤ N . (d) Prove that, if H ≤ G then H G.(a) Let g,x, y ∈ G and compute the conjugate of an element xyx− 1y− 1∈ U :

gxyx− 1y− 1g− 1 = gxg− 1gyg− 1gx− 1g− 1gy− 1g− 1 = ( gxg− 1)(gyg− 1)(gxg− 1)− 1(gyg− 1)− 1∈ U.

By 2.7.4b, G = U G.


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(b) We have, for x, y ∈ G, that ( G x)(G y)(G x− 1)(G y− 1) = G (xyx − 1y− 1) = G . Right multiplying theequation by G y and then by G x gives that ( G x)(G y) = ( G y)(G x), so G/G is abelian.

(c) If G/N is abelian, then, for x, y ∈ G we have (Nx )(Ny)(Nx − 1)(Ny − 1) = ( Nx )(Nx − 1)(Ny)(Ny − 1) =N (xx − 1yy− 1) = N . Hence xyx− 1y− 1

∈

N . Then U

⊂ N and therefore G

≤ N by virtue of being the

smallest subgroup containing U .

(d) Let g ∈ G and h ∈ H . Because G ≤ H , we have that ghg− 1h− 1∈ H . Therefore ghg− 1∈ H , so H G.

H 2.7.6 – Let M, N G. Prove that NM/M ∼= N/ (N ∩M ).By 2.6.5, N ∩ M N . It is trivial that M NM because M G and NM ≤ G. Therefore, with theknowledge that all quantities are meaningful, dene φ : NM/M → N/ (N ∩M ) by φ(nmM ) = n (N ∩M )for n ∈ N , m ∈ M . The map is well-dened: suppose n1m1M = n2m2M for n1 , n 2 ∈ N and m1 , m 2 ∈ M .In fact, this also means that n1M = n2M , so that n1 = n2m3 for some m3 ∈ M . However, also noticethat m3 = n− 12 n1 ∈ N , so m3 ∈ M ∩N . Then n1(N ∩M ) = n2m3(N ∩M ) = n2(N ∩M ) and the map iswell-dened as claimed.

φ is a homomorphism: let n1 , n 2 ∈ N and m1 , m 2 ∈ M . Then φ((n1m1M )(n2m2M )) = φ(n1Mn 2M ) =φ(n1n2M ) = n1n2(N ∩M ) = n1(N ∩M )n2(N ∩M ) = φ(n1m1M )φ(n2m2M ). Finally, the kernel of φ istrivial: suppose φ(nmM ) = N ∩M for n ∈ N and m ∈ M . This means that n ∈ N ∩M ≤ M , i.e. we havenmM = M . Therefore φ is an isomorphism and the result is proven.

H 2.7.7 – For a, b ∈ R, let τ ab : R → R be given by τ ab (x) = ax + b. Let G = {τ ab | a, b ∈ R, a = 0} and let N = {τ 1b ∈ G}. Prove that N G and that G/N ∼= R\{0} under multiplication.By 2.5.7, G is a group under composition, τ ab ◦τ cd = τ ac,ad + b and τ − 1ab = τ 1a ,− ba . Then, with τ ab ∈ G,

τ ab τ 1cτ − 1ab = τ a,ac + bτ 1a ,− ba = τ 1,ac ∈ N so that N G. Dene φ : G/N → R\{0} by φ(τ ab N ) = a. φ is well-dened: suppose τ ab N = τ cd N so thatτ cd = τ ab τ 1e for some e ∈ R. Then τ cd = τ a,ae + b so that c = a and φ(τ cd N ) = c = a as required. φ is ahomomorphism: φ(τ ab Nτ cd N ) = φ(τ ab τ cd N ) = φ(τ ac,ad + bN ) = ac = φ(τ ab )φ(τ cd ). The kernel of φ is trivial:

if φ(τ ab N ) = 1, then a = 1 so that τ ab ∈ N , i.e. τ ab N = N . Therefore φ is an isomorphism and the result isproven.

H 2.7.8 – Let D2n be the dihedral group with generators x, y such that x2 = yn = e and xy = y− 1x. (a)

Prove that y D 2n . (b) Prove that D2n / y ∼= Z2 .


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(a) y has order n , and D 2n has order 2 n, so y has index 2n/n = 2 in D2n . By 2.6.2, y D 2n .(b) The order of D2n / y is 2, and the unique group of order 2, up to isomorphism, is Z2 . More constructively,

φ : D2n / y → Z2 given by φ(x i yj Y ) = i is an isomorphism.

H 2.7.9 – Prove that Z (G) G. (see 2.5.13)

For g ∈ G, z ∈ Z (G), we have gzg− 1 = zgg− 1 = z ∈ Z (G), so Z (G) G.

H 2.7.10 – Prove that a group of order 9 is necessarily abelian.

Missing.

H 2.7.11 – Let G be non-abelian with order 6. Prove that G ∼= S 3 .The non-identity elements of G have order 2, 3 or 6 by Lagrange’s theorem. If G contained an element of

order 6, then G would be cyclic and hence abelian. Therefore if G is non-abelian, its non-identity elements

have orders 2 and/or 3. Let’s recall S 3 = {e, (12) , (13) , (23) , (123), (213)}. It contains 2 elements of order 3and 3 elements of order 2. Its multiplication table is:

S 3 e (123) (213) (12) (13) (23)

e e (123) (213) (12) (13) (23)

(123) (123) (213) e (13) (23) (12)(213) (213) e (123) (23) (12) (13)

(12) (12) (23) (13) e (213) (123)

(13) (13) (12) (23) (123) e (213)

(23) (23) (13) (12) (213) (123) e

Suppose G has no elements of order 3. Then all of its elements have order 2, but this forces G to be abelian

by 2.3.10. Therefore there must exist an element a ∈ G with order 3. Let b ∈ G be distinct from {e,a,a 2}.If the order of b is 3, then

{e,a,a 2 ,b ,b2

} contains no duplicates: if b2 = a2 then multiplying on the left by

a and on the right by b gives a = b, which is a contradiction; if b2 = a then squaring gives a2 = b4 = b,

which is a contradiction. Hence we have 5 elements of the group in the case that b has order 3. Consider

ab: the cases ab = e, ab = a, ab = a2 , ab = b and ab = b2 all yield immediate contradictions, so we must

conclude that ab is the sixth element of the group. Furthermore, the same argument has us conclude that

ba is none of {e,a,a 2 ,b ,b2}, so ab = ba. For elements u, v of an arbitrary group, uv = vu easily implies thatum vn = vn um . In our case, the fact that ab = ba gives us that G is abelian, so this is not the desired group.


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Therefore a and a2 are the only elements of order 3, and the remaining elements of G all have order 2:

G = {e,a,a 2 ,b,c,d} with b2 = c2 = d2 = e. Consider the product ab. Each of the possibilities ab = e,ab = a, ab = a2 , and ab = b gives an immediate contradiction, so ab ∈ {c, d}, and, by a similar argument,ac ∈ {b, d} and ad ∈ {b, c}. There are two distinct situations possible: {ab = c,ac = d,ad = b} and

{ab = d,ac = b,ad = c}. From those congurations, it’s easy to construct the entire multiplication table forthe two Gs:G1 e a a2 b c d

e e a a2 b c d

a a a 2 e c d b

a2 a2 e a d b c

b b d c e a2 a

c c b d a e a2

d d c b a2 a e

G2 e a a2 b c d

e e a a2 b c d

a a a 2 e d b c

a2 a2 e a c d b

b b c d e a a2

c c d b a2 e a

d d b c a a2 e

Comparing to the group table for S 3 , we see that φ : G1 → S 3 given by φ(e) = e, φ(a) = (123), φ(a2) = (213),φ(b) = (12), φ(c) = (13), φ(d) = (23) is a homomorphism. It is also a bijection, so φ is an isomorphism,

proving the equivalence of G1 to S 3 . On the other hand, ψ : G2 → S 3 given by ψ(e) = e, ψ(a) = (213),ψ(a2) = (123), ψ(b) = (12), ψ(c) = (13), ψ(d) = (23) is again a bijective homomorphism, giving the

equivalence of G2 to S 3 . These were the only two possible non-abelian groups of order 6, and both are

isomorphic to S 3 , so the claim is proven.

