3.41:

Let G be a group of order 2^mk where k is odd. Assume G has an element x of order 2^m. (In our newly learned language, we are assuming that G has a cyclic Sylow 2-subgroup.) We are to show that the set of element of odd order in G is a subgroup. If it is a subgroup, it is automatically normal, indeed characteristic, since automorphisms preserve order.

The author suggests to consider the action of G on itself by left multiplication. Let us consider G to be a subgroup of S_G, with x(g) = xg for x in G and g in G. Under this action x corresponds to a permutation of order 2^m. Then the least common multiple of the sizes of orbits of x is 2^m. Considering the action of <x> on G, one sees that every orbit has size equal to |<x>| = 2^m, since the stabilizer of any point is trivial. (x^k g = g implies x^k=1.) Moreover, x has no fixed points. Then x is a product of k 2^m-cycles. Since k is odd, this is an odd permutation. Let A_G be the alternating subgroup of S_G. Since x is odd, G ^ A_G is not equal to G. Since |S_G : A_G| = 2, we have A_GG = S_G, hence |G : G ^ A_G| = 2 by the second isomorphism theorem. Then |G ^ A_G| = 2^m-1k. Moreover, any element of odd order in G must be a product of cycles of disjoint cycles of odd length (since the order is the least common multiple of the cycle sizes), and hence must be even. Thus G ^ A_G contains all the elements of odd order in G. And furthermore. x² is even, hence is an element of G ^ A_G, and the order of x² is 2^m-1.

Now we finish the proof by induction on m. If m = 0, then all elements of G have odd order, hence they form a subgroup of G. If m is 1 or bigger, we apply the preceding argument to produce a subgroup H of index two in G containing all elements of odd order in G, and which contains an element of order 2^m-1. Then by the inductive hypothesis, the set of elements of odd order in H is a subgroup of H, hence a subgroup of G, and this set is equal to the set of elements of odd order in G. This completes the inductive step.

This problem is related to Hall's theorems about solvable groups. The set of elements of odd order is a Hall pi-subgroup, where pi is the set of primes not equal to 2. Such a subgroup is called a "p-complement." So this problem shows: if G has a cyclic Sylow 2-subgroup, then G has a normal 2-complement.