The problem is as follows:
Let $G$ be a group. $H \triangleleft G, K \triangleleft G$. To prove that:
both $G/H$ and $G/K$ are solvable $\iff$ $G/(H \cap K)$ is solvable.
This proposition is not hard to prove by making use of the lemma that "both subgroups and quotient groups of a solvable group are solvable".
As a beginner (for practice), I try to prove this proposition by using the notion of Abel series. I am able to construct the Abel series of $G/H$ (or $G/K$) from that of $G/(H \cap K)$. However, I failed to construct the Abel series of $G/(H \cap K)$ from those of $G/H$ and $G/K$. Could you offer me some hint?
You can use the commutator series for groups. A group $G$ is solvable if and only if for the series defined as $$G^{0} = G, G^{i+1} = [G^{i}, G^{i}]$$ there exists some $n$ such that $G^{n} = 0.$
So, $G/K$ is solvable iff for some $n$, $(G/K)^{n} = 0$ iff $G^{n} \subseteq K.$ Similarly, $G/H$ is solvable iff for some $m$, $G^{m} \subseteq H.$
So, if $G/K$ and $G/H$ are solvable, then find $n$ and $m$ as above, and then for $l = \max(n, m)$, $G^{l} \le, H, K$ and hence $G^{l} \le H \cap K$. Hence, $G/(H \cap K)$ is solvable.
Conversely, if $G/(H \cap K)$ is solvable, then for some $n$, $G^{n} \subseteq H \cap K$ and hence $G/K$ and $G/H$ are solvable. This finishes the proof.
Additionally, if you wish to use the abelian tower definition instead, it is not hard to prove that that definition is equivalent to the commutator series definition by using the fact that for a group $G$, a subgroup $H$ contains $[G,G]$ if and only if $G/H$ is abelian.