If $N$ is a normal subgroup of $G$ with $N$ and $G/N$ solvable, prove that $G$ is solvable I'm trying to prove that if $N$ is a normal subgroup of $G$, with $N$ and $G/N$ solvable, then $G$ is solvable.
Proving that $G/N$ is abelian would of course suffice, but I'm not sure if that's a necessary condition or not.  I suppose it's possible that I could find some normal subgroup $K$ in $G$ such that $N\subseteq K\subseteq G$ with both $G/K$ and $K/N$ abelian, but I can't see how to go about constructing it.
I've tried a couple things with the isomorphism theorems, and looking at properties preserved under homomorphisms and such, but nothing's panned out.
Can anyone help me here, I'm a bit stuck, thanks.
 A: This is a good to know that:

Theorem: $G$ is a group and $N\unlhd G$ then, $\forall n\in \mathbb N, \big(\frac{G}{N}\big)^{(n)}=\frac{G^{(n)}N}{N}$

wherein $G^{(1)}=G'$ and $G^{(i)}=\big(G^{(i-1)}\big)', i\geq 2$. Definition tells there are non-negative integer $n,m$ such that $N^{(n)}=\{e\}$ and $\big(\frac{G}{N}\big)^{(m)}=\{N\}$ because $N$ and $\frac{G}{N}$ both are soluble. Using the above theorem, we have $$\{N\}=\bigg(\frac{G}{N}\bigg)^{(m)}=\frac{G^{(m)}N}{N}\Longrightarrow G^{(m)}\leq N $$
$$G^{(m)}\leq N \Longrightarrow \big(G^{(m)}\big)^{(n)}\leq N^{(n)}=\{1\}\Longrightarrow G^{(m+n)}\leq N^{(n)}=\{1\} $$ or $G$ is soluble gorup.
A: You are a little bit wrong in your question. There are two definitions (but of course identical) of solvable group. One is using commutator and the other one is using the fact that $G/G'$ is an abelian group (NOT just that $G/N$ is abelian such mentioned in your question).
I will answer using the commutator definition one as it's more simple and straightforward.
Solvable group is a group that firstly it has a series defined as:

$G = G^{(0)} \gt G^{(1)} \gt G^{(2)} \gt ...$

where $G^{i+1} = [x, y]$ and where $x, y \in G^{(i)}$, so basically the next group in the series is the commutator group of its previous group. Note that $G^{(i+1)} / G^{(i)}$ is an abelian, so we know that the second definition of solvable group is using that fact.
Secondly the series must end at some point where $G^{(d)} = 1$:

$G = G^{(0)} \gt G^{(1)} \gt G^{(2)} \gt ... \gt G^{(d)} = 1$

That is the complete definition of solvable group using commutator.
Now to prove, first we need this lemma:

Homomorphic images of soluble groups are soluble.

So if we have a homomorphism $\Phi: G \rightarrow G/N$ and $G$ is solvable, so is $G/N$.
By assumption (the question) $G/N$ is solvable, so by definition of solvable group we have $(G/N)^{(d)} = 1$. By the above lemma and using the same homomorphism mentioned above then we have:
$\Phi(G^{(d)}) = (G/N)^{(d)} = 1 = N$
$G^{(d)} \leq N$
To continue, we need this second lemma:

Subgroups of solvable groups are solvable.

As by assumption N is a solvable group so we have:
$(G^{(d)})^{(e)} \leq N^{(e)} = 1$
$G^{(d+e)} = 1$
Thus $G$ is solvable.
A: If $~1=U_0\le U_1\le \cdots \le U_n=G/N~$ and $~1=V_0\le V_1\le \cdots \le V_m=N$ are subnormal series with abelian factor groups, then consider the series induced by "superimposing" the former over latter:
$$1=V_0\le V_1\le \cdots\le V_m=N=U_0'\le U_1'\le \cdots \le U_n'=G$$
where $U_i'$ are the unique subgroups of $G$ such that $U_i'/N=U_i$ (given by the lattice theorem).
