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

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.2 Exercise 4.2.13

Solution: $G$ 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 Exercise 4.2.11, $\pi(x)$ is a product of $k$ disjoint 2-cycles. Since $k$ is odd, $\pi(x)$ is an odd permutation. By Exercise 4.2.12, $\pi[G]$ has a subgroup of index 2.


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