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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 E B” denotes that A is a normal subgroup of B.

H 1.3.14* (Fermat’s Little Theorem) – Prove that if a ∈ Z then ap≡ a mod p.

Proceed by induction on (positive) integer a. The theorem holds for a = 1 because 1p = 1 ≡ 1 mod p.

Suppose that kp≡ k mod p and then compute, by the binomial theorem,

(k + 1)p= p X i=0 p i  ki1p−i= kp+ 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= kp+ 1 + p(. . .) is justified by lemma 2 (below) which states that p | p

n for n ∈ {1, . . . , p − 1}.

The case a = 0 is trivial: 0p= 0 ≡ 0 mod p. Let a ∈ Z+ as before. As was just proven, there exists c ∈ Z such that ap− a = cp. Now, if p > 2 (odd), we have (−a)p− (−a) = −ap+ 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 n r =


r!(n−r)! is an integer.

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

1 = 1 are both integers, as claimed. Suppose

that n−1r  is an integer for all r ∈ {0, . . . , n − 2}. Now n − 1 r  +n − 1 r − 1  = (n − 1)! r!(n − r − 1)! + (n − 1)! (r − 1)!(n − r)! = (n − 1)!  (n − r) + r r!(n − r)!  =n r  .

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

n 0  = n! 0!n!= 1 n n  = n! n!0! = 1 and the claim is proven.

Lemma 2 – Let n ∈ {1, . . . , p − 1}. Then p | np.

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 qai


divisors of n!(p−n)! and aitheir respective powers. As every factor of n(n−1) · · · 2·1·(p−n)(p−n−1) · · · 2·1

is 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 qi is p. The factor of p in the numerator is then preserved upon taking the

quotient, and p | np. Note that it makes no sense to talk about p dividing p

n unless p

n ∈ 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. Set x = 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 left

by 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 = aibi for all a, b ∈ G. Prove that G

is abelian.

Let N be the smallest of the three consecutive integers. Then we have that (ab)N = aNbN, (ab)N +1 =

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

the second equation implies

ab = aN +1bN +1b−Na−N = aN +1ba−N hence baN = aNb. Inverting the second equation and right multiplying it to the third equation gives

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

Therefore aN +1b = baN +1= baNa = aNba 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 finite. Prove the existence of an N ∈ Z such that aN = e for all a ∈ G.

Let a ∈ G. As G is finite and closed under multiplication, the set {a0, a1, a2, . . .} is finite. Hence there exist


can write m = kn + r with k, r ∈ Z and 0 ≤ r < n. Now we have the statement akn+r= anor a(k−1)n+r= e. Enumerate G as {a1, . . . , an}. For each ai, there exists an Ni = (ki− 1)ni+ ri, computed by the above

method, such that aNi

i = e. Let N =Q Ni. Then aNi = (a Ni

i )

N/Ni = eN/Ni = 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 identity commutes with everything. Suppose ab 6= 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 6= e. (2) If ab = a, then b = e so ba = a = ab, which is a contradiction. Hence ab 6= a. (3) If ab = b, then a = e so ba = b = ab, which is a contradiction. Hence ab 6= 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 6= a in the group. Elements of this type can be counted in pairs {a, a−1}. There are therefore an even number 2k of elements with a2 6= e. The identity satisfies e2 = e, so there are |G| − 2k − 1 ≥ 0 non-identity elements a

satisfying a2= e. If |G| is even, then |G| − 2k − 1 is non-zero and hence there exists a non-identity element

a ∈ 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.


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. Hence ab ∈ H ∩ K. H ∩ K is closed under inverses: let a ∈ H ∩ K. Then a−1∈ H, K, and hence a−1∈ H ∩ K. By lemma 2.3, H ∩ K ≤ G.

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

Define f : {aH | a ∈ G} → {Ha | a ∈ G} by f (aH) = Ha−1. This map is well-defined: suppose a1, a2 ∈ G

are 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, Ha is the image of a−1H. The

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

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

proves injectivity.

H 2.5.3 – Let G have no proper subgroups. Prove that |G| is prime.

Let g ∈ G with g 6= e. hgi is a subgroup of G. Because g 6= e, hgi is not the trivial subgroup {e}. Hence, because G has no proper subgroups, it must be that hgi = G which gives |g| = |G|. Suppose |G| = mn with m, n > 1 (so m, n < |G|). Then (gm)n= e which implies that |gm| = n < |G|. The subgroup hgmi 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 finite indices in G. Give an upper bound for the index of H ∩ K. Missing.

H 2.5.7 – With a, b ∈ R, let τab: R → R be given by τab= ax + b. Let G = {τab| a 6= 0}. Prove that G is a


(τab◦ τcd)(x) = a(cx + d) + b = acx + ad + b = τac,ad+b(x). As R is a field and a, c 6= 0, τac,ad+b ∈ G so that

G is closed under composition. The operation is associative:

(τab◦ τcd) ◦ τef = τac,ad+b◦ τef = τace,acf +ad+b= τab◦ τce,cf +d= τab◦ (τcd◦ τef).