H 2.7.12 – Let G be abelian and let N ≤ G. Prove that G/N is abelian.Because G is abelian, N is normal in G and G/N is a group. Let g1 , g2 ∈ G. Then ( g1N )(g2N ) = ( g1g2)N =(g2g1)N = ( g2N )(g1N ) so that G/N is abelian.

H 2.8.1 – Are the following maps automorphisms? (a) G = Z under addition, T : x → −x. (b) G = R+under multiplication, T : x → x2 . (c) G cyclic of order 12, T : x → x3 . (d) G = S 3 , T : x → x− 1 .(a) Yes. Let a, b ∈ Z. T is a homomorphism: we have T (a + b) = −(a + b) = ( −a)+ ( −b) = T (a) + T (b). T isinjective: if T (a) = T (b), then

−a =

−b so a = b. T is surjective: a = T (

−a). Hence T is an automorphism.

(b) Yes. Let x, y ∈ R+ . T is a homomorphism: we have T (xy) = ( xy)2 = x2y2 = T (x)T (y). T is injective: if T (x) = T (y), then x2 = y2 so x = y because we restrict to positive reals. T is surjective: x = T (√ x) where√ x is the unique, positive square root which exists for all positive x.(c) No. Say G = g . Then g4 = e, but T (g4) = g12 = e, so the kernel of T is non-trivial. Hence T is notinjective and thus not an isomorphism.


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(d) No. T [(12)(13)] = T [(213)] = (123), while T [(12)]T [(13)] = (12)(13) = (213).

H 2.8.2 – Let H ≤ G, and let T be an automorphism of G. Prove T (H ) ≤ G.

For h1 , h2 ∈ H , we have T (h1)T (h2) = T (h1h2) ∈ T (H ) and T (h1)− 1 = T (h− 11 ) ∈ T (H ). By lemma 2.3,T (H ) ≤ G.

H 2.8.3 – Let N G, and let T be an automorphism of G. Prove T (N ) G.

By 2.8.2, T (N ) ≤ G. Let g ∈ G and n ∈ N . Then gT (n)g− 1 = T (T − 1(g)nT − 1(g− 1)) = T (n ) ∈ T (N ) forsome n = T − 1(g)nT − 1(g− 1) ∈ N because N is normal.

H 2.8.4 – Prove that Inn( S 3) ∼= S 3 .The center of S 3 is trivial, as we can easily check: (12)(13) = (213) while (13)(12) = (123), so (12) ∈ Z (S 3)and (13) ∈ Z (S 3). (23)(123) = (13) while (123)(23) = (12) so (13) ∈ Z (S 3) and (123) ∈ Z (S 3). Finally,(12)(213) = (13) while (213)(12) = (23), so (213) ∈ Z (S 3). Therefore Z (S 3) = {e} and, by lemma 2.19,Inn( S 3) ∼= S 3 /Z (S 3) ∼= S 3 .

H 2.8.5 – Prove that Inn( G) Aut( G).

Let g ∈ G, T w : x → wxw− 1

∈ Inn( G) and φ ∈ Aut( G). Then ( φT w φ− 1

)(g) = φ(T w (φ− 1

(g))) =φ(wφ− 1(g)w− 1) = φ(w)gφ(w− 1) = φ(w)gφ(w)− 1 = T φ (w ) (g), where T φ (w ) : x → φ(w)xφ(w)− 1 . HenceφT w φ− 1 = T φ (w ) ∈ Inn( G), which proves that Inn( G) Aut( G).

H 2.8.6 – Let G = {e,a,b,ab} be a group of order 4 with a2 = b2 = e and ab = ba. Determine Aut( G).An automorphism of G xes the identity, and permutes the three elements of order 2. Then it is clear that

Aut( G) is isomorphic to a subgroup of S 3 . Furthermore, we can exhibit elements of order 2 and 3 in Aut( G)

which proves that Aut( G) ∼= S 3 in its entirety.For example, φ : G → G given by φ(a) = b, φ(b) = a and φ(ab) = ab is an automorphism of order2. To see that φ is a homomorphism, we just need to check that φ(ab) = ab = φ(b)φ(a) = φ(a)φ(b),

φ(aab) = a = bab = φ(a)φ(ab) and φ(bab) = b = aab = φ(b)φ(ab). All other possible products automatically

work because the group is abelian. Therefore φ is a homomorphism, and we see easily that φ2(a) = φ(b) = a

and φ2(b) = φ(a) = b, so φ2 = id and hence φ is an order 2 automorphism.


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Additionally, ψ : G → G given by ψ(a) = b, ψ(b) = ab and ψ(ab) = a is an automorphism of order 3. Wecan check that ψ is a homomorphism: ψ(ab) = a = bab = ψ(a)ψ(b), ψ(aab) = ab = ψ(ab)ψ(a) = ψ(a)ψ(ab),

and ψ(bab) = b = aba = ψ(b)ψ(ab). To check the order of ψ, we raise it to powers: ψ2(a) = ψ(b) = ab,

ψ2(b) = ψ(ab) = a and ψ2(ab) = ψ(a) = b. Therefore ψ2 = id, but ψ3(a) = ψ(ab) = a, ψ3(b) = ψ(a) = b

and ψ3(ab) = ψ(b) = ab so, in fact, ψ

3= id.

As a result, we know that Aut( G) must be isomorphic to a subgroup of S 3 which can accomodate elements

of orders 2 and 3. This forces the subgroup to have order at least 6 = lcm(2 , 3), but that’s the order of S 3itself. Hence Aut( G) ∼= S 3 .

H 2.8.7 – Let C ≤ G. C is “characteristic” if φ(C ) ⊂ C for all φ ∈ Aut( G). (a) Prove that a characteristic subgroup is normal. (b) Prove that the converse of (a) is false.

(a) Suppose C is characteristic and let g ∈

G. gCg− 1 is the image of C under the inner automorphism

T g : x → gxg− 1 . Because T g ∈ Aut( G), we have that gCg− 1 = T g (C ) ⊂ C . This holds for arbitrary g ∈ G,so C G.

(b) A normal subgroup, N G, is xed by all inner automorphisms, by denition. In order for the converse

to fail to hold, there need to be automorphisms outside of Inn( G) which don’t x N . Consider G = V 4 , the

Klein group of 2.8.6, and N = a = {e, a} which is normal because it is of index 2. Then take the order 2automorphism φ of G, with φ(a) = b, φ(b) = a and φ(ab) = ab. We have φ( a ) = {e, b} = a . Hence N isnormal, but not characteristic because φ doesn’t x it.

H 2.8.8 – Prove that the commutator subgroup G = aba− 1b− 1 | a, b ∈ G (see 2.7.5) is characteristic.Let φ ∈ Aut( G) and let a1b1a− 11 b− 11 · · ·an bn a− 1n b− 1n ∈ G . Because φ is a homomorphism, we have

φ(a1b1a− 11 b− 11 · · ·an bn a− 1n b− 1n ) = φ(a1)φ(b1)φ(a1)− 1φ(b1)− 1 · · ·φ(an )φ(bn )φ(an )− 1φ(bn )− 1∈ G .

Therefore φ(G ) ⊂ G , so G is characteristic.

H 2.8.9 – Let N G and let M be a characteristic subgroup of N . Prove M G.

Every inner automorphism xes N but, a priori , could move M ≤ N around. Notice that if T g ∈ Inn(G),it restricts to an automorphism T g |N of N because T g (N ) = N . Now M is characteristic in N , so we haveT g |N (M ) = M . This is simply a restriction of a map to the domain N , so, in fact, T g (M ) = M as well.Hence M is xed by every inner automorphism, i.e. M G.

Some remarks: If N had not been normal, then inner automorphisms T g wouldn’t restrict to automorphisms


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of N (because T g might throw some elements of N aeld), and the argument breaks down. Also, if N were

normal but M merely normal (not characteristic) in N , the argument would also break down: T g restricts

to an automorphism of N which is not, in general, an inner automorphism of N (not unless g ∈ N , in fact).Thus the restriction T g |N would not x M if it were only normal but not characteristic. Therefore thisexercise represents a strengthening of the conditions of 2.6.14, where we see that normality of subgroups isnot a transitive property.

