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Suppose that $H$ is a subgroup of $S_n$ of odd order. Prove that H is a subgroup of $A_n$.

How can I solve this problem without using Cayley's Theorem? So far, I understand that H contains both even and odd permutations but how would this make a difference? Thanks.

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Let $h\in H$ and consider its decomposition in disjoint cycles: $h=\sigma_1 \cdots \sigma_m$. Since $H$ has odd order, $h$ must have odd order, by Lagrange's theorem. This means that all $\sigma_k$ have odd order because the order of $h$ is the lcm of the orders of the $\sigma_k$. But a cycle of odd order can be written as a product of an even number of transpositions, as in $(12345)=(12)(13)(14)(15)$ (depending on how you interpret composition). So, $h$ can be written as a product of an even number of transpositions and so belongs to $A_n$.

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  • $\begingroup$ Haven't studied normal groups yet. $\endgroup$ – user93383 Oct 3 '13 at 2:03
  • $\begingroup$ I have only studied permutation & cyclic groups. $\endgroup$ – user93383 Oct 3 '13 at 2:24

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