The set is closed under inverses: τab−1= τ1

a,−ba ∈ G. Finally, there is an identity element e = τ1,0. Therefore

G is a group. Note that G is non-abelian because τab◦ τcd= τac,ad+b6= τac,cb+d= τcd◦ τab.

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

of H in G.

Because Q is a field, we have that, for σab, σcd∈ H, σab◦ σcd= σac,ad+b∈ H and σab−1 = σ1 a,−


a ∈ H. By

lemma 2.3, H ≤ G.

Now let τab, τcd ∈ G be such that Hτab = Hτcd. Then τab◦ τcd−1 = τa c,b−


c ∈ H. Therefore


c ∈ Q is the

condition on this situation. In particular, τaband τacare in the same right coset regardless of b and c, which

tells us that the right cosets are indexed by just one real parameter.

{Hτab} = {{τcd∈ 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 τab, τcd∈ G be such that τabH = τcdH. Then τab−1◦ τcd= τc a,


a ∈ H. Again, this is the statement

that ac ∈ Q, so two elements of G are equivalent left-modulo H if the ratio of their first parameters is rational. Therefore consider the cosets τabH and Hτab. If τcd ∈ τabH then ca ∈ Q. Hence ac ∈ Q which gives that

τcd∈ Hτab. Therefore τabH ⊂ Hτab. The reverse inclusion is identical, so τabH = Hτab.

(b) Consider G = S3= {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 2.5.12 – 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 the equation ga = ag on the left and right by g−1 to find 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) = T


N (a) ≤ G.

H 2.5.14 – Let G = hgi 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 n0such

that gn0 ∈ H (remember H is closed under inverses, so it can’t have just negative powers of g). The claim is

that H = hgn0i. It is clear, because H is a group containing gn0, that hgn0i ≤ H. Let gk ∈ H be arbitrary.

By the division algorithm, we can write k = qn0+ r with q ∈ Z and r ∈ {0, 1, . . . , n0− 1}. Now, by closure,

gkg−qn0 = gr ∈ H. By our assumption that n

0 is the smallest positive exponent in H, we must conclude

that r = 0. Therefore n0| k and we see that every element of H is a power of gn0, whence H ≤ hgn0i. This

proves the claim, so H = hgn0i 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 = hgi. The claim is that gm is also a generator if and only if m is relatively prime to n. If m is relatively prime to n, then there exist a, b ∈ Z such that am + bn = 1. Now g = gam+bn = (gm)a(gn)b = (gm)a which tells us that g ∈ hgmi so G = hgi ≤ hgmi ≤ G. Hence m, n

relatively prime implies that gmgenerates G. On the other hand, if G = hgmi, then g ∈ hgmi, so that for

some 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 = hgmi 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.


e = am= aq|a|+r = (a|a|)qar= ar. As r < |a| and |a| is the minimal exponent taking a to the identity, we must 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 anba−n= b2n: suppose this is

true for n and compute

an+1ba−(n+1) = a(anba−n)a−1= ab2na−1= b2n+1.

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

Therefore a5ba−5= b32, but, because a5= a−5= e, the left hand side is simply b. We have, finally, 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 finite, 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 E 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(ah2a−1) ∈ H(aHa−1) so (Ha)(Ha−1) ⊂ H(aHa−1); conversely,

h1(ah2a−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 aHa−1 contains e = aea−1, so in fact H(aHa−1) = He = H which implies that aHa−1 ⊂ H, i.e. that H E G.

H 2.6.2 – Let H ≤ G have index 2. Prove that H E G.

Let g ∈ G\H. The right cosets of H in G may be enumerated as {H, Hg}. Because distinct cosets are disjoint, Hg 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 would be gH = H, but that isn’t the case: ge = g 6∈ H while ge ∈ gH.), so that gH = Hg. By lemma 2.10, H E G because every one of its right cosets in G is also a left coset in G.


Let n1, n2∈ 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 = n0h−1 ∈ N H, again using the existence of n0 ∈ N such that

n0= h−1n−1h. Therefore, by lemma 2.3, N H is a group.

H 2.6.4 – Let M, N E G. Prove that M ∩ N E 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 and

gxg−1∈ N so that gxg−1 ∈ M ∩ N . Therefore g(M ∩ N )g−1⊂ M ∩ N and hence M ∩ N E G.

H 2.6.5 – Let H ≤ G and N E G. Prove that H ∩ N E H.

Let x ∈ H ∩ N and h ∈ H. We have hxh−1 ∈ H because it is a product of three elements of H. We also have that hxh−1∈ N because N is normal and x ∈ N . Therefore hxh−1∈ H ∩ N so H ∩ N E 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 E 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 with

everything, and i2= j2= k2= ijk = −1. First observe that Q

8is non-abelian: ij = (ij)(−k2) = (−ijk)k =

k while ji = j−1i−1= (ij)−1= k−1= −k.

We can enumerate all the subgroups of Q8by considering subgroups generated by combinations of elements.