H 2.8.10 – Let G be nite and let φ ∈ Aut( G) x only the identity. Prove that every g ∈ G may be written as g = x− 1φ(x) for some x ∈ G.Consider the map ψ : G → G given by ψ(x) = x− 1φ(x). This map is injective: let x, y ∈ G be such thatψ(x) = ψ(y). Then x− 1φ(x) = y− 1φ(y). Multiplying on the left by y, on the right by φ(x− 1), and using

the fact that φ is a homomorphism, we have yx− 1 = φ(yx− 1). By our assumption on φ, we must conclude

that yx− 1 = e, so that x = y. This proves injectivity. Because G is nite, ψ is also necessarily surjective(however, ψ is not a homomorphism in general, and this is immaterial). Therefore for every g ∈ G, thereexists an x ∈ G such that g = ψ(x) = x− 1φ(x).

H 2.8.11 – Let G be nite, and let φ ∈ Aut( G) x only the identity. Suppose additionally that φ2 = id .Prove that G is abelian.

Let g ∈ G. By 2.8.10, there exists x ∈ G such that g = x− 1φ(x). Now φ(g) = φ(x− 1)φ2(x) = φ(x)− 1x =(x− 1φ(x)) − 1 = g− 1 . Suppose G were not abelian, with a, b

∈ G such that ab

= ba. Then φ(ab) = b− 1a− 1

while φ(a)φ(b) = a− 1b− 1 . If we were to have that φ(ab) = φ(a)φ(b), then inverting b− 1a− 1 = a− 1b− 1 would

produce the contradiction ab = ba. Therefore, because φ is assumed to be a homomorphism, G must be

abelian.

Remark: In fact, if G is a group then G is abelian if and only if ρ : g → g− 1 is an automorphism of G. ρis trivially a bijection (uniqueness of inverses gives injectivity, and g = ρ(g− 1) gives surjectivity). If G is

abelian, then ρ is a homomorphism because ρ(gh) = h− 1g− 1 = g− 1h− 1 = ρ(g)ρ(h). If G is not abelian,

then ρ is not a homomorphism by the above argument. The restriction to nite G in the problem statement

enables us to use the obscure (in my opinion) result 2.8.10.

H 2.8.12* – Let G be nite, and let φ ∈ Aut( G) be such that φ(x) = x− 1 for at least three quarters of the elements of G. Prove that φ(x) = x− 1 for all x in G and that G is abelian.

Missing.


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H 2.8.13 – Give an example of a non-abelian nite group G with an automorphism that maps exactly three

quarters of the elements of G to their inverses.

Missing.

H 2.8.14* – Let G be nite with |G| > 2. Prove Aut( G) is non-trivial.If G is abelian, then inversion ρ : g → g− 1 is an automorphism (see 2.8.11). If ρ = id, then g = g− 1 forall g ∈ G, i.e. all non-identity elements have order 2. In that case, construct the map φ : G → G whichtransposes two non-identity elements and xes everything else. φ is an automorphism (this needs to be

checked!). Therefore, if G is abelian, either ρ or φ is a non-trivial automorphism of G.

If G is non-abelian, take an element a ∈ Z (G). Then there exists g ∈ G with ag = ga, and thereforeaga − 1 = g. Thus the inner automorphism T a : g → aga − 1 of conjugation by a does not x g, so T a = id butT a ∈ Aut( G).

H 2.8.15* – Let G have even order 2n . Suppose that exactly half of the elements of G have order 2 and the

rest form a subgroup H of order n. Prove that |H | is odd and that H is abelian.If H were of even order, then by 2.3.11 or Cauchy’s theorem, it would contain an element of order 2. It is

assumed that H is the collection of elements with order different from 2, so |H | must be odd.Let x ∈ H . Then xh ∈ H for any h ∈ H . Furthermore, xh has order 2, because it is outside of H . The mapφ : G → G given by φ(g) = xgx− 1 = xgx is an inner automorphism of G. Because H is normal (index 2), φxes H and hence restricts to an automorphism φ|H of H . Now we see that hφ(h) = hxhx = ( hx )2 = e, sothat φ|H (h) = h− 1 . Therefore inversion is an automorphism on H , and, as seen in 2.8.11, this gives that H is abelian.

H 2.8.16* – Let φ(n) be the Euler φ-function. Let a ∈ Z, with a > 1. Prove that n | φ(an −1).Consider Z∗a n − 1 , the multiplicative (modulo an −1) group of integers coprime to ( an −1). The number of elements in Z∗a n − 1 is φ(an

−1). Furthermore, a is coprime to an

−1 because ( an − 1)a + (

−1)(an

−1) = 1 so

that a ∈ Z∗a n − 1 . The order of a is n because an ≡ 1 mod (an −1) while, for 1 < k < n , 1 < a k < a n −1.By Lagrange, this order must divide the order of the group, and therefore n | φ(an −1).

H 2.9.1 – Let g ∈ G. Dene λg : G → G by λg (x) = gx. Prove that λg is a bijection and that λg λh = λgh .λg is surjective: if y ∈ G, then y = gg− 1y = λg (g− 1y). λg is injective: if x, y ∈ G, then λg (x) = λg (y) implies


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that gx = gy so x = y. Finally, if g , h ∈ G, then λ gh (x) = ghx while λ g (λh (x)) = λg (hx ) = ghx. Thereforeλgh = λ g λh . Hence each λg is a permutation of G and the λg form a subgroup (under composition) of the

set of bijections G → G.

H 2.9.2 – Let λg be dened as in 2.9.1. Dene τ g : G → G by τ g (x) = xg. Prove that, for g, h ∈ G, we have λg τ h = τ h λg .

Let x ∈ G. We have λg (τ h (x)) = λg (xh ) = gxh while τ h (λg (x)) = τ h (gx) = gxh. Hence λg τ h = τ h λg .

H 2.9.3 – Let λg and τ g be dened as in 2.9.2. If θ : G → G is a bijection such that λg θ = θλg for all g ∈ G, prove that θ = τ h for some h ∈ G.For x

∈ G, λg (θ(x)) = gθ(x) while θ(λg (x)) = θ(gx). Note that we don’t assume θ to be a homomorphism.

However, we see that θ(gx) = gθ(x) for all x, g ∈ G. If we pick g = x− 1 , then we nd that θ(e) = x− 1θ(x) forall x ∈ G. Solving for θ(x) yields θ(x) = xθ(e), which is the desired result. Specically, we have θ = τ θ (e ) .

H 2.9.4 – Let H ≤ G. (a) Show that gHg− 1 ≤ G for every g ∈ G. (b) Prove that W = g∈G gHg− 1 is a normal subgroup of G.

(a) Let h1 , h2 ∈ H and let g ∈ G. We have gh1g− 1gh2g− 1 = gh1h2g− 1∈ gHg − 1 , so gHg− 1 is closed undermultiplication. Additionally, the inverse of gh1g− 1 is gh− 11 g− 1∈ gH g

− 1 . By lemma 2.3, gHg− 1 ≤ G.(b) W ≤ G because it is the intersection of subgroups. Suppose w ∈ W so that, for every g ∈ G, there existsh ∈ H such that w = ghg− 1 . Let x ∈ G, and consider xwx − 1 . For any g ∈ G, there exists h ∈ H suchthat w may be written as w = ( x− 1g)h(x− 1g)− 1 = x− 1ghg− 1x. Consequently, xwx− 1 = xx − 1ghg− 1xx − 1 =

ghg− 1∈ gHg− 1 . This can be done for arbitrary x, g ∈ G, so xwx− 1∈ W and W G.

H 2.9.5 – Let |G| = p2 . Prove that G has a normal subgroup of order p.Lemma 2.21 states that if G is a nite group, and H ≤ G is a proper subgroup such that |G| [G : H ]!, thenH must contain a non-trivial normal subgroup of G. The lemma is proven by considering the action of left

multiplication by G on the set G/H of left cosets of H in G.

