Exercise 8.4.5

Proof.

As $H\ne \{e\}$, there exists a permutation $\sigma \in H,\sigma \ne e$. Then ${\sigma }^2\in H\cap A_n$, so ${\sigma }^2=e$. Moreover $\sigma$ is an odd permutation, otherwise $\sigma \in H\cap A_n$, and then $\sigma =e$.

Let $\tau$ be any permutation in $H\setminus \{e\}$. With the same reasoning, $\tau$ is an odd permutation, and so is $\sigma$. Hence ${\sigma }^{-1}\tau \in H\cap A_n$, hence ${\sigma }^{-1}\tau =e$, so $\tau =\sigma$. $H$ has no other element than $e,\sigma$.

The order of $\sigma$ is 2, so in the decomposition of $\sigma$ in disjoint cycles, since the order of $\sigma$ is the lcm of the orders of these cycles, all the cycles have order 2.

As $\sigma \notin A_n$, $\sigma$ is a product of an odd number of disjoint 2-cycles. □