If a finite group has order 2k where k is odd, then it has a subgroup of index 2

Prove that if |G| = 2k where k is odd, then G has a subgroup of index 2.


G contains an element x of order 2 by Cauchy’s Theorem. Let \pi : G \rightarrow S_G be the left regular representation of G. By this previous exercise, \pi(x) is a product of k disjoint 2-cycles. Since k is odd, \pi(x) is an odd permutation. By this previous exercise, \pi[G] has a subgroup of index 2.

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