As p2 is not prime, we know by 2.5.3 that G has a proper subgroup H . By Lagrange, this subgroup must

have order p. By the fundamental theorem of arithmetic, |G| = p2 does not divide [ G : H ]! = p!, so theconditions of lemma 2.21 are satised. Therefore H must contain a non-trivial normal subgroup K of G. By

Lagrange, this subgroup must have order p, i.e. K = H . Thus H G and |H | = p.


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H 2.9.6* – Let |G| = p2 and let H G with |H | = p. Prove that H ≤ Z (G).If G contains an element of order p2 , then G is cyclic and hence abelian. In that case, Z (G) = G so

H ≤ Z (G). Otherwise, all non-identity elements of G have order p. Let x ∈ H . In fact, because H hasprime order, it is cyclic and H =

x . Let y

∈ G. By lemma 2.21 (as used in 2.9.5),

y G. Lagrange’s

theorem applied to x ∩y (as a subgroup of x or y ) tells us that it has order 1 or p. In other words,x = y = x ∩y or x ∩y = {e}. In the former case, x = y gives that x and y commute. In the

latter case, x ∩y = {e} invites the application of 2.6.12, which again gives that xy = yx. As x ∈ H andy ∈ G were arbitrary, this shows that H ≤ Z (G).

H 2.9.7* – Let |G| = p2 . Prove that G is abelian.If G contains an element of order p2 , then G is cyclic and hence abelian. Suppose G contains no element

of order p2 . Then, by Lagrange, every element g ∈

G has order p. By lemma 2.21 (as used in 2.9.5),

g is

normal and, by 2.9.6, g ⊂ Z (G). In particular, g ∈ Z (G). This argument applies to arbitrary g ∈ G, soZ (G) = G, i.e. G is abelian.

H 2.9.8 – Let |G| = 2 p. Prove G has a subgroup H of order p and that H G.If x ∈ G has order 2 p, then x2 has order p, so H = x2 is a subgroup of order p. Otherwise, suppose thereis no element in G of order 2 p. If x ∈ G has order p, then H = x is a subgroup of order p. Otherwise,suppose G has no elements of order p or 2 p. Then all non-identity elements of G have order 2. Therefore G

is abelian by 2.3.10, and we reach a contradiction: if G is abelian, then Cauchy’s theorem for abelian groupsimplies the existence of an element x ∈ G of order p. Therefore G necessarily has a subgroup H of order p.The index of H is 2, and so it is normal by 2.6.2.

H 2.9.9 – Let |G| = pq , where p = q are both prime. Suppose H, K G with |H | = p and |K | = q . Prove that G is cyclic.

Let H = h and K = k so h p = e and kq = e. If x ∈ h ∩k then |x| | p and |x| | q , so |x| = 1 andconsequently

h

∩k =

{e

}. By 2.6.12, hk = kh. Now consider the element hk. Its order must be 1, p,

q or pq . As k ∈ h , we cannot have k = h− 1 , so hk = e. Compute ( hk) p = h pk p = k p = e because q p.Similarly, ( hk )q = hqkq = hq = e because p q . Therefore |hk | = pq , and G = hk is cyclic.

H 2.9.10* – Let |G| = pq , where p > q are both prime. (a) Prove that G has a subgroup of order p and a subgroup of order q . (b) Prove that if q ( p−1) then G is cyclic. (c) Prove that, given two primes p, q with


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q ( p−1), there exists a non-abelian group of order pq . (d) Prove that any two non-abelian groups of order pq are isomorphic.

Missing.

H 2.10.1, 2.10.2 – Decompose into products of disjoint cycles:

(a)1 2 3 4 5 6 7 8 9

2 3 4 5 1 6 7 9 8 (b)

1 2 3 4 5 6

6 5 4 3 1 2

(a) (12345)(6)(7)(89).

(b) (1625)(34).

H 2.10.3 – Express as products of disjoint cycles: (a) (15)(16789)(45)(123) . (b) (12)(123)(12) .

Herstein composes permutation from left to right, while I compose permutations from right to left. This

problem statement has been modied accordingly. Disjoint cycles commute, so this is often irrelevant.

(a) Consider the action of each cycle in order. Starting with 123456789, (123) sends this into 312456789.

(45) sends this into 312546789. (16789) sends this into 912543678. (15) sends this into 412593678. Hence

(15)(16789)(45)(123) = (123678954).

(b) Starting with 123, (12) sends this into 213. (123) sends this into 321. (12) sends this into 231. Therefore

(12)(123)(12) = (132).

H 2.10.4 – Prove that (12 · · ·n)− 1 = ( n, n −1, n −2, . . . , 2, 1)First note that a transposition (2-cycle) is its own inverse. It’s straightforward to check by hand Herstein’s

assertion, pg. 67, that ( a1 , a 2 , . . . , a m ) = ( a1 , a m )(a1 , a m − 1) · · ·(a1 , a 2). Therefore(12 · · ·n)− 1 = [(1 n)(1, n −1) · · ·(12)]

− 1 = (12) · · ·(1n) = (1 ,n ,n −1, . . . , 2) = ( n, n −1, n −2, . . . , 2, 1).

H 2.10.5 – Find the cycle structure of all the powers of (12 · · ·8)Let σ = (12 · · ·8). We’ll compute what σ i does to 12345678. The sequence goes

12345678 σ−→ 81234567 σ

−→ 78123456 σ

−→ 67812345σ

−→ 56781234 σ

−→ 45678123 σ

−→ 34567812 σ

−→ 23456781


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Then we have σ = (12345678), σ2 = (1357)(2468), σ3 = (14725836), σ4 = (15)(26)(37)(48), σ5 = (16385274),

σ6 = (1753)(2864), σ7 = (18765432) and σ8 = e.

H 2.10.6 – (a) What is the order of an n-cycle? (b) What is the order of the product of disjoint cycles of

lengths m1 , m 2 , . . . , m k ? (c) How do you nd the order of a given permutation?

(a) Let σ be an n -cycle. If σk (i) = i for some k < n , then σ can be decomposed into smaller cycles because

i has an orbit which is a cardinality k subset of {1, 2, . . . , n }. Furthermore, by the pigeonhole principle,σn (i) = i. Hence the order of an n -cycle is n .

(b) Let σ1 , . . . , σ k be disjoint cycles. As disjoint cycles commute, ( i σi )m = i σ

mi . The order of i σi is

therefore N = lcm i (m i ), the least common multiple of the orders (lengths, by (a)) of all the σi . It is clear

that ( i σi )N = e because m i | N for all i. For any smaller exponent K , there is at least one σi with m i K

so that (i

σi)K = e.

(c) Decompose the permutation into disjoint cycles. The order of the permutation is the least common

multiple of the lengths of the disjoint cycles.

H 2.10.7 – Compute a− 1ba where (a) a = (12)(135) , b = (1579) . (b) a = (579) , b = (123) .

(a) a = (1352) and a− 1 = a3 = (1253). Compute a− 1ba by applying each permutation in order.

a(123456789) = 251436789, b(251436789) = 951426387, and a− 1(951426387) = 192456387. Therefore

a− 1ba = (2379).

(b) a− 1 = a2 = (597). a(123456789) = 123496587, b(123496587) = 312496587, and a− 1(312496587) =

312456789. Therefore a− 1ba = (123) = b. We could have seen this immediately because a and b are disjoint.

H 2.10.8 – (a) For x = (12)(34) and y = (56)(13) , nd a permutation a such that a− 1xa = y. (b) Prove

there is no a such that a− 1(123)a = (13)(578) . (c) Prove there is no a such that a− 1(12)a = (15)(34)

(a) We must have xa = ay . Then x(a(1)) = a(y(1)) = a(3), x(a(3)) = a(y(3)) = a(1), x(a(2)) = a(y(2)) =

a(2), and so on. We see that a(2) and a(4) must be xed by x, so a(2) , a (4) ∈ {5, 6}. On the other hand,x should transpose a(3) with a(1) and a(5) with a(6). There is freedom in creating a, but the following

choice works: a(1) = 1, a(3) = 2, a(5) = 3, a(6) = 4, a(2) = 5, and a(4) = 5. This may also be written as

a = (253)(46). This a works, though it is not unique. For example, a = (1263)(45) is another solution.

