Here we go for the last time with the group $G=GL_2(\mathbb F_3)$.
We know that it has three $2$-Sylow subgroups. So let $P\in\operatorname{Syl}_2(G)$.
I was searching for its center but I don't know where to start. The only thing I noticed is that since $P$ is a $2$-group then $Z(P)\neq1$. Moreover since $Q_8<P$ then $P$ can't be abelian, hence $Z(P)<P$. Hence $|Z(P)|\in\{2,4,8\}$. But $P$ contains al least $6$ subgroup of order $4$ and $4$ subgroups of order $8$. So I'm a little bit lost.
EDIT I know that $P$ has $1$ element of order $1$, one central involution $-\mathbb I$ and $4$ non-central involutions (hence $5$ element of order $2$), $6$ elements of order $4$ and $4$ elements of order $8$. From this I deduced that $P$ has at least $6$ subgroups of order $4$ and $4$ subgroups of order $8$. If I prove (for example) that there are no other groups of such orders ($8$ and $4$) then observing that $-\mathbb I$ must stay in $Z(P)$ I could deduce something.
Can someone help me please? Thanks a lot