The largest normal subgroup of odd order in the centraliser For a group $G$ denote by $O_{2'}(G)$ the largest normal subgroup of $G$ of odd order. 
Now let $N = O^{2'}(G)$ be the smallest normal subgroup of $G$ such that $G/N$ has odd order, and $H \le G$ be some subgroup of odd order. Suppose $H \le C_G(N)$, then $H \le O_{2'}(C_G(N)) = 1$
1) Why is $O_{2'}(C_G(N)) = 1$ and
2) Why is $H \le O_{2'}(C_G(N))$?
 A: For $S \in {\rm Syl}_{2}(G),$ we have that $Z(S) \in {\rm Syl}_{2}(C_{G}(S)),$ from which it follows by Schur-Zassenhaus ( or Burnside's transfer theorem) that $C_{G}(S) = Z(S) \times K$ for some subgroup $K$ of odd order. Then we have that $C_{G}(S)$ already has a normal $2$-complement. Now $S \leq N,$ so that $C_{G}(N) \leq C_{G}(S).$ Hence $C_{G}(N)$ also has a normal $2$-complement. Since $H$ is an odd order subgroup of $C_{G}(N),$ and $C_{G}(N)$ has a normal $2$-complement, we do have $H \leq O_{2^{\prime}}(C_{G}(N)).$ However, I do not see that we can conclude that $O_{2^{\prime}}(C_{G}(N)) = 1$, unless we already know that $O_{2^{\prime}}(G) = 1.$ Just consider the case that $G$ is an Abelian group which is not a $2$-group. 
I believe that the correct statement is that every odd order subgroup of $C_{G}(N)$ is contained in $O_{2^{\prime}}(G),$ which is a true statement. Perhaps the proof you are reading had already reduced to the case that $O_{2^{\prime}}(G) = 1.$
 To continue that proof that $H \leq O_{2^{\prime}}(G),$ having already established that $H \leq O_{2^{\prime}}(C_{G}(N))$, as we have done, note that $N \lhd G,$ so that $C_{G}(N) \lhd G$ and $O_{2^{\prime}}(C_{G}(N)) {\rm char } C_{G}(N) \lhd G.$ 