(b) Let x = (123) and y = (13)(578). Applying the same analysis as above, we see that x(x(a(1))) = a(1)


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while x(a(1)) = a(3) = a(1). Hence a(1) must be brought on an orbit of length 2 by x, but it is clear that xdoes this to no symbol. The cycles of x are all of length 1 or 3. Therefore no such a exists. A more direct

way to see this is that |a− 1xa | = |x| = 3 while |y| = 6 by 2.10.6.(c) Let x = (12) and y = (15)(34). We have x(a(1)) = a(5), x(a(2)) = a(2), x(a(3)) = a(4), x(a(4)) = a(3),

and x(a(5)) = a(1). Only a(2) is xed here, but x must x three of its ve inputs. This is a contradiction,

and so it is impossible to construct such an a.

H 2.10.9 – For what m is an m-cycle an even permutation?

The m-cycle σ = ( a1 · · ·am ) = ( a1am ) · · ·(a1a3)(a1a2) may be written as the product of m −1 transposi-tions. While this decomposition into transpositions is not unique, the parity (even/odd) of the number of

transpositions is (see pg. 67). Then σ is an even permutation if and only if m −1 is even, i.e. if m is odd.

H 2.10.10 – What are the parities of the following permutations? (a) (12)(123) . (b) (45)(123)(12345) . (c)

(25)(14)(13)(12) .

(a) (12)(123) = (12)(13)(12) is odd, as it is the product of three transpositions. Alternately, because sgn is

multiplicative, sgn((12)(123)) = ( −1)(+1) = −1.(b) (45)(123)(12345) = (45)(13)(12)(15)(14)(13)(12) is the product of seven transpositions, so it is odd.

Again, we can also compute sgn((45)(123)(12345)) = ( −1)(+1)(+1) = −1.

(c) (25)(14)(13)(12) is the product of four transpositions, so it is even.

H 2.10.11 – Prove that S n = (12 · · ·n), (12) .Let σ = (12 · · ·n), τ = (12) and U = σ, τ . The result of applying σ to n symbols is to rotate them all tothe right by one spot. Similarly, applying σ− 1 effects a left rotation by one. Compute στσ− 1 by considering

its action on 12 · · ·n:123 · · ·n

σ − 1

−→ 234 · · ·n1 τ

−→ 324 · · ·n1 σ

−→ 1324· · ·n.

We see that στ σ− 1

= (23). Computing σ2

τσ− 2

, a pattern becomes clear which suggests that σk

τσ− k

=(k + 1 , k + 2). The base case k = 1 was just shown, so assume this result to be true. Then σk +1 τσ− (k+1) =


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σ(k + 1 , k + 2) σ− 1 . Observe the action of this permutation:

1 2 3 · · · k + 1 k + 2 k + 3 · · · n −1 nσ− 1 2 3 4 · · · k + 2 k + 3 k + 4 · · · n 1

(k + 1 , k + 2) 2 3 4

· · · k + 3 k + 2 k + 4

· · · n 1

σ 1 2 3 · · · k + 1 k + 3 k + 2 · · · n −1 nfrom which we see that σ(k + 1, k + 2) σ− 1 = ( k + 2 , k + 3). Therefore the claim is proven that σk τσ− k =

(k + 1 , k + 2). In particular, ( k, k + 1) ∈ U for k ∈ {1, 2, . . . , n −1}.Let’s investigate the product ( ab)(bc)(ab). This permutation sends abc → bac → bca → cba, so (ab)(bc)(ab) =(ac). This is useful because we can write (1 k) = (1 , k −1)(k −1, k)(1, k −1) where we use the just-discoveredfact that ( k −1, k) ∈ U . Iterating this procedure, starting with k = 3 (because (1 , k −1) = (12) is given to bein U ), we nd that (1 k) ∈ U for k ∈ {2, 3, . . . , n }. For instance, (13) = (12)(23)(12) and (14) = (13)(34)(13).

Now the arbitrary transposition ( ab) we see to be in U because ( ab) = (1 a)(1b)(1a). Therefore, as everypermutation may be decomposed into transpositions, every permutation is contained in U . Now S n ≤ U ≤S n , so this nalizes the proof that U = S n .

H 2.10.12* – Prove that, for n ≥ 3, the subgroup U n generated by the 3-cycles is An .Because (i) 3-cycles are even permutations, (ii) inverses of 3-cycles are again 3-cycles, and (iii) products of

even permutations are again even, we have that U n ≤ An . Every even permutation is the product of aneven number of transpositions. Let a,b,c,d ∈ {1, 2, . . . , n } be distinct. We have ( ab)(cd) = ( adc)(abc) and(ab)(ad) = ( adb). Every product of two transpositions is in one of these two forms (the rst case has no indexshared, the second case has one index shared), and both forms may be rewritten as a product of 3-cycles.

Hence any even permutation σ may be written as the product of 3-cycles, i.e. σ ∈ U n . Therefore An ⊂ U n ,and so An = U n .

H 2.10.13* – Let N An contain a 3-cycle. Prove that N = An .

The spirit of this proof is to show that if N contains a 3-cycle then it contains all other 3-cycles. Then by

2.10.12, the result would be proven. Let ( abc)

∈ N . Note that ( abc)2 = ( abc)− 1 = ( acb)

∈ N . Using the

normality of N , we can compute things like ( cde)− 1(abc)(cde) = ( abe) ∈ N . This allows us to swap out anindex, with the ultimate goal of generating an arbitrary 3-cycle from our original ( abc). Continuing in this

fashion, we see that, if a,b,c,d,e ∈ {1, 2, . . . , n } are distinct and f ∈ {1, 2, . . . , n }\{a,b,d,e}, then(abf )− 1(bcd)− 1(cde)− 1(abc)(cde)(bcd)(abf ) = ( def ) ∈ N.

This conguration of indices is only possible for n ≥ 5. Thankfully, smaller cases are almost trivial. A1


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and A2 don’t accomodate 3-cycles, so they are irrelevant. A3 = (123) has order 3 so it contains no non-trivial subgroups. A4 is actually susceptible to the above argument, although the long formula breaks down

because there aren’t 5 symbols to choose from. Instead, we see that to go from any one 3-cycle in A4 to

another, the only steps which might be required are swapping up to two indices and squaring (as noted

above, ( abc)2

= ( acb)). Swapping the indices is done by conjugation, e.g. ( cde)− 1

(abc)(cde) = ( abe), asexplored above. Therefore in A4 it is also the case that a normal subgroup containing a 3-cycle must contain

every 3-cycle. By 2.10.12, An ≤ N , so N = An .

H 2.10.14* – Prove that A5 has no non-trivial normal subgroups. (i.e. it is “simple”)

Missing, although a proof may be found in 2.11.6c.

H 2.10.15 – Assume that A5 is simple. Prove that if H ≤ A5 is proper, then |H | ≤ 12.A5 has order 60, so, by Lagrange’s theorem, |H | ∈ {2, 3, 4, 5, 6, 10, 12, 15, 20, 30}. A subgroup of order30 has index 2, so it would be normal by 2.6.2. Because A5 is simple, this is disallowed, so |H | = 30.Suppose |H | = 20 so [A5 : H ] = 3. By lemma 2.21, because |A5| = 60 3! = [A5 : H ]!, H contains anon-trivial normal subgroup of A5 . However, as A5 contains no non-trivial normal subgroups, we come to a

contradiction. Therefore |H | = 20. The same argument rules out |H | = 15 because 60 4!. Therefore |H |must be 12 or smaller. Lemma 2.21 says nothing about an order 12 subgroup because 60 | 5! = 60.

H 2.11.1 – (a) In S n , prove that there are n !r (n − r )! distinct r-cycles. (b) Find the number of conjugates of (12 · · ·r ) in S n . (c) Prove that, if σ ∈ S n commutes with (12 · · ·r ), then σ = τ (12 · · ·r ) i with i ∈{0, 1, . . . , r −1} and τ ∈ S n leaving all of {1, 2, . . . , r } xed.(a) An r-cycle permutes r out of n symbols, and there are nr =

n !r !(n − r )! ways of choosing those symbols.