First of all, we have the trivial subgroups h1i = {1} and hi, j, ki = Q8of orders 1 and 8 respectively. Both are

trivially normal. Next, the subgroup h−1i = {−1, 1} is of order 2. Finally, the subgroups hii, hji, hki of order 4 round out the list. We can see that to be the case by noting that −1 is redundant as a generator if we include any 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, hi, ji = {1 = i4, −1 = i2, i, −i = i3, j, −j = j3, k = ij, −k = ij3} = Q8.

It is easy to see that h−1i is normal because its elements commute with everything: ih−1i = {i, −i} = h−1ii, 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, hii, hji, hki are all normal. Therefore all subgroups of Q8 are normal, but, as displayed above,


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 finite and let H ≤ G be the only subgroup of G of order |H|. Prove that H E G. The map f : H → gHg−1 given by f (h) = ghg−1 is a bijection. Injectivity: f (h1) = f (h2) implies

gh1g−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 subgroup

has the same cardinality as H, by assumption. If H is the only subgroup of order |H|, then conjugation fixes H, i.e. gHg−1= H for all g ∈ G. This is the condition that H E G. The restriction to finite G is likely just

to allow Herstein’s use of the word “order”.

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

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

H 2.6.12* – Let M, N E 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 E G and n ∈ G, there exists m0 ∈ M such that m0 = nmn−1. Hence nmn−1m−1 = m0m−1 ∈ M . On the other hand, because N E G and m ∈ G, there exists n0 ∈ N such that n0 = mn−1m−1. Then nmn−1m−1 = nn0 ∈ N . Now we must have that

nmn−1m−1∈ M ∩ N which is assumed to be trivial. Hence nmn−1m−1= e or, multiplying on the right by

m and then n, nm = mn.

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

Say T = hti. By 2.5.14, a subgroup S of T is again cyclic, so let S = htmi. Because T is normal, there

exists a k0 such that gtkg−1 = tk0 ∈ T (k0 depends on k and g). Then consider an element (tm)k of S.

g(tm)kg−1= (gtkg−1)m= (tk0)m= (tm)k0 ∈ S. Therefore gSg−1

= S and S E G.

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


H 2.6.15 – Let N E G. For a ∈ G, prove that |N a| in G/N divides |a| in G.

We have (N a)|a|= N (a|a|) = N e = N , the identity in G/N . By 2.5.16, |N a| divides |a|.

H 2.6.16 – Let G be finite and let N E 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 = (N x)[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. Now x = 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) = 2x

. (c) G = R under addition, φ : G → G given by φ(x) = x + 1. (d) G as in (c), φ : G → G given by φ(x) = 13x. (e) G any 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 ∈ G

such that φ(x) = x2= 1. Hence ker φ = {−1, 1}.

(b) No: φ(2 · 3) = 26= 64 while φ(2)φ(3) = 22· 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) = 13x + 13y = φ(x) + φ(y). The identity in G is 0, so the kernel is the set of those x ∈ G such that φ(x) = 13x = 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 fixed 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 finite and let n ∈ Z be relatively prime to |G|. Prove that every x ∈ G may be written as g = xn with x ∈ G.


Herstein assumes, additionally, that G is abelian and hints to consider the map φ : G → G given by φ(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 finite G and choice of n coprime to |G|. In that context, it makes sense to talk about a well-defined, unique nth root of a group element.

H 2.7.4 – Let U ⊂ G. Let hU i ≤ G be the smallest subgroup of G containing U . (a) Prove that such a hU i exists. (b) If gug−1∈ U for all g ∈ G and u ∈ U , prove that hU i E G.

(a) Let A be the collection of all subgroups V of G which contain U . G ∈ A, so A is not empty. hU i = T

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 containing U also contains hU i. In this sense, hU i fits the criterion.

(b) By lemma 3 (below), hU i is exactly the set of all finite products of elements of U and their inverses. Note that if gug−1= u0 ∈ U then taking the inverse gives gu−1g−1= u0−1∈ hU i. Now an arbitrary element

of hU i may be written as uk1

1 · · · uknn (with n ≥ 0 and ki= ±1) so conjugation gives

guk1 1 · · · u kn n g −1= guk1 1 g −1g · · · g−1gukn n g −1= u0k1 1 · · · u 0kn n ∈ hU i, where u0i= guig−1. Therefore hU i E G. Lemma 3 – Define V = {uk1 1 u k2

2 · · · uknn ∈ G | n ∈ {0, 1, · · · }, ui ∈ U, ki = ±1}, the set of all finite (or

empty, in which case the result is e) products of elements from U or their inverses. Then V = hU i.

Proof: As hU i is a subgroup containing U , it also contains all inverses of elements of U . Furthermore, it is closed under multiplication, so V ⊂ hU i immediately. Notice that V is both trivially closed under multiplication and non-empty (because when n = 0 we see that e ∈ V ). Furthermore, the inverse of uk1

1 · · · uknn is u−kn n· · · u −k1

1 which is again in V . By lemma 2.3, V ≤ G. Now U ⊂ V , so, by (a), hU i ≤ V .

Therefore V = hU i and the claim is proven.

