$2$-Sylow of a subgroup of $A_n$

Question

I don't know how to start to prove this:
Let $H$ be a subgroup of $S_n$ and let $T$ be a subgroup of $H$ such that $T$ is a $2$-Sylow of $H$.
Show that $H$ is conntained in $A_n\iff$ $T\le A_n$.
Maybe this is trivial but I'm having problems to solve this exercise...

Answer 1

The forward implication is immediate.

For the reverse implication, suppose $T\le A_n$.

We want to show $H\le A_n$.

Suppose instead that $h\in H$, where $h$ is an odd permutation.

Our goal is to derive a contradiction.

Since $h$ is an odd permutation, and since cycles of odd order are even permutations, at least one of the cycles in the disjoint cycle decomposition of $h$ must have even order, hence $h$ also has even order.

Since $h$ has even order, some odd power of $h$, say $h^m$, has order which is a power of $2$, so $h^m$ must be an element of some $2$-Sylow subgroup of $H$.

Since the $2$-Sylow subgroups of $H$ are conjugate to each other, it follows that $h^m$ is conjugate to some element of $T$.

Since $T\le A_n$, all elements of $T$ are even permutations, hence since conjugation preserves parity, $h^m$ is an even permutation.

But $h$ is an odd permutation, hence since $m$ is odd, $h^m$ is an odd permutation, contradiction.