Beyond this choice, there are ( r −1)! ways of permuting all but one of the symbols in the r -cycle. Thus, forinstance, we count the distinct 3-cycles (123) and (132) but, ignoring movement of the r th symbol, we don’t

overcount (312) = (123) as unique. Thus the number of distinct r -cycles in S n is (r −1)! nr = n !r (n − r )! .(b) All permutations of the same cycle decomposition are conjugate to one another. In particular, all r -cycles

in S n have the same cycle decomposition {1, 1, . . . , 1, r}, with n −r ones. Hence (12 · · ·r ) is conjugate toevery one of the n !r (n − r )! r-cycles in S n .

(c) It is clear that (12 · · ·r ) commutes with its powers as well as with τ , which is disjoint from it. Thereforeif A = {τ (12 · · ·r )i | i ∈ {0, 1, . . . , r −1}, τ xing {1, 2, . . . , r }} is the set of interest, then A ⊂ N ((12 · · ·r )),where N (a) is the normalizer of a, i.e. the set of elements commuting with a. The cardinality of A is


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r (n − r )! because there are ( n − r )! ways of permuting the n − r elements {r + 1 , r + 2, . . . , n }. We alsoknow that the normalizer satises (# conjugates of a) = |S n |/ |N (a)|. In the case of a = (12 · · ·r ), this gives

n !r (n − r )! = n!/ |N (a)|, so |N (a)| = r (n −r )!. Therefore A = N (a).

H 2.11.2 – (a) Find the number of conjugates of (12)(34) in S n for n ≥ 4. (b) Determine N ((12)(34)) .(a) For σ ∈ S n , we have σ(12)(34) σ− 1 = ( σ(1)σ(2))( σ(3)σ(4)). If σ were to commute with (12)(34), then wewould need ( σ(1)σ(2)) = (12), ( σ(3)σ(4)) = (34) or ( σ(1)σ(2)) = (34), ( σ(3)σ(4)) = (12). The action of σ on

the n −4 other symbols is irrelevant. Enumerate these σ by considering rst σ(1) ∈ {1, 2, 3, 4} with 4 choices.Once σ(1) is picked, σ(2) is immediately determined and two choices remain for σ(3). After picking σ(3),

σ(4) is also determined. Hence there are 4 ·2 ·(n −4)! = 8( n −4)! distinct σ which commute with (12)(34).Then the number of conjugates of (12)(34) is |S n |/ |N ((12)(34)) | = n!/ (8(n −4)!) = 18 n(n −1)(n −2)(n −3).Alternately, count in the following way: the conjugates of (12)(34) are those permutations with the same

cycle structure ( ab)(cd). We need to pick 2 elements from n to constitute a, b, and 2 elements from n −2 toconstitute c, d. Then we will have overcounted by a factor of 2, getting, for instance, (12)(34) and (34)(12)

which are equal. Therefore the number of conjugates is n2n − 2

2 / 2 = 18 n(n −1)(n −2)(n −3).

(b) All commuting elements σ are described in (a).

H 2.11.3 – Prove that S p contains ( p −1)! + 1 elements satisfying x p = e.The identity element satises e p = e. By 2.11.1, the number of p-cycles in S p is p!/ ( p

·0!) = ( p

−1)! which,

by 2.10.6a, have order p. Therefore we have exhibited ( p −1)! + 1 elements satisfying x p = e. Are thereothers? No. Consider a non-identity element σ ∈ S p . The order of σ is equal to the least common multipleof the lengths of the cycles in its disjoint cycle decomposition by 2.10.6b. If it were the case that σ p = e,

then the order of σ must be p because p is prime, so that the cycles in its decomposition must have length 1

or p. Because there are only p symbols to choose from in S p , one p-cycle already exhausts them all and we

therefore exclude possibilities such as σ being the product of two or more p-cycles. Thus for σ p = e to hold,

we must have that σ is a p-cycle, and so our original count is acceptable.

H 2.11.4 – Let G be nite and let a ∈ G have exactly two conjugates. Prove that G is not simple.We have that |G|/ |N (a)| = (# conjugates of a) = 2, so that N (a) has index 2. By 2.6.2, N (a) is normal.The only oddball case to consider is if N (a) = {e}, so that G is of order 2 and is therefore abelian. If thiswere the case, then N (a) = G, which is a contradiction.


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H 2.11.5 – (a) Find α, β ∈ A5 such that α and β are conjugate in S 5 but not in A5 . (b) Find all conjugacy classes in A5 and the cardinalities of each.

(a) Let α = (12345) and β = (21345). They are both in A5 because they are 5-cycles which, by 2.10.9, are

even permutations. Furthermore, we easily see that (12) α (12) = β so that conjugation by (12)

∈ A5 brings

α into β . Now suppose σ, τ are such that both σασ − 1 = β and τ ατ − 1 = β . Then we have τ − 1σασ − 1τ = α ,

i.e. that τ − 1σ ∈ N (α). For this particular α = (12345), we are very familiar with its normalizer. From2.11.1c, we see that N (α ) = {(12345) i | i ∈ {0, 1, 2, 3, 4}}. Notice that N (α) ≤ A5 . Now put σ = (12) andlet τ ∈ S 5 be any other solution to τ ατ − 1 = β . We have that τ − 1(12) ∈ A5 . If τ ∈ A5 , then we would have(12) ∈ A5 , a contradiction. Therefore τ ∈ A5 so we have proved that this α and β are conjugate in S 5 butnot in A5 .

Notice that the same can’t be done for 3-cycles. That is, all 3-cycles are conjugate in A5 . Take, for example,

α = (123) and β = (234). We have (1234)(123)(4321) = (234), but if σ is another element of N ((123)),

then, by the above comments, σ− 1(1234) = (123) i (45) j with i ∈ {0, 1, 2} and j ∈ {0, 1}. Solving for σyields σ = (1234)(123) i (45) j = (14)(13)(12)(123) i (45) j which may be in A5 if i = 0 and j = 1. Then wehave σ = (14)(13)(12)(45) = (12345) ∈ A5 . Indeed, (12345)(123)(54321) = (234), so the two 3-cycles areconjugate in A5 . More generally, α = ( abc) is conjugate to β = ( bcd) by (abcdf ) ∈ A5 and α = ( abc) isconjugate to β = ( cdf ) by (acfbd ) ∈ A5 . In summary, all 3-cycles are conjugate in A5 , with

e(abc)e− 1 = ( abc) (abcdf )(abc)(abcdf )− 1 = ( bcd) (acfbd )(abc)(acfbd )− 1 = ( cdf ).

(b) The possible cycle structures for elements of A5 are

{1, 1, 1, 1, 1

},

{1, 2, 2

},

{1, 1, 3

}, and

{5

}. The rst

cycle structure has only one element: the identity. The conjugacy class of the identity is trivially C (e) = {e}with only one member.

Pick the element (12)(34) with structure {1, 2, 2}. Let’s investigate its conjugacy class: in 2.11.2a, weconsidered the elements which commute with (12)(34), however we now restrict our attention to those

commuting elements which live in A5 . The full normalizer in S 5 can be seen by the remarks of 2.11.2a to

be N S 5 ((12)(34)) = {e, (12) , (34) , (12)(34) , (13)(24) , (14)(23) , (1324), (1423)}. Half of these elements are oddpermutations, so the normalizer of interest is N A 5 ((12)(34)) = {e, (12)(34) , (13)(24) , (14)(23) }. Now theconjugacy class of (12)(34) has 60 / 4 = 15 elements, i.e. |C A 5 ((12)(34)) | = 15. At this point, it’s naturalto wonder if all elements of structure {1, 2, 2} share this conjugacy class. In S 5 , they certainly do, but herewe only allow conjugation by even permutations. As seen in (a), this is a nontrivial restriction. In fact, it

is true that all elements of structure {1, 2, 2} are conjugate to one another under A5 . This may be provenby a direct argument of the type given for 3-cycles at the end of the discussion of (a). However, we will

momentarily prove it indirectly by ignoring the issue for now.

