Does a maximal subgroup have to contain the center?

Question

Let $G$ be a non-abelian group with center $Z(G)$, and let $H$ be a maximal subgroup of $G$ which is non-abelian. Is it true that $Z(G)\leq H$?

I am trying to prove that it is true by arguing that if not, then the group $J:=\langle H,Z(G)\rangle$ is going to be a proper subgroup of $G$ containing $H$, which would contradict the maximality of $H$. It is clear that $H$ is a proper subgroup of $J$, but I am not sure how to show that $J$ is a proper subgroup of $G$, i.e that $J\neq G$.

It may be that the statement is not true in general, but in particular I want to prove it when $G=SL(2,q)$.

Any help would be appreciated!

Answer 1

A fairly straightforward example is $H\times C_p$ where $H$ is a non-Abelian simple group and $C_p$ is cyclic of prime order $p$. The centre is $C_p$ and $H$ is maximal.

Answer 2

Note that your statement is true if $H$ is assumed to be abelian: if $G$ is a non-abelian group, $H$ a abelian maximal subgroup, then $Z(G) \subseteq H$. Proof: if $G=HZ(G)$ then clearly $G$ would be abelian, hence you must have $H=HZ(G)$.
Although you cannot always interchange maximal and abelian (a maximal abelian subgroup can be something different than an abelian maximal subgroup), the statement remains true if you do that in this case.