Let $G$ be a finite group and $P$ a Sylow $2$-subgroup. If $g$ is an element of $G$ with order a power of $2$ then it lies in some conjugate of $P$. I get this.
But if also $g\in C_G(P)$, then it is true that $g\in P$. Why? I can't see that immediately.
If $g\in C_G(P)$, then the subgroup $Q$ generated by $P$ and $g$ is a $2$-group (every element of $Q$ can be written in the form $g^n p$ for some $n\in\mathbb{Z}$ and $p\in P$ since $g$ commutes with $P$, and $(g^np)^{2^m}=1$ for $m$ sufficiently large since $g$ and $p$ commute and both have order a power of $2$). Since $P$ is $2$-Sylow and $Q\supseteq P$, this implies $Q=P$, so $g\in P$.