Let $G$ be a finite group, and $L$ a maximal subgroup of $G$. If $L$ is non-abelian and simple, then in $G$ there exists at most two minimal normal subgroups.
What I got: Suppose we have three minimal normal subgroups $P,Q,R$. Then as $P\cap R = Q \cap R = 1$ we have $P \in C_G(R)$ and $Q \in C_G(R)$, therefore $PQ \in C_G(R)$. So I know about the centraliser of $R$ that it is i) normal in $G$, because $R$ is normal, and ii) by the above, that it is non-trivial and not a minimal normal subgroup, because $P\cap Q = 1$. So I guess I have to show somehow that the centraliser indeed has these properties for a maximal subgroup $L$ with these properties to get a contradiction, but here I have no idea how to proceed?