Minimal Normal Subgroups of Solvable Groups (6.1.31)

Dummit and Foote Abstract Algebra, section 6.1, exercise 31:

A group $M$ is called a minimal normal subgroup of $G$ when $M \trianglelefteq G$ and every proper nontrivial subgroup of $M$ is nonnormal in $G$. Prove that when $G$ is finite and solvable, any minimal subgroup $M$ is an elementary abelian $p$-group for some prime $p$. [Examine $M$'s characteristic subgroups $M'$ and $\langle~x^p~|~x∈P~\rangle$]

Proof: Let $M^p$ denote $\langle~x^p~|~x∈P~\rangle$. If $M$ is trivial, the case holds, so assume $M≠1$.

Lemma 1: $G^n~\text{char}~G$ for any finitely generated group $G$ and $n∈\mathbb{Z}$. Proof: $$x∈G^n⇔x=y_1^n...y_r^n⇔φ(x)=φ(y_1^n...y_r^n)=φ(y_1)^n...φ(y_r)^n⇔φ(x)∈G^n~\square$$

Since any characteristic subgroup of $M$ is concomitantly normal in $G$, we must not have any proper nontrivial characteristic subgroups of $M$. Assume $M'=M$; since subgroups of a solvable group are solvable by Proposition 10(1), and yet the derived chain of $M$ terminates before reaching the identity, we have a contradiction. So assume $M'=1$, and now $M$ is abelian.

Suppose $M^p=1$ for some prime $p$. Then all the generators of $M^p$ are trivial, which is to say $x^p=1$ for all $x∈M$ and now $M$ is elementary abelian. So assume that $M^p=M$ for any prime $p$. Inductively prove that $M^{n}=M$ for all $n∈\mathbb{Z^+}$, an absurd conclusion for $M^{|M|}=M$. The case is clear when $n=1$, so proceed to the inductive step. The case is assumed true for prime $n$, so we have prime $q$ dividing $n≠q$. We have $(M^{n/q})^q=M$ by induction, revealing that for any element of the abelian $M$ we have a representation of the form $(m_{1,1}^{n/q}...m_{1,r_1}^{n/q})^q(m_{2,1}^{n/q}...m_{2,r_2}^{n/q})^q...(m_{s,1}^{n/q}...m_{s,r_s}^{n/q})^q=$$m_{1,1}^n...m_{1,r_1}^nm_{2,1}^n...m_{2,r_2}^n...m_{s,1}^n...m_{s,r_s}^n∈M^n$.$~\square$