Show that any two elements of order 2 in a group G ($|G| = 2m$), are conjugate in G

Let G be a group such that $|G| = 2m$ where $m$ is an odd integer. Show that any two elements of G of order 2 are conjugate in G.

Hint: If x and y are elements of G of order 2, then $\langle x \rangle$ and $\langle y \rangle$ are Sylow 2-subgroups of G. Why?

So, I know that two elements are called conjugate if $x=aya^{-1}$ for any $a∈G$. Also, by the Sylow third theorem I find that $n_2\equiv 1\pmod2$ and $n_2\mid m$. So basically, to solve the question, I have to, like, prove that, if say $H=\langle x \rangle$ and $K=\langle y \rangle$, then $K=aHa^{-1}$. And how do I even prove that H and K are Sylow 2-subgroups? I understand that I somehow have to use Sylow theorem but I am completely and utterly confused as to what, where and when.

Edit: Maybe I can somehow use the Thompson Transfer Lemma? Somehow feels similar, I guess? Proving the Thompson Transfer Lemma

• Remember three things: A 2-Sylow subgroup is a maximal subgroup of order a power of $2$ dividing $|G|$. Any two 2-Sylow subgroups are conjugate to each other. A group of order 2 has a unique non identity element.. Conclude. – Ravi Nov 19 '17 at 18:37