M417 Homework 7 Solutions Spring 2004
1. If f: G -> H is a homomorphism of groups, and if B is a normal subgroup of H, show that f -1(B) is a normal subgroup of G.

Let g be an element of G, and x an element of f -1(B). Then f(gxg-1) = f(g)f(x)(f(g))-1 is in B, since f(x) is in B and since B is normal in H. Thus gxg-1 is in f -1(B), so g(f -1(B))g-1 is contained in f -1(B) for every g in G. By the lemma from class, this means that g(f -1(B))g-1 = f -1(B) for every g in G, so f -1(B) is a normal subgroup of G.

2. If f: G -> H is a surjective homomorphism of groups, and if A is a normal subgroup of G, show that f(A) is a normal subgroup of H.

Let h = f(g) be an element of H (we know there is a g for every h since f is onto), and let f(x) be an element of f(A). Then hf(x)h-1 = f(g)f(x)(f(g))-1 = f(gxg-1), and this is in f(A), since gxg-1 is in A (this being because x is in A and A is normal in G). Thus hf(A)h-1 is contained in f(A) for every h in H, so f(A) is a normal subgroup of H.

3. Give an example showing that the previous problem requires the hypothesis that f be surjective.

We know that < (12) > is not a normal subgroup of Sn, for n > 2 (from an example in class, or directly by the fact that (13)(12)(13) = (23) is not in < (12) >). Let f: G -> H be the inclusion homomorphism of G = < (12) > in H = Sn. Then A = G is normal in G, but f(A) is not normal in H.

Theorem 6.1 on p. 122 is sometimes called Little Cayley. Here is Big Cayley: Let A be a subgroup of a group G, and let P be the group of bijections from the set G/A of left cosets of A in G, to itself (i.e., P is the group of permutations of G/A). Then there is a homomorphism f: G -> P whose kernel is contained in A.

4. Show that Little Cayley is a corollary of Big Cayley.

We must show that G is isomorphic to a subgroup of a group of permutations. Take A = < e >, and apply Big Cayley. Then f: G -> P is a homomorphism with ker(f) contained in and thus equal to < e > = A. Thus f is injective, so G is isomorphic to f(G), which is a subgroup of the permutation group P.

5. Let A be a proper subgroup of a finite group G, such that p = |G : A| is the smallest prime dividing dividing |G|.
• Use Big Cayley to prove that A is normal.

By Lagrange's Theorem we know |ker(f)|r = |A| for some positive integer r, and by hypothesis we know |A|p = |G|. But by Big Cayley we know have a homomorphism f: G -> P where f(G) is a subgroup of the group P of permutations of G/A. Since |G/A| = p, we know |P| = p!, and by Lagrange we know pr = |f(G)| divides |P|. Thus pr divides p!, so r divides (p-1)!. But every prime factor of r is at least p by hypothesis, and every prime factor of (p-1)! is of course less than p. Thus r cannot have any prime factors, so r = 1, so |ker(f)| = |ker(f)|r = |A|. Since by Big Cayley A contains ker(f), it follows that A = ker(f). Since kernels are normal, A is normal.

• Now show that any subgroup of index 2 in a finite group G is normal. Is this still true even if G is infinite? Yes, as we see below.

Since 2 is the least prime, any subgroup of index 2 in a finite group must have index equal to the least prime dividing the order of the group. But say G is an infinite group, and A is a subgroup of index 2. Since A has index 2 in G, there are only two cosets of A in G, with A being one of them. Thus the complement of A in G, denoted G-A, is the other coset, and this is true whether we're speaking of left or right cosets. Let g be in G. If g is in A, then gA = A = Ag. If g is not in A, then gA is not equal to A, hence gA must be the other coset, G-A. Likewise, Ag = G-A. Thus gA = Ag, for every element g of G, so A is normal in G.