If G is finite solvable then it has a normal subgroup of special form

Question

Question

Show that if $G$ is a finite solvable group then it contains a normal subgroup of the form $c_p\times...\times c_p$ with $p$ a prime.

Attempt

Since $G$ is solvable it contains a fully invariant abelian subgroup $H$.$H$ is abelian and finite so $H\simeq c_{n_1}\times...\times c_{n_k}$ where $n_{1}|...|n_k$.Let $p|n_1,...,n_k$ then $c_p\times...\times c_p \lhd H \unlhd G(\ast)$.

Is this correct so far?Does $( \ast)$ imply that $c_p\times...\times c_p \lhd G$?

Answer 1

Let $M$ be a minimal normal subgroup. Since $M$ is solvable its derived group must eventually become trivial and since $M \gt 1$, we see that $M' \lt M$. Since $M'$ char $M \lhd G$, it follows that $M' \lhd G$, and the minimality of $M$ implies $M'=1$, so $M$ is abelian. Now pick a prime $p$ that divides $|M|$. Then the set $S=\{x \in M : x^p=1 \}$ is a non-trivial (use Cauchy's Theorem!) subgroup of $M$ and it is even characteristic. Hence $S \lhd G$. And by the minimality of $M$, we get $M=S$, that is $M \cong C_p \times \cdots \times C_p$. This completes the proof.