H 2.7.5 – Let U = {xyx−1y−1 | x, y ∈ G}. Define G0 = hU i (the commutator subgroup of G). (a) Prove

that G0E G. (b) Prove that G/G0 is abelian. (c) If G/N is abelian, prove that G0 ≤ N . (d) Prove that, if H ≤ G0 then H E 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, G0 = hU i E G.


(b) We have, for x, y ∈ G, that (G0x)(G0y)(G0x−1)(G0y−1) = G0(xyx−1y−1) = G0. Right multiplying the equation by G0y and then by G0x gives that (G0x)(G0y) = (G0y)(G0x), so G/G0 is abelian.

(c) If G/N is abelian, then, for x, y ∈ G we have (N x)(N y)(N x−1)(N y−1) = (N x)(N x−1)(N y)(N y−1) = N (xx−1yy−1) = N . Hence xyx−1y−1 ∈ N . Then U ⊂ N and therefore G0 ≤ N by virtue of being the

smallest subgroup containing U .

(d) Let g ∈ G and h ∈ H. Because G0≤ H, we have that ghg−1h−1∈ H. Therefore ghg−1 ∈ H, so H E G.

H 2.7.6 – Let M, N E G. Prove that N M/M ∼= N/(N ∩ M ).

By 2.6.5, N ∩ M E N . It is trivial that M E N M because M E G and N M ≤ G. Therefore, with the knowledge that all quantities are meaningful, define φ : N M/M → N/(N ∩ M ) by φ(nmM ) = n(N ∩ M ) for n ∈ N , m ∈ M . The map is well-defined: suppose n1m1M = n2m2M for n1, n2 ∈ N and m1, m2∈ M .

In fact, this also means that n1M = n2M , so that n1 = n2m3 for some m3 ∈ M . However, also notice

that m3= n−12 n1∈ N , so m3∈ M ∩ N . Then n1(N ∩ M ) = n2m3(N ∩ M ) = n2(N ∩ M ) and the map is

well-defined as claimed.

φ is a homomorphism: let n1, n2 ∈ N and m1, m2 ∈ M . Then φ((n1m1M )(n2m2M )) = φ(n1M n2M ) =

φ(n1n2M ) = n1n2(N ∩ M ) = n1(N ∩ M )n2(N ∩ M ) = φ(n1m1M )φ(n2m2M ). Finally, the kernel of φ is

trivial: suppose φ(nmM ) = N ∩ M for n ∈ N and m ∈ M . This means that n ∈ N ∩ M ≤ M , i.e. we have nmM = M . Therefore φ is an isomorphism and the result is proven.

H 2.7.7 – For a, b ∈ R, let τab: R → R be given by τab(x) = ax + b. Let G = {τab| a, b ∈ R, a 6= 0} and let

N = {τ1b∈ G}. Prove that N E G and that G/N ∼= R\{0} under multiplication.

By 2.5.7, G is a group under composition, τab◦ τcd= τac,ad+b and τab−1= τ1

a,−ab. Then, with τab∈ G,

τabτ1cτab−1 = τa,ac+bτ1

a,−ba = τ1,ac ∈ N

so that N E G. Define φ : G/N → R\{0} by φ(τabN ) = a. φ is well-defined: suppose τabN = τcdN so that

τcd = τabτ1e for some e ∈ R. Then τcd = τa,ae+b so that c = a and φ(τcdN ) = c = a as required. φ is a

homomorphism: φ(τabN τcdN ) = φ(τabτcdN ) = φ(τac,ad+bN ) = ac = φ(τab)φ(τcd). The kernel of φ is trivial:

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


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


(a) hyi has order n, and D2n has order 2n, so hyi has index 2n/n = 2 in D2n. By 2.6.2, hyi E D2n.

(b) The order of D2n/hyi is 2, and the unique group of order 2, up to isomorphism, is Z2. More constructively,

φ : D2n/hyi → Z2 given by φ(xiyjhY i) = i is an isomorphism.

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

For g ∈ G, z ∈ Z(G), we have gzg−1= zgg−1= z ∈ Z(G), so Z(G) E 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 ∼= S3.

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 S3= {e, (12), (13), (23), (123), (213)}. It contains 2 elements of order 3

and 3 elements of order 2. Its multiplication table is:

S3 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, a2}.

If the order of b is 3, then {e, a, a2, 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, a2, b, b2}, so ab = ba. For elements u, v of an arbitrary group, uv = vu easily implies that


Therefore a and a2 are the only elements of order 3, and the remaining elements of G all have order 2: G = {e, a, a2, 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 configurations, it’s easy to construct the entire multiplication table for the two Gs: G1 e a a2 b c d e e a a2 b c d a a a2 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 a2 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 S3, we see that φ : G1→ S3given 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 S3. On the other hand, ψ : G2 → S3 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 S3. These were the only two possible non-abelian groups of order 6, and both are

isomorphic to S3, 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 7→ −x. (b) G = R+ under multiplication, T : x 7→ x2. (c) G cyclic of order 12, T : x 7→ x3. (d) G = S3, T : x 7→ 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 is injective: 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= y2so 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 = hgi. Then g4 6= e, but T (g4) = g12= e, so the kernel of T is non-trivial. Hence T is not


(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 E G, and let T be an automorphism of G. Prove T (N ) E 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 (n0) ∈ T (N ) for some n0 = T−1(g)nT−1(g−1) ∈ N because N is normal.