Elements with structure {1, 1, 3} are 3-cycles. As per 2.11.1c, N S 5 ((123)) = {(123) i (45) j | i ∈ {0, 1, 2}, j ∈


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{0, 1}}. We only consider N A 5 = N S 5 ∩A5 = {(123) i | i ∈ {0, 1, 2}}. From this we see that |C ((123)) | =|A5|/ |N A 5 ((123)) | = 60/ 3 = 20 is the size of the conjugacy class of (123). Furthermore, by the discussion atthe end of (a), we know that, for any 3-cycle σ ∈ S 5 , C A 5 (σ) = C A 5 ((123)) because all 3-cycles are conjugatein A5 .

The only remaining cycle structure is {5}, the 5-cycles. We saw above that (12345) and (21345) are notconjugate in A5 , so they spawn distinct conjugacy classes. Specically, N A 5 ((12345)) = {(12345) i | i ∈{0, 1, 2, 3, 4}} and N A 5 ((21345)) = {(21345) i | i ∈ {0, 1, 2, 3, 4}} as discussed in (a), so that |C A 5 ((12345)) | =|C A 5 ((21345)) | = 60/ 5 = 12.Now we have demonstrated 5 distinct conjugacy classes of total size 1 + 15 + 20 + 12 + 12 = 60 = |A5|.Therefore this constitutes a complete collection of conjugacy classes. As promised, this indirectly proves

that all permutations of cycle type {1, 2, 2} are conjugate to one another in A5 .

H 2.11.6 – Let N G. (a) Let a ∈ N and prove that every conjugate of a in G is also in N . (b) Prove that |N | = [G : N (a)] for some choices of a ∈ N . (c) Prove that A5 is simple.(a) Let g ∈ G. Then gag− 1∈ N because N is normal.(b) For a, b ∈ N , a ∼ b if a = gbg− 1 for some g ∈ G is trivially an equivalence relation. These equivalenceclasses (“conjugacy classes”) partition N . The conjugacy class of element a ∈ N has order [G : N (a)] bytheorem 2.h, where N (a) still refers to the normalizer of a with respect to G. If we take a single representative

a from each conjugacy class and add up [ G : N (a)], we will nd |N |.Alternately, consider the conjugacy classes C (a) = {gag − 1 | g ∈ G} of G. If C (a) ∩N = {e}, then thereexist g ∈ G and n ∈ N such that gag− 1 = n ∈ N . Now a = g− 1ng ∈ N , so in fact gag− 1∈ N for all g ∈ Gbecause N is normal. Therefore C (a) ⊂ N or C (a) ∩N = {e}. Apply the class equation to G ∩N = N tond |N | = a |C (a) ∩N |. Those a producing C (a) disjoint from N may be excluded from the sum becausethey contribute 0.

(c) Let N A5 be non-trivial. The order of N must be one of {2, 3, 4, 5, 6, 10, 12, 15, 20, 30}. The ve distinctconjugacy classes of A5 have sizes 1, 12, 12, 15, 20 as per 2.11.5b. By (b), the possible orders under 30 for

N are then {1, 13, 16, 21, 25, 28} (notice that N must contain the identity, so the conjugacy class of size 1must always be included). This list does not intersect with the list of orders allowed by Lagrange’s theorem.Therefore A5 has no non-trivial normal subgroups.

H 2.11.7 – Let n ∈ Z+ . Prove that if |G| = pn then G has a subgroup of order m for all 0 ≤ m < n .This statement is trivial for n = 1. Assume the result to be true for n −1. Because G is a p-group, it has a


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non-trivial center Z . Therefore p | |Z | so, by Cauchy’s theorem, there exists an element a ∈ Z with order p.The subgroup A = a is normal in G because it is a subgroup of the center, so consider the quotient groupḠ = G/ a of order pn − 1 . By induction, Ḡ contains a subgroup H̄ of order pn − 2 .

Dene the quotient map φ : G

→ Ḡ by φ(g) = g A and let g1 , g2

∈ G. φ is well-dened: if g1 A = g2 A then

g− 12 g1 ∈ A, so φ(g1) = g1 A = g2g− 12 g1 A = g2 A = φ(g2). φ is a homomorphism: φ(g1g2) = g1g2 A =(g1 A )(g2 A ) = φ(g1)φ(g2). Now dene H = {g ∈ G | φ(g) ∈ H̄ }. Because φ is a homomorphism, H is asubgroup of G. φ restricts to a homomorphism ψ = φ|H : H → Ḡ. We see that ker( ψ) = A and im( ψ) = H̄ ,so by theorem 2.d, H̄ ∼= H/A . Now it is clear that |H | = |H̄ | · |A| = pn − 1 so that G has a subgroup of order pn − 1 . By induction, this subgroup contains subgroups of orders pm for all 0 ≤ m < n , so the result isproven.

As a tangential result to be used in 2.11.8, we can prove that H G, i.e. that a group of order pn has a

normal subgroup of order pn − 1 . We will proceed again by inducting on n . We can use n = 1 as the trivial

base case (though 2.9.5 shows explicitly that the claim holds for n = 2). Assume the result for n−1 and againconsider the subgroup H = {g ∈ G | φ(g) ∈ H̄ }. Let g ∈ G and h ∈ H . Then φ(ghg− 1) = φ(g)φ(h)φ(g)− 1 isin H̄ by induction (because H̄ is a normal subgroup of order n −2 in order n −1 Ḡ)

Lemma 4 – Let G be nite and let p be the smallest prime dividing |G|. Let H ≤ G be of index p. Prove that H G.

Suppose that H is not normal, so that N (H ) = G. We still have H ≤ N (H ), so we must have N (H ) = H by order considerations. Let G act by conjugation on G/H , i.e. dene φ : G

→ S (G/H ) by (φ(g))( γH ) =

gγg− 1H for γ ∈ G. φ is a homomorphism: ( φ(g)φ(h))( γH ) = ghγh − 1g− 1H = φ(gh). Consider its kernel:ker φ = {k ∈ G | kγk − 1H = γH for all γ ∈ G}. We see that if k ∈ ker φ then kγk− 1γ − 1∈ H for all γ ∈ G.Then let γ ∈ H and it must be that kγk− 1 ∈ H , so k ∈ N (H ). However, we assumed that N (H ) = H , sothis implies that k ∈ H . Therefore ker φ ≤ H .Theorem 2.d tells us that im( φ) ∼= G/ ker φ so that [ G : ker φ] = |im(φ)| divides p! because im(φ) is a subgroupof S (G/H ), a group of order p!. Counting in another way, we have that [ G : ker φ]· |ker φ| = |G|, which is thestatement that [ G : ker φ] | |G|. Now there exist a,b,c,d ∈ Z such that a[G : ker φ] = p!, b[G : ker φ] = |G|and cp!+ d|G| = ( p!, |G|). Plugging the rst two into the third yields ( ac + bd)[G : ker φ] = ( p!, |G|). Therefore[G : ker φ] | ( p!, |G|). Because p| p! and p||G|, and no smaller (non-unit) integer divides |G|, it must be that( p!, |G|) = p. Finally, we can conclude our argument: we have that [ G : ker φ] | p, and, because ker φ ≤ H ,[G : ker φ] = [G : H ][H : ker φ] = p[H : ker φ]. Then [G : ker φ] = 1 is clearly forbidden, while [ G : ker φ] = p

implies that H = ker φ which is normal as it is the kernel of a homomorphism. This is a contradiction of

our assumption that H isn’t normal. Thus it must be the case that H is normal.


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H 2.11.8 – Let n ∈ Z+ . Prove that if |G| = pn then there exists r ∈ Z and subgroups N i , i ∈ {0, 1, . . . , r }such that {e} = N 0 ≤ N 1 ≤ ·· · ≤ N r − 1 ≤ N r = G where N i N i+1 and N i +1 /N i is abelian. (i.e. p-groups are “solvable”)

Put r = n. By 2.11.7, N n = G has a subgroup N n − 1 of order pn − 1 . As this subgroup has index p, lemma 4

gives that it is normal. Furthermore, N n /N n − 1 has order p, so it is cyclic and therefore abelian. Iterating,

we construct N i− 1 from N i by taking a normal subgroup of order pi − 1 from the order pi group N i . This

generates the desired tower of subgroups. Note that this is not the only such tower.

