TITLE: Subgroups of finite solvable groups. Solvable?
QUESTION [9 upvotes]: I am attempting to prove that, given a non-trivial normal subgroup $N$ of a finite group $G$, we have that $G$ is solvable iff both $N$, $G/N$ are solvable. I was able to show that if $N,G/N$ are solvable, then $G$ is; also, that if $G$ is solvable, then $G/N$ is. I am stuck showing that $N$ must be solvable if $G$ is.
It seems intuitive that any subgroup of a finite solvable group is necessarily solvable, as well. Is this true in general? For normal subgroups? How can I go about proving this result?
Edit: By solvable, I mean we have a finite sequence $1=G_0\unlhd ... \unlhd G_k=G$ such that $G_{j+1}/G_j$ is abelian for each $1\leq j<k$.

REPLY [6 votes]: It's easy to show that a group $G$ is solvable iff one of the following holds

$G$ has a solvable series, (which is the definition you gave)
The $n$-th derived subgroup $G^{(n)}$ vanish for some $n$.

Then for any $H\le G$, we have $H^{(n)}\subset G^{(n)}$ therefore $H$ is solvable too.