Uniqueness of minimal normal subgroup of a primitive group Let $N$ be a solvable minimal normal subgroup of a primitive group $G \le S^\Omega$.  I want to show that $N$ is the only minimal normal subgroup of $G$.   (This is part of Theorem 11.5 in Wielandt's Finite Permutation Groups, where I have a question on one step of the proof).  
I can show that $N$ is abelian and transitive, hence regular.  For a fixed prime $p$ dividing $|N|$, define $H:=\{m^p: m \in N\}$.  It can be shown that $H=1$. Hence $N$ is elementary abelian and $|N|$ is a prime power.  
Suppose there exists another subgroup $M$ such that $M \ne N, M \trianglelefteq G$, and $M$ is minimal. Then $M \cap N=1$ by minimality of $N$. Why should $M$ be in the centralizer of $N$? (Suppose we could show $M \le Z_{S^\Omega}(N)$.  Since $Z_{S^\Omega}(N)=N$,  $M \le N$.  Then $M=1$ by minimality of $N$, which proves the assertion.) Why does $M \cap N=1$ imply that $M$ lies in the centralizer of $N$?

I think when $M\cap N=1$ then for all $m\in M$ and for all $n\in N$ we have $$m^{-1}n^{-1}mn\in M\cap N$$ and so $$mn=nm$$ Does this help you?
