TITLE: Showing that if $N \le G$ is finite minimal normal with every simple homomorphic image abelian, then it is abelian itself
QUESTION [5 upvotes]: I've been working on the following problem, with no success so far:
"Let $N$ be a finite minimal normal subgroup of a group $G$, and suppose $N$  has the property that every simple homomorphic image is abelian.  Show that $N$ is abelian."
I am told to use the result of the previous problem, "If $H \le G$, then $G' \le H$ iff $H \triangleleft G$ and $G/H$ is abelian."
($G'$ is the derived subgroup of $G$)
Here are some of my thoughts so far:
It seems the natural subgroup to consider here would be $N \le G$.  Since $G' \triangleleft G$ but $N$ is minimal normal, $G' \not\le N$, which implies that (since we assume $N \triangleleft G$) $G/N$ is not abelian.  If this leads us to the proof, I certainly don't see how it does so.
Generally, for showing that a subgroup is abelian, we could show that $N'$ is trivial.  We could also establish it as the homomorphic image of some abelian group or subgroup, but in this problem it seems it would be a preimage, but then abelian-ness may not be preserved (consider the zero homomorphism from a non-abelian group).
I would really appreciate a HINT on where to go with this.  I would also appreciate whatever commentary and discussion anyone has to offer on how to go about thinking about this (and the same holds for all of my questions), because I feel this is really where I lack experience.  Having attended actual lectures very little, I feel like I don't really know 'how to think' about problems.
Thanks for the help, as always.

REPLY [7 votes]: Argue like this.
Consider a maximal normal subgroup $M$ of $N$. (So $M$ is a proper subgroup of $N$, it is normal in $N$, and it is not properly contained in any proper normal subgroup of $N$.)
Then $N/M$ is a simple homomorphic image of $N$, and as such is abelian.
By the result you quote, $N' \le M < N$.
First spoiler

Now $N'$ is characteristic in $N$, which is normal in $G$, hence...

Second spoiler

 ... hence $N' < N$ is normal in $G$. Since $N$ is minimal normal, $N' = 1$, and $N$ is abelian.