I just proved that subgroups of solvable groups are solvable. So given that $G$ is solvable there is $1=G_0 \unrhd G_1 \unrhd \cdots \unrhd G_s=G$ where $G_{i+1}/G_i$ is abelian and for $N$ a normal subgroup we know it is solvable so there is $1=N_0 \unrhd N_1 \unrhd \cdots \unrhd N_{r}=N$ where $N_{i+1}/N_i$ is abelian.
I'm trying to construct the chain for the entire group $G/N$ using the Lattice Isomorphism Theorem but am stuck. Is it possible to do this?
The "standard" proof:
Consider, for each $i$, the subgroups $G_iN/N$ of $G/N$. It is straight-forward to show that $G_iN$ is normal in $G_{i+1}N$ (and thus $G_iN/N$ is normal in $G_{i+1}N/N$).
Now $\dfrac{G_{i+1}N/N}{G_iN/N} \cong G_{i+1}N/G_iN$.
Taking any $x,y \in G_{i+1}$, and $n,n' \in N$, we see that:
$[xn(G_iN),yn'(G_iN)] = [xn,yn']G_iN$, and since $N \lhd G$, $[xn,yn'] \in [x,y]N$, so that:
$[xn,yn']G_iN = [xn,yn']NG_i = ([xn,yn']N)G_i = $
$([x,y]N)G_i = [x,y]NG_i = [x,y]G_iN = ([x,y]G_i)N$.
Since $G_{i+1}/G_i$ is abelian, $[x,y]G_i = G_i$ so $[xn,yn']G_iN = G_iN$,so that
$G_{i+1}N/G_iN$ is abelian, as desired.
(here, $[x,y] = xyx^{-1}y^{-1}$ and we are using the fact that for a group $G$, $G$ is abelian iff $[x,y] = e$ for all $x,y \in G$).
It should be noted that some of the quotients $G_{i+1}N/G_iN$ may be trivial, resulting in a shorter subnormal series for $G/N$.
(Corrected) Note that since the chain starts at $1<N$, there exists a maximun $i$ such that $G_i<N$. Multiplying and quotienting out that part of the chain gives a chain of quotients $1\lhd NG_i/N\lhd \cdots \lhd G/N$ and $(NG_{i+1}/N)/(NG_i/N)\simeq NG_{i+1}/NG_i\hookrightarrow G_{i+1}/G_i$ is abelian. The argument is essentially the same as the following: suppose that $\eta:G\to H$ is onto and $G$ is solvable. Then $H$ is solvable.
There's another part to the story: if $G/N$ and $N$ are solvable, show that $G$ is solvable too.
