TITLE: Use Sylow Theory to show existence of certain subgroup
QUESTION [6 upvotes]: I am studying for a qualifying exam but have gotten stuck on this problem:
Let $G$ be a finite group, $S$ a Sylow $2$-subgroup of $G$, $T\leq S$ with $|S/T| = 2$, and $g\in G$ with $|g|=2$. Suppose that $hgh^{-1} \not \in T$ for all $h\in G$. Show there exists a subgroup $H$ of $G$ with $|G/H| = 2$ and $g\not\in H$.
I have considered the action of $G$ on $G/T$ by left multiplication. If we let $|G| = 2^km$, where $2\not \mid m$, then $|T| = 2^{k-1}$ and so $|G/T| = 2m$. We are looking for a subgroup of order $2^{k-1}m$. Then I looked at the kernel of the action, since $g$ is not in the kernel. However, the kernel is not a large enough subgroup.
Any suggestions for this problem would be appreciated!

REPLY [2 votes]: It looks like you are well on your way.
Hints:

Show that $g$ acts on $G/T$ as a product of $m$ disjoint 2-cycles.
The action of $G$ on $G/T$ gives us a homomorphism from $G$ to the symmetric group $S_{2m}$. Does $S_{2m}$ have a useful index two subgroup? Pull it back!