$G$ is a finite solvable group. $M$ is a minimal nontrivial normal subgroup of $G$. Prove that $M$ is abelian Hint is to construct a subgroup $N \subset M$ of prime index, and prove that commutators of M lie in N.
There is an answer here but I'm not so familiar with commutators as I've only read about them in Dummit and Foote (we never did them in class). Also it's not clear to me how to create a subgroup of prime index.
 A: Take $\;M'=[M,M]\,=\;$ the commutator subgroup of  $\;M\;$ . This is a characteristic subgroup of a normal subgroup and thus normal in the whole group $\;G\;$. But $\;M\;$ is minimal normal (non-trivial), and of course also solvable, so it must be $\;M'=1\iff M\;$ is abelian.
A: A subgroup $H$ of $G$ is called characteristic, denoted by $H \ char \ G$, if $φ(H) =H$ holds for any $φ \in Aut(G)$.
Let $M$ be a minimal normal subgroup of $G$. You need to check:


*

*$M$ is solvable.

*There is a subgroup $N$ with $N<M$.

*$N \ char \ M$ which implies $N=1$, so that $M$ is abelian.

*If $p$ divides $|M|$, then $pM$ must be trivial in $G$.


Therefore, $M$ is an elementary abelian $p$-group.
A: The other two answers helped me figure it out but I wanted to post a complete answer.
Consider the subgroup $M'$ generated by commutators of elements of $M$.
Note that $M'$ is characteristic in $M$; for any automorphism $\phi$ and a generator $xyx^{-1}y^{-1}$:
$$\phi(xyx^{-1}y^{-1}) = \phi(x)\phi(y)\phi(x)^{-1}\phi(y)^{-1} \in M'$$
$M'$ is characteristic in $M$ and $M \trianglelefteq G$ so $M' \trianglelefteq G$.
Since $M$ is minimal normal, $M' = \{1\}$ or $M' = M$.
Note that $M$ is solvable because $G$ is solvable.
By proposition 7 from section 5.4 of Dummit and Foote (page 169), $M/M'$ is the largest abelian quotient of $M$.
If $M' = M$ then $\{1\}$ is the largest abelian quotient, and $M$ is not solvable.
So $M' = \{1\}$ and therefore $M$ is abelian.

To see why the proposition is true (writing out the proof from Dummit and Foote), first note that $M/M'$ is abelian:
$$(xM')(yM') = xyM' = yx(x^{-1}y^{-1}xy)M' = yxM'$$
Now suppose $H \trianglelefteq M$ and $M/H$ is abelian.
Then:
$$1H = x^{-1}y^{-1}xyH$$
So $[x, y] \in H$ for all $x, y \in M$, so $M' \leq H$.
So $M/M'$ is the largest abelian quotient of $M$.
