I have recently begun working my way through Serge Lang's Algebra, and on page 19 there is stated that for a group $G$ with a normal subgroup $H$, $G$ is solvable if and only if $H$ and $G/H$ are solvable. However, only the proof of $G$ solvable $\implies H$ solvable is given, with the rest of the statements to be proven as an exercise.
This is my attempt at proving $G$ solvable $\implies G/H$ solvable:
We have $G$ solvable, so that we can find a tower of subgroups $$ G = G_0 \supset G_1 \supset\cdots\supset G_m = 1 $$ so that the conditions on solvability are met. We then take the quotient so that $$ G/H = (G/H)\_0\supset (G/H)\_1 \supset\cdots\supset (G/H)\_m = H $$ where $(G/H)\_j = G_j/H$.
It is then immediately verified that all factor groups in this tower are abelian, as $(G_i/H)/(G_{i+1}/H) = G_i/G_{i+1}$, which is abelian by assumption.
To prove that each $G_{i+1}/H$ is normal in $G_i/H$, take $xH\in G_i/H$, $x^\prime H\in G_{i+1}/H$ and write $$ (xH)(x^\prime H)(xH)^{-1} = xHx^\prime Hx^{-1}H = xx^\prime x^{-1}H = x^\prime H $$ As a last step, we refine with the trivial group at the end to obtain the desired tower so that $G$ solvable does indeed imply $G/H$ solvable. The converse seems obvious as well to me.
What I am struggling with, however, is the proof that $H$ solvable $\implies G$ solvable. Here is how far I got:
We write the abelian tower of $H$, and then refine with $G$ on the left since $H$ is a subgroup of $G$ so that we have $$ G = G_0 \supset H = H_0 \supset H_1 \supset\cdots\supset H_m = 1 $$ By assumption, $H$ is normal in $G$ and each $H_{i+1}$ is normal in $H_i$. Likewise, each $H_i/H_{i+1}$ is abelian by assumption, so it only remains to prove that $G/H$ is abelian. This does however not seem like a viable angle of attack on this problem.
Is the rest of my reasoning valid? How else would one show $H$ solvable $\implies G$ solvable?
