Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $N$?

share|cite|improve this question
up vote 4 down vote accepted

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?

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.