H 2.8.4 – Prove that Inn(S3) ∼= S3.

The center of S3 is trivial, as we can easily check: (12)(13) = (213) while (13)(12) = (123), so (12) 6∈ Z(S3)

and (13) 6∈ Z(S3). (23)(123) = (13) while (123)(23) = (12) so (13) 6∈ Z(S3) and (123) 6∈ Z(S3). Finally,

(12)(213) = (13) while (213)(12) = (23), so (213) 6∈ Z(S3). Therefore Z(S3) = {e} and, by lemma 2.19,

Inn(S3) ∼= S3/Z(S3) ∼= S3.

H 2.8.5 – Prove that Inn(G) E Aut(G).

Let g ∈ G, Tw : x 7→ wxw−1 ∈ Inn(G) and φ ∈ Aut(G). Then (φTwφ−1)(g) = φ(Tw(φ−1(g))) =

φ(wφ−1(g)w−1) = φ(w)gφ(w−1) = φ(w)gφ(w)−1 = Tφ(w)(g), where Tφ(w) : x 7→ φ(w)xφ(w)−1. Hence

φTwφ−1= Tφ(w)∈ Inn(G), which proves that Inn(G) E 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 fixes the identity, and permutes the three elements of order 2. Then it is clear that Aut(G) is isomorphic to a subgroup of S3. Furthermore, we can exhibit elements of order 2 and 3 in Aut(G)

which proves that Aut(G) ∼= S3 in its entirety.

For example, φ : G → G given by φ(a) = b, φ(b) = a and φ(ab) = ab is an automorphism of order 2. 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


Additionally, ψ : G → G given by ψ(a) = b, ψ(b) = ab and ψ(ab) = a is an automorphism of order 3. We can 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 6= id, but ψ3(a) = ψ(ab) = a, ψ3(b) = ψ(a) = b

and ψ3(ab) = ψ(b) = ab so, in fact, ψ3= id.

As a result, we know that Aut(G) must be isomorphic to a subgroup of S3 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 S3

itself. Hence Aut(G) ∼= S3.

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 Tg: x 7→ gxg−1. Because Tg∈ Aut(G), we have that gCg−1 = Tg(C) ⊂ C. This holds for arbitrary g ∈ G,

so C E G.

(b) A normal subgroup, N E G, is fixed by all inner automorphisms, by definition. In order for the converse to fail to hold, there need to be automorphisms outside of Inn(G) which don’t fix N . Consider G = V4, the

Klein group of 2.8.6, and N = hai = {e, a} which is normal because it is of index 2. Then take the order 2 automorphism φ of G, with φ(a) = b, φ(b) = a and φ(ab) = ab. We have φ(hai) = {e, b} 6= hai. Hence N is normal, but not characteristic because φ doesn’t fix it.

H 2.8.8 – Prove that the commutator subgroup G0= haba−1b−1| a, b ∈ Gi (see 2.7.5) is characteristic. Let φ ∈ Aut(G) and let a1b1a−11 b


1 · · · anbna−1n b−1n ∈ G0. Because φ is a homomorphism, we have

φ(a1b1a−11 b −1

1 · · · anbna−1n b −1

n ) = φ(a1)φ(b1)φ(a1)−1φ(b1)−1· · · φ(an)φ(bn)φ(an)−1φ(bn)−1 ∈ G0.

Therefore φ(G0) ⊂ G0, so G0 is characteristic.

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

Every inner automorphism fixes N but, a priori, could move M ≤ N around. Notice that if Tg ∈ Inn(G),

it restricts to an automorphism Tg|N of N because Tg(N ) = N . Now M is characteristic in N , so we have

Tg|N(M ) = M . This is simply a restriction of a map to the domain N , so, in fact, Tg(M ) = M as well.

Hence M is fixed by every inner automorphism, i.e. M E G.


of N (because Tg might throw some elements of N afield), and the argument breaks down. Also, if N were

normal but M merely normal (not characteristic) in N , the argument would also break down: Tg 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 Tg|N would not fix M if it were only normal but not characteristic. Therefore this

exercise represents a strengthening of the conditions of 2.6.14, where we see that normality of subgroups is not a transitive property.

H 2.8.10 – Let G be finite and let φ ∈ Aut(G) fix 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 that ψ(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 finite, ψ is also necessarily surjective (however, ψ is not a homomorphism in general, and this is immaterial). Therefore for every g ∈ G, there exists an x ∈ G such that g = ψ(x) = x−1φ(x).

H 2.8.11 – Let G be finite, and let φ ∈ Aut(G) fix 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 6= 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 7→ g−1 is an automorphism of G. ρ is 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 finite 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 finite, 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.


H 2.8.13 – Give an example of a non-abelian finite group G with an automorphism that maps exactly three quarters of the elements of G to their inverses.