H 2.11.9 – Let n ∈ Z+ . Let |G| = pn and let H ≤ G be proper. Prove that N (H ) = H .As H ≤ N (H ), this is the statement that a proper subgroup of a p-group is properly contained in itsnormalizer. Induct on the exponent n . For n = 1, the result holds because the only non-trivial subgroup is

{e}

, which is normalized by all of G. Therefore assume that, in group of orders pm , m < n , all subgroups

are properly contained in their normalizers.

In the group G of order pn , consider a proper subgroup H and let Z = Z (G) be the center of G, which is

non-trivial because G is a p-group. If Z is not a subgroup of H , then, because Z ≤ N (H ) (regardless of H ), we have H < Z, H properly while Z, H ≤ N (H ). Therefore we may restrict our attention to thecase that Z ≤ H . As the center is always a normal subgroup, consider the map φ : G → G/Z given byφ(g) = gZ . This map is a well-dened homomorphism (see 2.11.7, for instance). Because |Z | > 1, the groupG/Z is a p-group of order strictly less than pn . Note that φ(H ) < G/H properly, for if gZ = hZ for some

h

∈ H , g

∈ G, then gh− 1

∈

Z

≤ H implies that g

∈ H . Therefore gZ

∈ φ(H ) for any g

∈ G

\H . By

induction, φ(H ) is properly contained in its normalizer N (φ(H )) ≤ G/Z . Therefore there exists x ∈ G\H such that ( xZ )(hZ )(xZ )− 1 ∈ φ(H ) for all h ∈ H . But then for each h ∈ H there exists h1 ∈ H such thatxhx − 1Z = h1Z , implying that xhx − 1h− 11 ∈ Z ≤ H , i.e. that xhx − 1∈ H . Thus we have an x ∈ N (H ) whilex ∈ H , so the claim is proven.

H 2.11.10 – Let n ∈ Z+ . Let |G| = pn and let H ≤ G have order pn − 1 . Prove that H G.H has index p, so by lemma 4, H G.

H 2.11.11* – Let n ∈ Z+ . Let N G be non-trivial. Prove that N ∩Z = {e}.Let a1 , . . . , a k be representatives of the conjugacy classes of G, ordered such that a1 , . . . , a m ∈ N andam +1 , . . . , a k ∈ N . Recall, as was argued in 2.11.6b, that the conjugacy classes C (a i ) have either C (a i ) ⊂ N or C (a i ) ∩N = {e}. First arrange the {a1 , . . . , a m } so that the rst r represent conjugacy classes of size 1


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(i.e. elements in N ∩Z ) and the latter m −r represent classes of size larger than 1. Then we can write theclass equation for G ∩N = N as

|N | =m

i=1|C (a i ) ∩N | = |N ∩Z |+

m

i = r|C (a i )| = |N ∩Z |+

m

i = r

|G||N (a i )|

.

As |N | < p n , every term in the sum is divisible by p, hence |N ∩ Z | = |N | − |G || N (a i ) | is divisible by p,implying that N ∩Z is non-trivial, as desired.As an aside, the equivalence |C (a)| = |G|/ |N (a)| may be proven by considering the map f : G/N (a) → C (a)given by f (xN (a)) = xax − 1 . f is well-dened: if xN (a) = yN (a) for x, y ∈ G, then x− 1y ∈ N (a) whencex− 1yay − 1x = a. This gives that yay− 1 = xax − 1 , so f (xN (a)) = f (yN (a)). f is injective: by almost the

same argument, if f (xN (a)) = f (yN (a)) for x, y ∈ G then xax − 1 = yay− 1 so that x− 1yay − 1x = a, orxN (a) = yN (a). Because f is a map between nite sets, this proves that f is a bijection, so |C (a)| =|G|/ |N (a)|.

H 2.11.12 – If G/Z is cyclic, prove that G is abelian.

By lemma 2.19, G/Z ∼= Inn(G). Say Inn( G) = φx where φx (g) = xgx− 1 . Now let a, b ∈ G. Forsome r, s ∈ Z, we have φa = φrx and φb = φsx . Consider the commutator aba− 1b− 1 . We can cast this asφa (b)b− 1 = φrx (b)b− 1 = xr bx− r b− 1 . Once again, rewrite this as x r φb(x− r ) = xr φsx (x− r ) = xr xs x− r x− s = e.

Therefore aba− 1b− 1 = e for arbitrary a, b ∈ G so G is abelian.Alternately, let G/Z = xZ and let a, b ∈ G. Then there exist r, s ∈ Z with aZ = xr Z and bZ = xs Z .Therefore a = x

r

z1 and b = xs

z2 for some z1 , z2 ∈ Z . Now ab = xr

z1xs

z2 = x(r + s )

z1z2 while ba =xs z2xr z1 = x( r + s ) z1z2 by virtue of z1 and z2 being in the center. Therefore ab = ba for arbitrary a, b ∈ G,so G is abelian.

H 2.11.13 – If |G| = 15, prove that G is cyclic.By Cauchy’s theorem, there exists f ∈ G with order 5. By lemma 4, N = f G because [G : N ] = 3 isthe smallest prime dividing 15. Consider the action of G on N by conjugation, ψ : G → Aut( N ) given byψ(g) = σg where σg (x) = gxg− 1 . Notice that N must be normal so that G may act on it by conjugation. ψ

is a homomorphism: ψ(gh)(x) = σgh = ghxh − 1g− 1 = ( ψ(g) ◦ψ(h))( x). Because N is cyclic, we can easilycount the order of Aut( N ). If N = x for some x ∈ N , then every automorphism η ∈ Aut( N ) sends xto another generator, and that choice of η(x) uniquely determines the entire map because η(xn ) = η(x)n

suffices to compute the image of any other element xn ∈ N . By 2.5.15, there are φ(5) = 4 (Euler totient

function) generators of N , so 4 choices of where to send x and hence |Aut( N )| = 4.Now, by Lagrange’s theorem, |ψ(G)| divides |Aut( N )| = 4. Also, by theorem 2.d, |ψ(G)| divides |G| = 15 (in


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particular, |ψ(G)| · |ker ψ| = |G|). Therefore |ψ(G)| = 1, i.e. ker ψ = G so that the inner automorphism forevery g ∈ G xes every element of N . Let t ∈ G be an element of order 3 (guaranteed to exist by Cauchy’stheorem). Then tf t − 1 = f because ψ(t) = id, so tf = f t . Now we see that tf has order lcm(3 , 5) = 15, so,

in fact, G = tf is cyclic.

H 2.11.14 – If |G| = 28, prove that G has a normal subgroup of order 7.By Cauchy’s theorem, there exists x ∈ G with order 7. By lemma 2.21, a subgroup of order 7 in G must benormal because |G| = 28 24 = 4!. Therefore x is a normal subgroup of G with order 7.

H 2.11.15 – If |G| = 28 and G contains a normal subgroup F of order 4, prove that G is abelian.By 2.11.14, there exists S G with

|S

| = 7. By Lagrange’s theorem, S

∩ F =

{e

}, so SF has order

|S ||F |/ |S ∩F | = 28 and therefore SF = G. Because F is normal by assumption, 2.6.12 gives that sf = f sfor all s ∈ S , f ∈ F . Furthermore, groups of order 4 and 7 are both necessarily abelian. Then considerg1 = s1f 1 ∈ G and g2 = s2f 2 ∈ G. We have g1g2 = s1f 1s2f 2 = s2f 2s1f 1 = g2g1 and hence G is abelian.

H 2.12.1 – Let G = S 4 . Exhibit a 2-Sylow subgroup and a 3-Sylow subgroup of G.

|G| = 24 = 2 3 · 3, so a 3-Sylow subgroup has order 3 while a 2-Sylow subgroup has order 8. A 3-Sylowsubgroup is (123) = {e, (123), (213)}. A 2-Sylow subgroup is (1234), (14)(23) ∼= D4 . As dened in 2.7.8,a presentation of D4 is x, y | x

2

= y4

= e and xy = y− 1

x . We see that ((14)(23))2

= (1234)4

= e and(14)(23)(1234) = (123) = (4321)(14)(23) so that x = (14)(23) and y = (1234) does provide a set of generators

for an order 8 subgroup of S 4 isomorphic to D 4 .

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