Showing that if $N \le G$ is finite minimal normal and every simple homomorphic image is abelian, then N is elementary abelian

This is part 2 of this question. Unsurprisingly, I'm having some difficulty with it. Hints are much appreciated. Here's the setup again:

"Let $N$ be a finite minimal normal subgroup of a group $G$, and suppose that $N$ has the property that every simple homomorphic image is abelian. Show that all nonidentity elements of $N$ have equal prime order."

As a hint I'm given "show that for some prime $p$, the subset $S=\{x \in N \,|\, x^p=1\}$ contains a nonidentity element."

Basically as far as I've gotten is showing that $N$ has prime power order:

Let $\phi: G \to H$ be a nontrivial homomorphism. Since $\phi(N)$ is simple and abelian, it is a cyclic group of prime order, say $p$. If $x \in N$, then $|\phi(x)|$ divides $|x|$, so $|x|$ is a multiple of $p$. Furthermore, it is a power of $p$ since if it had some other prime factor, say $q$, then $q$ divides $|N|$ and so by Cauchy's Theorem we have an element of order $q$, which $p$ does not divide. Therefore $|N|$ is a power of $p$.

How we prove that it cannot have elements of prime power order greater than $p$ is what is eluding me. I will continue thinking and check back tomorrow to see where I have gone astray...

Note: Before even posting I now realize that's not even a valid argument: elements of order $q$ could exist, but they must lie in $\ker(\phi)$... Should I think further about this?

• You cannot say that for every homomorphism $\phi$ from $G$ that $\phi(N)$ is simple and abelian. Rather, the condition is that for every $\phi:N\to H$ such that $\phi(N)$ is simple we have $\phi(N)$ is abelian. I think a place to start is to note that for every normal subgroup $M\lhd N$, $N/M$ is a homomorphic image. When is $N/M$ simple? – user714630 May 3 '13 at 3:27
• Don't you already know that $N$ is abelian? What are some characteristic subgroups of finite abelian groups? – Jack Schmidt May 3 '13 at 3:36
• Where are you using the minimal part of minimal normal? – Alexander Gruber May 3 '13 at 3:44
• Hint: A minimal normal subgroup is characteristically simple, so if it is finite then it is a product of isomorphic simple groups. – Nicky Hekster May 3 '13 at 12:49

The trick is to first show that $N$ is abelian and after that to show that all non-identity elements have order $p$ for some fixed prime $p$.
Note that any non-trivial group will have a simple quotient, so the condition that any simple quotient is abelian means that $N'$ is a proper subgroup of $N$. But $N'$ is a characteristic subgroup of $N$, so it is normal in $G$, and by assumption of minimality of $N$, this means that $N' = \{1\}$ so $N$ is abelian.
Now, since $N$ is abelian, any Sylow subgroup is characteristic, so by minimality, we conclude that $N$ must be a $p$-group for some prime $p$.
But in any abelian $p$-group the set of elements of order dividing $p$ is a non-trivial characteristic subgroup, so by minimality, this subset is $N$ itself, which precisely means that $N$ is elementary abelian.