Let $G$ be a finite group and $P$ a Sylow $p$-subgroup of $G$. Let $Q$ be a subgroup of $P$. If $Z(Q)$ is a Sylow $p$-subgroup of $C_G(Q)$, is $C_G(Q)$ a internal direct product of $Z(Q)$ and $O_{p'}(C_G(Q))$?
Notation: $O_{p'}(G)$ is the largest normal subgroup of $G$ of order prime to $p$.
I think the answer is yes, but I can't do it.
It is easy to see that $Z(Q)$ and $O_{p'}(C_G(Q))$ are normal subgroups of $C_G(Q)$, and $Z(Q) \cap O_{p'}(C_G(Q)) = 1$.
And I know the following formula : $$ |AB||A \cap B| = |A||B| $$ if $A, B$ are finite subgroups of a group $G$.
How can I prove that $C_G(Q) = Z(Q) O_{p'}(C_G(Q))$?
Thanks!
Since $Z(Q)$ is a Sylow $p$-subgroup of $C_G(Q)$ and of course normal in $C_G(Q)$, the Schur-Zassenhaus Theorem guarantees a complement $H$: $C_G(Q)=HZ(Q)$, and $H \cap Z(Q)=1$. Since $Z(Q) \subseteq Z(C_G(Q))$, it follows that $H \unlhd C_G(Q)$. hence, $O_{p^{'}}(C_G(Q))=H$ and you are done.