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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 $N$?

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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?

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Thanks, that works because by hypothesis $M$ and $N$ are both normal in $G$, whence $m^{-1}n^{-1}mn$ is in both $M$ and $N$. –  user70056 Mar 30 '13 at 1:59
    
@user70056: You are Welcome. –  Babak S. Mar 30 '13 at 2:04

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