Subgroups of finite solvable groups. Solvable? 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 $1\leq j<k$.
 A: 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. 
A: With your definition, to show that if $G$ is solvable then $N$ is solvable, let
$$ 1 =G_0 \triangleleft G_1\triangleleft\cdots\triangleleft G_{m-1}\triangleleft G_m=G$$
be such that $G_{i+1}/G_{i}$ is abelian for each $i$.
Note: We do not need to assume that $N$ is normal; the argument below works just as well for any subgroup of $G$, not merely normal ones.
Let $N_i = G_i\cap N$. Note that since $G_i\triangleleft G_{i+1}$, then $N_i\triangleleft N_{i+1}$: indeed, if $x\in N_i$ and $y\in N_{i+1}$, then $yxy^{-1}\in N$ (since $x,y\in N$) and $yxy^{-1}\in G_i$ (since $G_i\triangleleft G_{i+1}$), hence $yxy^{-1}\in N\cap G_i = N_i$. 
So we have a sequence
$$1 = N_0\triangleleft N_1\triangleleft\cdots\triangleleft N_{m} = N.$$
Thus, it suffices to show that $N_{i+1}/N_i$ is abelian for each $i$.
Note that $N_{i} = N\cap G_i = (N\cap G_{i+1})\cap G_i = N_{i+1}\cap G_i$. 
Now we simply use the isomorphism theorems:
$$\frac{N_{i+1}}{N_i}  =\frac{N_{i+1}}{N_{i+1}\cap G_i} \cong \frac{N_{i+1}G_i}{G_i} \leq \frac{G_{i+1}}{G_i}$$
since $N_{i+1},G_i$ are both subgroups of $G_{i+1}$ and $G_i$ is normal in $G_{i+1}$, so $N_{i+1}G_i$ is a subgroup of $G_{i+1}$. 
But $G_{i+1}/G_i$ is abelian by assumption,  hence $N_{i+1}/N_i$ is (isomorphic to) a subgroup of an abelian group, hence also abelian, as desired.