H 2.8.14* – Let G be finite with |G| > 2. Prove Aut(G) is non-trivial.

If G is abelian, then inversion ρ : g 7→ g−1 is an automorphism (see 2.8.11). If ρ = id, then g = g−1 for all g ∈ G, i.e. all non-identity elements have order 2. In that case, construct the map φ : G → G which transposes two non-identity elements and fixes 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 6∈ Z(G). Then there exists g ∈ G with ag 6= ga, and therefore aga−1 6= g. Thus the inner automorphism Ta: g 7→ aga−1 of conjugation by a does not fix g, so Ta 6= id but

Ta∈ 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 6∈ H. Then xh 6∈ H for any h ∈ H. Furthermore, xh has order 2, because it is outside of H. The map φ : G → G given by φ(g) = xgx−1= xgx is an inner automorphism of G. Because H is normal (index 2), φ fixes H and hence restricts to an automorphism φ|H of H. Now we see that hφ(h) = hxhx = (hx)2= e, so

that 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∗an−1, the multiplicative (modulo an− 1) group of integers coprime to (an− 1). The number of

elements in Z∗an−1 is φ(an− 1). Furthermore, a is coprime to an− 1 because (an−1)a + (−1)(an− 1) = 1 so

that a ∈ Z∗an−1. The order of a is n because an ≡ 1 mod (an− 1) while, for 1 < k < n, 1 < ak < an− 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. Define λg: G → G by λg(x) = gx. Prove that λg is a bijection and that λgλh= λgh.


that gx = gy so x = y. Finally, if g, h ∈ G, then λgh(x) = ghx while λg(λh(x)) = λg(hx) = ghx. Therefore

λgh= λgλh. 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 defined as in 2.9.1. Define τg : G → G by τg(x) = xg. Prove that, for g, h ∈ G, we

have λgτh= τhλg.

Let x ∈ G. We have λg(τh(x)) = λg(xh) = gxh while τh(λg(x)) = τh(gx) = gxh. Hence λgτh= τhλg.

H 2.9.3 – Let λg and τg be defined as in 2.9.2. If θ : G → G is a bijection such that λgθ = θλg for all

g ∈ G, prove that θ = τh 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 find that θ(e) = x−1θ(x) for all x ∈ G. Solving for θ(x) yields θ(x) = xθ(e), which is the desired result. Specifically, we have θ = τθ(e).

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


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 under

multiplication. Additionally, the inverse of gh1g−1 is gh−11 g−1∈ gHg−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 exists h ∈ H such that w = ghg−1. Let x ∈ G, and consider xwx−1. For any g ∈ G, there exists h ∈ H such that 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 E 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 finite group, and H ≤ G is a proper subgroup such that |G| - [G : H]!, then H 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 the

conditions of lemma 2.21 are satisfied. 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 E G and |H| = p.


H 2.9.6* – Let |G| = p2 and let H E 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 has prime order, it is cyclic and H = hxi. Let y ∈ G. By lemma 2.21 (as used in 2.9.5), hyi E G. Lagrange’s theorem applied to hxi ∩ hyi (as a subgroup of hxi or hyi) tells us that it has order 1 or p. In other words, hxi = hyi = hxi ∩ hyi or hxi ∩ hyi = {e}. In the former case, hxi = hyi gives that x and y commute. In the latter case, hxi ∩ hyi = {e} invites the application of 2.6.12, which again gives that xy = yx. As x ∈ H and y ∈ 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), hgi is normal and, by 2.9.6, hgi ⊂ Z(G). In particular, g ∈ Z(G). This argument applies to arbitrary g ∈ G, so Z(G) = G, i.e. G is abelian.

H 2.9.8 – Let |G| = 2p. Prove G has a subgroup H of order p and that H E G.

If x ∈ G has order 2p, then x2has order p, so H = hx2i is a subgroup of order p. Otherwise, suppose there

is no element in G of order 2p. If x ∈ G has order p, then H = hxi is a subgroup of order p. Otherwise, suppose G has no elements of order p or 2p. 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 groups implies 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 6= q are both prime. Suppose H, K E G with |H| = p and |K| = q. Prove that G is cyclic.

Let H = hhi and K = hki so hp = e and kq = e. If x ∈ hhi ∩ hki then |x| | p and |x| | q, so |x| = 1 and

consequently hhi ∩ hki = {e}. By 2.6.12, hk = kh. Now consider the element hk. Its order must be 1, p, q or pq. As k 6∈ hhi, we cannot have k = h−1, so hk 6= e. Compute (hk)p = hpkp = kp 6= e because q - p. Similarly, (hk)q = hqkq = hq 6= e because p - q. Therefore |hk| = pq, and G = hhki 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


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.


