# What could be said about centralizers of normal subgroups if $G$ contains a simple, non-abelian maximal subgroup

Let $G$ be a finite group containing a maximal simple and non-abelian group, is it true that the centraliser of each normal subgroup is either trivial or a minimal normal subgroups?

EDIT: Maybe to give some context. If $G$ contains a maximal subgroup which is core-free, then this property holds for the centraliser of each normal subgroup. For a proof, let $G$ be a group and $L$ be a maximal core-free group, meaning that the intersection of all conjugates of $L$ is trivial, in particular $L$ contains no normal subgroup of $G$. Now let $N \unlhd G$ be non-trivial, as $L \cap N = 1$ we have $G = LN$. Also $C_G(N) \unlhd G$ and so $C_G(N) \cap L \unlhd L$ and therefore $L \le N_G(C_G(N)\cap L)$. As $N \le C_G(C_G(N))$ we have $$N \le C_G(C_G(N) \cap L) \le N_G(C_G(N)\cap L)$$ which implies $G = LN \le N_G(C_G(N)\cap L)$, i.e. $C_G(N) \cap L) \unlhd G$, and as $L$ is core-free it follows that $C_G(N) \cap L = 1$. Now let $M \le C_G(N)$ be a non-trivial normal subgroup of $G$, then again $M \cap L = 1$ and so $G = LM$. By Dedekind's Identity we have with $M \le C_G(N) \le LM$ $$C_G(N) = M(C_G(N) \cap L) = M$$ hence either $C_G(N) = 1$ or it is a minimal normal subgroup of $G$.

Now I am conjecturing that if $L$ is simple and non-abelian, a similar conclusion could be drawn, but I am stuck in the proof.

## 1 Answer

No.

Take $G=A_5\times Z_2$ then $C_G(1\times Z_2)=G$