Abstract

We sketch the solutions of just a few problems in Herstein’s book.

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Sketches of some problems in Herstein’s abstract

algebra

Ching-Lueh Chang

March 12, 2008

2.5.19 If H is a subgroup of finite index in G, prove that there is only a finitenumber of distinct subgroups of the form aHa−1.

Proof. Observe

aHa−1 = aHHa−1 = aH−1Ha−1 = (Ha−1)−1(Ha−1).

The second inequality follows from H = H−1.

2.5.20 If H is of finite index in G prove that there is a subgroup N of G,contained in H, and of finite index in G such that aNa−1 = N for all a ∈ G.Can you give an upper bound for the index of this N is G?

Proof. The set N = ∩g∈Gg−1Hg provably forms a normal subgroup of G.Clearly, N ⊆ H.

Let a ∈ G be arbitrary and Hg1, ...,Hgn be all the distinct right cosets ofH. Then

Na

= (∩g∈Gg−1Hg)a= (∩g∈Gg−1HHg)a= (∩g∈Gg−1H−1Hg)a= (∩g∈G(Hg)−1(Hg))a= (∩n

i=1(Hgi)−1(Hgi))a= (∩n

i=1(Hgi)−1(Hgia).

We observe that Hg1a, . . . , Hgna are right cosets of H and are distinct, hencethey form a permutation of Hg1, . . . ,Hgn. This and the above equation showthat N has at most n! right cosets.

2.5.24 Let G be a finite group whose order is not divisible by 3. Suppose that(ab)3 = a3b3 for all a, b ∈ G. Prove that G must be abelian.

Proof. We assume o(G) > 1 for otherwise the statement is trivial. Let g ∈ Gbe any non-identity element. We show that g3 6= e where e is the identity of G.

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This is because if g3 = e, then g2 6= e by the assumption that g is non-identity.Hence o(g) = 3 and 3|o(G), a contradiction.

We now show that the function f : G → G mapping x ∈ G to x3 is one-to-one. This is because if x3 = y3, then (xy−1)3 = x3y−3 = e. But we have seenthat the identity is the only element that, when taken to a power of 3, yieldsthe identity. Hence xy−1 = e and x = y.

Now since G is finite, the one-to-one f must be a bijection. Expanding(ab)3 = a3b3, we see that

baba = aabb (1)

for all a, b ∈ G. Now for any a, b ∈ G, (a−1ba)3 = a−1b3a by direct expansionand cancelation, but (a−1ba)3 = a−3b3a3 by the assumption of the problem.Hence a−1b3a = a−3b3a3 and therefore a2b3 = b3a2 for all a, b ∈ G. This implies

a2b = ba2 (2)

for all a, b ∈ G because f is a bijection. Eqs. (1)–(2) complete the proof.

2.5.25 Let G be an abelian group and suppose that G has elements of ordersm and n, respectively. Prove that G has an element whose order is the leastcommon multiple of m and n.

Proof. Let a and b be elements of orders m and n, respectively. Write lcm(m,n) =Πt

i=1pαi

i where the pi’s are distinct primes and αi > 0 for each i. For any

i ∈ {1, . . . , t}, pαi

i divides either m or n. If say, pαi

i |m, then am/(pαi

i ) is of orderpαi

i . In this way we obtain elements x1, . . . , xt of orders pα1

1 , . . . , pαt

t , respectively.Now consider elements in G of the form

xβ11 · · ·xβt

t (3)

where βi ∈ N, i ∈ {1, . . . , t}.Our next goal is to show that the single element (x1 · · ·xt) generates all

elements of the form in Eq. (3). For this purpose, we prove that for all βi ∈ N,i ∈ {1, . . . , t}, the system

k = β1 mod pα11

...k = βt mod pαt

t

has a solution in k. This follows by setting

k =t∑

i=1

βicipα11 · · · pαi−1

i−1 pαi+1

i+1 · · · pαtt

where ci satisfies

cipα11 · · · pαi−1

i−1 pαi+1

i+1 · · · pαtt = 1 mod pαi

i .

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It is not hard to see that elements of the form in Eq. (3) form a subgroupH of G. Hence for each i, the order pαi

i of the subgroup generated by xi divideso(H) by the Lagrange theorem. This gives o(H) ≥ Πt

i=1pαii . Clearly o(H) ≤

Πti=1p

αii also holds. Since the element (x1 · · ·xt) generates exactly H, it has

order o(H).

2.5.26 If an abelian group has subgroups of orders m and n, respectively, thenshow it has a subgroup whose order is the least common multiple of m and n.

Proof. Again write lcm(m,n) = Πti=1p

αi

i . For each i ∈ {1, . . . , t}, pαi

i divideseither m or n. Thus, Sylow’s theorem implies the existence of a subgroup Hi of Gof order pαi

i . Now consider K = Πti=1Hi, which can be verified to be a subgroup

of G. Since Hi is a subgroup of K for each i and the pi’s are distinct primes,theLagrange theorem implies Πt

i=1pαi

i |o(K). This and the fact that o(K) ≤ Πti=1p

αi

i

complete the proof.

2.5.38 Let G be a finite abelian group in which the number of solutions in Gof the equation xn = e is at most n for every positive integer n. Prove that Gmust be a cyclic group.

Proof. For each positive integer k, let rk be the number of elements of order k.It can be verified that rk = 0 whenever k 6 |o(G). If there is an ` ∈ G of orderk, then `1, . . . , `k are distinct solutions to xk = e. Hence they are exactly allthe solutions to xk = e. But among these k solutions to xk = e (which triviallyinclude all elements of order k), at most φ(k) of them are of order k because `h

is not of order k whenever h is not coprime to k.Now o(G) =

∑k rk =

∑k|o(G) rk ≤

∑k|o(G) φ(k) = o(G), forcing all inequal-

ities to be equalities. Thus for each k|o(G), exactly φ(k) elements are of orderk and G is therefore cyclic.

2.6.7 Is the converse of Problem 6 true? If yes, prove it. If no, give an exampleof a non-abelian group all of whose subgroups are normal.

Proof. The 8-element quaternion group is non-abelian, yet all of its subgroupsare normal. See http://en.wikipedia.org/wiki/Quaternion_group

2.6.12 Suppose that N and M are two normal subgroups of G and that N∩M ={e}. Show that for any n ∈ N,m ∈ M, nm = mn.

Proof. Observe nm ∈ Nm = mN and hence nm = mn′ for some n′ ∈ N.Similarly, nm ∈ nM = Mn, implying nm = m′n for some m′ ∈ M. We now havemn′ = m′n, or equivalently, n′n−1 = m−1m′. Now t = n′n−1 = m−1m′ ∈ N∩Mand hence t = e.

2.6.14 Prove, by an example, that we can find three groups E ⊆ F ⊆ G, whereE is normal in F, F is normal in G, but E is not normal in G.

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Proof. Thanks to ntnusliver@ptt for pointing out that dihedral groups consituteexamples desired for this problem. Consider regulars 4-gons under reflectionand rotation. All 8 results constitute G. Take F to be even rotations with orwithout reflections; take E to be reflections or not. The rest can be verified.

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