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 modified 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, a2, . . . , am) = (a1, am)(a1, am−1) · · · (a1, a2). Therefore

(12 · · · n)−1= [(1n)(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 σi does to 12345678. The sequence goes

12345678−→ 81234567σ −→ 78123456σ −→ 67812345σ



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, m2, . . . , mk? (c) How do you find 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, . . . , σk be disjoint cycles. As disjoint cycles commute, (Qiσi) m =Q iσ m i . The order of Q iσi is

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

that (Q

iσi)N = e because mi| N for all i. For any smaller exponent K, there is at least one σiwith mi- K so that (Q

iσi)K 6= 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), find 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 fixed 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)


while x(a(1)) = a(3) 6= a(1). Hence a(1) must be brought on an orbit of length 2 by x, but it is clear that x does 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 fixed here, but x must fix three of its five 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 Sn= h(12 · · · n), (12)i.

Let σ = (12 · · · n), τ = (12) and U = hσ, τ i. The result of applying σ to n symbols is to rotate them all to the right by one spot. Similarly, applying σ−1 effects a left rotation by one. Compute στ σ−1by considering its action on 12 · · · n:

123 · · · nσ


−→ 234 · · · n1−→ 324 · · · n1τ −→ 1324 · · · n.σ

We see that στ σ−1 = (23). Computing σ2τ σ−2, a pattern becomes clear which suggests that σkτ σ−k =


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

1 2 3 · · · k + 1 k + 2 k + 3 · · · n − 1 n σ−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 n

from which we see that σ(k + 1, k + 2)σ−1 = (k + 2, k + 3). Therefore the claim is proven that σkτ σ−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 (1k) = (1, k − 1)(k − 1, k)(1, k − 1) where we use the just-discovered fact that (k − 1, k) ∈ U . Iterating this procedure, starting with k = 3 (because (1, k − 1) = (12) is given to be in U ), we find that (1k) ∈ 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) = (1a)(1b)(1a). Therefore, as every permutation may be decomposed into transpositions, every permutation is contained in U . Now Sn≤ U ≤

Sn, so this finalizes the proof that U = Sn.

H 2.10.12* – Prove that, for n ≥ 3, the subgroup Un 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 Un ≤ An. Every even permutation is the product of an

even 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 first case has no index shared, 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 ∈ Un. Therefore An ⊂ Un,

and so An= Un.

H 2.10.13* – Let N E 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 an index, 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.


and A2 don’t accomodate 3-cycles, so they are irrelevant. A3= h(123)i has order 3 so it contains no

non-trivial subgroups. A4is 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), as

explored above. Therefore in A4it 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 order

30 has index 2, so it would be normal by 2.6.2. Because A5 is simple, this is disallowed, so |H| 6= 30.

Suppose |H| = 20 so [A5 : H] = 3. By lemma 2.21, because |A5| = 60 - 3! = [A5 : H]!, H contains a

non-trivial normal subgroup of A5. However, as A5contains no non-trivial normal subgroups, we come to a

contradiction. Therefore |H| 6= 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 Sn, prove that there are r(n−r)!n! distinct r-cycles. (b) Find the number of conjugates

of (12 · · · r) in Sn. (c) Prove that, if σ ∈ Sn commutes with (12 · · · r), then σ = τ (12 · · · r)i with i ∈

{0, 1, . . . , r − 1} and τ ∈ Sn leaving all of {1, 2, . . . , r} fixed.

(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, for instance, we count the distinct 3-cycles (123) and (132) but, ignoring movement of the rth symbol, we don’t overcount (312) = (123) as unique. Thus the number of distinct r-cycles in Sn is (r − 1)! nr =r(n−r)!n! .

(b) All permutations of the same cycle decomposition are conjugate to one another. In particular, all r-cycles in Sn have the same cycle decomposition {1, 1, . . . , 1, r}, with n − r ones. Hence (12 · · · r) is conjugate to

every one of the r(n−r)!n! r-cycles in Sn.

(c) It is clear that (12 · · · r) commutes with its powers as well as with τ , which is disjoint from it. Therefore if A = {τ (12 · · · r)i| i ∈ {0, 1, . . . , r − 1}, τ fixing {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


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


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 Sn for n ≥ 4. (b) Determine N ((12)(34)).

(a) For σ ∈ Sn, we have σ(12)(34)σ−1= (σ(1)σ(2))(σ(3)σ(4)). If σ were to commute with (12)(34), then we

would 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 first 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 |Sn|/|N ((12)(34))| = n!/(8(n − 4)!) = 18n(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 to constitute 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 n2 n−2

2 /2 = 1

8n(n − 1)(n − 2)(n − 3).

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

H 2.11.3 – Prove that Sp contains (p − 1)! + 1 elements satisfying xp= e.

The identity element satisfies ep= 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 xp = e. Are there

others? No. Consider a non-identity element σ ∈ Sp. The order of σ is equal to the least common multiple

of the lengths of the cycles in its disjoint cycle decomposition by 2.10.6b. If it were the case that σp = 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 Sp, 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 σp= 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 finite 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 this were the case, then N (a) = G, which is a contradiction.


