We Abstract sketch the solutions of just a few problems in Hersteins book.
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Sketches of some problems in Hersteins abstract algebra
Ching-Lueh Chang March 12, 2008
2.5.19 If H is a subgroup of nite index in G, prove that there is only a nite number of distinct subgroups of the form aHa1 . Proof. Observe aHa1 = aHHa1 = aH 1 Ha1 = (Ha1 )1 (Ha1 ). The second inequality follows from H = H 1 . 2.5.20 If H is of nite index in G prove that there is a subgroup N of G, contained in H, and of nite index in G such that aN a1 = N for all a G. Can you give an upper bound for the index of this N is G? Proof. The set N = gG g 1 Hg provably forms a normal subgroup of G. Clearly, N H. Let a G be arbitrary and Hg1 , ..., Hgn be all the distinct right cosets of H. Then Na = (gG g 1 Hg )a = (gG g 1 HHg )a = (gG g 1 H 1 Hg )a = (gG (Hg )1 (Hg ))a = (n (Hgi )1 (Hgi ))a i=1 = (n (Hgi )1 (Hgi a). i=1 We observe that Hg1 a, . . . , Hgn a are right cosets of H and are distinct, hence they form a permutation of Hg1 , . . . , Hgn . This and the above equation show that N has at most n! right cosets. 2.5.24 Let G be a nite group whose order is not divisible by 3. Suppose that (ab)3 = a3 b3 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 G be any non-identity element. We show that g 3 = e where e is the identity of G.
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This is because if g 3 = e, then g 2 = 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-toone. This is because if x3 = y 3 , then (xy 1 )3 = x3 y 3 = e. But we have seen that the identity is the only element that, when taken to a power of 3, yields the identity. Hence xy 1 = e and x = y. Now since G is nite, the one-to-one f must be a bijection. Expanding (ab)3 = a3 b3 , we see that baba = aabb (1)
for all a, b G. Now for any a, b G, (a1 ba)3 = a1 b3 a by direct expansion and cancelation, but (a1 ba)3 = a3 b3 a3 by the assumption of the problem. Hence a1 b3 a = a3 b3 a3 and therefore a2 b3 = b3 a2 for all a, b G. This implies a2 b = ba2 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 orders m and n, respectively. Prove that G has an element whose order is the least common multiple of m and n. Proof. Let a and b be elements of orders m and n, respectively. Write lcm(m, n) = i t=1 p where the pi s are distinct primes and i > 0 for each i. For any i i i {1, . . . , t}, p divides either m or n. If say, p |m, then am/(pi ) is of order i i i 1 t p . In this way we obtain elements x1 , . . . , xt of orders p , . . . , p , respectively. t 1 i Now consider elements in G of the form x1 xt t 1 (3)
i i i
(2)
where i 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 i N, i {1, . . , . t}, the system k = 1 mod p1 1 . . . k = t mod pt t has a solution in k. This follows by setting
t
k=
i=1
i i+1 i ci p1 pi1 pi+1 pt t 1 1
where ci satises
i i+1 ci p1 pi1 pi+1 pt = 1 mod pi . t 1 1 i
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It is not hard to see that elements of the form in Eq. (3) form a subgroup H of G. Hence for each i, the order pi of the subgroup generated by xi divides i o(H ) by the Lagrange theorem. This gives o(H ) t=1 pi . Clearly o(H ) i i t=1 pi also holds. Since the element (x1 xt ) generates exactly H, it has i i order o(H ). 2.5.26 If an abelian group has subgroups of orders m and n, respectively, then show it has a subgroup whose order is the least common multiple of m and n. Proof. Again write lcm(m, n) = t=1 p . For each i {1, . . . , t}, p divides i i i either m or n. Thus, Sylows theorem implies the existence of a subgroup Hi of G i of order p . Now consider K = t=1 Hi , which can be veried to be a subgroup i i of G. Since Hi is a subgroup of K for each i and the pi s are distinct primes,the i i Lagrange theorem implies t=1 p |o(K ). This and the fact that o(K ) t=1 p i i i i complete the proof. 2.5.38 Let G be a nite abelian group in which the number of solutions in G of the equation xn = e is at most n for every positive integer n. Prove that G must be a cyclic group. Proof. For each positive integer k, let rk be the number of elements of order k. It can be veried that rk = 0 whenever k |o(G). If there is an G of order k, then 1 , . . . , k are distinct solutions to xk = e. Hence they are exactly all the solutions to xk = e. But among these k solutions to xk = e (which trivially include all elements of order k ), at most (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) (k ) = o(G), forcing all inequalities to be equalities. Thus for each k |o(G), exactly (k ) elements are of order k and G is therefore cyclic. 2.6.7 Is the converse of Problem 6 true? If yes, prove it. If no, give an example of a non-abelian group all of whose subgroups are normal. Proof. The 8-element quaternion group is non-abelian, yet all of its subgroups are 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 N m = mN and hence nm = mn for some n N. Similarly, nm nM = M n, implying nm = m n for some m M. We now have mn = m n, or equivalently, n n1 = m1 m . Now t = n n1 = m1 m N M and hence t = e. 2.6.14 Prove, by an example, that we can nd three groups E F G, where E is normal in F, F is normal in G, but E is not normal in G.
i i
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Proof. Thanks to ntnusliver@ptt for pointing out that dihedral groups consitute examples desired for this problem. Consider regulars 4-gons under reection and rotation. All 8 results constitute G. Take F to be even rotations with or without reections; take E to be reections or not. The rest can be veried.
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