Minimal normal subgroups. Let $G$ a finite group with a proper non trivial normal subgroup $N$. Suppose that $N$ is solvable. Is it possible to find a normal abelian $p$-group $K$$\vartriangleleft$ $G$ (for some $p$ prime divisor of |$G$|)? How can I do?
Every kind of suggestion is appreciated. Thanks to everyone for the help!
 A: A minimal normal subgroup of a finite solvable group $N$ is an elementary abelian $p$-group for some prime $p$. If $N_1$ and $N_2$ are two of these (i.e. for the same fixed prime $p$), then by minimality $N_1 \cap N_2 = 1$, so $[N_1,N_2]=1$. Hence any two of them commute, and the subgroup $K$ of $N$ generated by all minimal normal $p$-subgroups of $N$ is itself elementary abelian, and it is characteristic. So $K$ char $N$ and $N \unlhd G$, hence $K \unlhd G$.
A: Look at $N$.
Consider its derived series, and smallest member, say $N^{(k)}\neq 1$ (but $N^{(k+1)}=1$).
Then $N^{(k)}$ is abelian (its derived subgroup is $1$).
Consider Sylow-$p$ subgroup $P$ of $N^{(k)}$ (for $p$ dividing $|N^{(k)}|$); it is unique, so characteristic in $N^{(k)}$. 
So we have following chain of subgroups 
$$1 < P < N^{(k)} < N < G.$$
($P$ char in $N^{(k)}$) and ($N^{(k)}$ char in $N$) implies ($P$ char in $N$).
($P$ char in $N$) and ($N$ normal in $G$) implies $P$ is normal in $G$. q.e.d.

needless to say: ($H$ char $K$) means $H$ is characteristic in $K$.