H 2.11.5 – (a) Find α, β ∈ A5such that α and β are conjugate in S5 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) 6∈ 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. From 2.11.1c, we see that N (α) = {(12345)i | i ∈ {0, 1, 2, 3, 4}}. Notice that N (α) ≤ A

5. Now put σ = (12) and

let τ ∈ S5 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 6∈ A5 so we have proved that this α and β are conjugate in S5 but

not 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 σ

yields σ = (1234)(123)i(45)j = (14)(13)(12)(123)i(45)j which may be in A

5 if i = 0 and j = 1. Then we

have σ = (14)(13)(12)(45) = (12345) ∈ A5. Indeed, (12345)(123)(54321) = (234), so the two 3-cycles are

conjugate in A5. More generally, α = (abc) is conjugate to β = (bcd) by (abcdf ) ∈ A5 and α = (abc) is

conjugate to β = (cdf ) by (acf bd) ∈ A5. In summary, all 3-cycles are conjugate in A5, with

e(abc)e−1= (abc) (abcdf )(abc)(abcdf )−1= (bcd) (acf bd)(abc)(acf bd)−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 first

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, we considered 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 S5 can be seen by the remarks of 2.11.2a to

be NS5((12)(34)) = {e, (12), (34), (12)(34), (13)(24), (14)(23), (1324), (1423)}. Half of these elements are odd

permutations, so the normalizer of interest is NA5((12)(34)) = {e, (12)(34), (13)(24), (14)(23)}. Now the

conjugacy class of (12)(34) has 60/4 = 15 elements, i.e. |CA5((12)(34))| = 15. At this point, it’s natural

to wonder if all elements of structure {1, 2, 2} share this conjugacy class. In S5, they certainly do, but here

we 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 proven

by 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, NS5((123)) = {(123)


{0, 1}}. We only consider NA5 = NS5 ∩ A5 = {(123)

i | i ∈ {0, 1, 2}}. From this we see that |C((123))| =

|A5|/|NA5((123))| = 60/3 = 20 is the size of the conjugacy class of (123). Furthermore, by the discussion at

the end of (a), we know that, for any 3-cycle σ ∈ S5, CA5(σ) = CA5((123)) because all 3-cycles are conjugate

in A5.

The only remaining cycle structure is {5}, the 5-cycles. We saw above that (12345) and (21345) are not conjugate in A5, so they spawn distinct conjugacy classes. Specifically, NA5((12345)) = {(12345)

i | i ∈

{0, 1, 2, 3, 4}} and NA5((21345)) = {(21345)

i| i ∈ {0, 1, 2, 3, 4}} as discussed in (a), so that |C

A5((12345))| =

|CA5((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 E G. (a) Let a ∈ N and prove that every conjugate of a in G is also in N . (b) Prove that |N | =P[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 equivalence classes (“conjugacy classes”) partition N . The conjugacy class of element a ∈ N has order [G : N (a)] by theorem 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 find |N |.

Alternately, consider the conjugacy classes C(a) = {gag−1 | g ∈ G} of G. If C(a) ∩ N 6= {e}, then there exist 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 ∈ G because N is normal. Therefore C(a) ⊂ N or C(a) ∩ N = {e}. Apply the class equation to G ∩ N = N to find |N | =P

a|C(a) ∩ N |. Those a producing C(a) disjoint from N may be excluded from the sum because

they contribute 0.

(c) Let N E A5be non-trivial. The order of N must be one of {2, 3, 4, 5, 6, 10, 12, 15, 20, 30}. The five distinct

conjugacy 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 1 must 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.


non-trivial center Z. Therefore p | |Z| so, by Cauchy’s theorem, there exists an element a ∈ Z with order p. The subgroup A = hai is normal in G because it is a subgroup of the center, so consider the quotient group


G = G/hai of order pn−1. By induction, ¯G contains a subgroup ¯H of order pn−2.

Define the quotient map φ : G → ¯G by φ(g) = ghAi and let g1, g2∈ G. φ is well-defined: if g1hAi = g2hAi then

g2−1g1∈ A, so φ(g1) = g1hAi = g2g−12 g1hAi = g2hAi = φ(g2). φ is a homomorphism: φ(g1g2) = g1g2hAi =

(g1hAi)(g2hAi) = φ(g1)φ(g2). Now define H = {g ∈ G | φ(g) ∈ ¯H}. Because φ is a homomorphism, H is a

subgroup of G. φ restricts to a homomorphism ψ = φ|H: H → ¯G. We see that ker(ψ) = A and im(ψ) = ¯H, so by theorem 2.d, ¯H ∼= H/A. Now it is clear that |H| = | ¯H| · |A| = pn−1 so that G has a subgroup of

order pn−1. By induction, this subgroup contains subgroups of orders pmfor all 0 ≤ m < n, so the result is


As a tangential result to be used in 2.11.8, we can prove that H E 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 again consider the subgroup H = {g ∈ G | φ(g) ∈ ¯H}. Let g ∈ G and h ∈ H. Then φ(ghg−1) = φ(g)φ(h)φ(g)−1 is

in ¯H by induction (because ¯H is a normal subgroup of order n − 2 in order n − 1 ¯G)

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

Suppose that H is not normal, so that N (H) 6= 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. define 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, so this 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 subgroup of S(G/H), a group of order p!. Counting in another way, we have that [G : ker φ] · | ker φ| = |G|, which is the statement 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 first 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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