Trouble showing that quotient of solvable group is solvable

Question

As the title suggests, I'm trying to demonstrate that a quotient of a solvable group is solvable. What I have so far is this:

Let $G$ be a solvable group with normal series $N_{0} \trianglelefteq \cdots \trianglelefteq N_{\ell}$, where $N_{k} / N_{k - 1}$ is abelian for $k = 1, \ldots, \ell$. Moreover, let $K \trianglelefteq G$. Then the sequence $M_{k} = N_{k} K / K$ is a normal series for $G / K,$ and $M_{k} / M_{k - 1} \cong N_{k} K / N_{k - 1}K$. We need to show that $M_{k} / M_{k - 1}$ is abelian, but I've no clue how to go about it. My first intuition was to say that if $x, y \in N_{k} K$, then $(xy) N_{k - 1} K = (yx) N_{k - 1} K$ iff $(xy)^{-1} (yx) \in N_{k - 1}K$, and try to show this expression is in $N_{k - 1} K$ by adding in expressions of the form $(g g^{-1})$ as appropriate to get the right parts to cancel, but I'm skeptical that this is the right way to go about it because I cannot see how to apply the fact that $N_{k} / N_{k - 1}$ is abelian, which I would imagine is pretty central to the claim. I also don't see an obvious way to make convenient cancellations because both $x$ and $y$ are elements of $N_{k} K$, which I don't really see how to collapse into elements of $N_{k - 1}K$.

I'm pretty thoroughly stuck, and this is the last part of the problem I need to figure out. Any help is appreciated. Thanks!

Answer 1

Plan:

1. Let's fix an index $i$. Because $K\unlhd G$, we know that $N_iK=\{nk\mid n\in N_i, k\in K\}$ for all $i$.
2. Consider the mapping $f:N_i\to N_iK/N_{i-1}K$ defined by $f(n)=nN_{i-1}K$ that maps an element $n\in N_i$ to its coset modulo $N_{i-1}K$. Verify that this is a homomorphism of groups.
3. Show that the homomorphism $f$ in the previous step is surjective.
4. Show that $N_{i-1}\subseteq \operatorname{ker} f$.
5. Show that we also have a surjective homomorphism (induced by $f$) $$N_i/N_{i-1}\to N_iK/N_{i-1}K.$$
6. Why does step 5 imply that $N_iK/N_{i-1}K$ is abelian?