Factoring by subgroups of subgroups

Question

Let $G$ be a group and $H$ be a subgroup such that $G/H$ is abelian. Let $H$ properly contain $K$, and $K$ is normal in $G$. Is $G/K$ also necessarily abelian?

One obvious counterexample can be found by letting $S$ be any non-abelian group, then $S/S$ is clearly abelian and then it is not required that all quotients of $S$ by proper subgroups be abelian ($G/\langle e \rangle \approx G$). Are there non-trivial counterexamples?

Answer 1

Of course there are nontrivial counterexamples.

If you want an example where all of $G$, $H$, and $K$ are nontrivial, take the dihedral group of order $16$, which is the group of rigid motions of a regular octagon. An explicit description is $$G = \langle r,s\mid r^8 = s^2 = e, sr=r^7s\rangle.$$

Let $H$ be the subgroup generated by $r$. This is normal, and $G/H$ is cyclic of order $2$, hence abelian.

Now let $K$ be the subgroup generated by $r^4$. This is the center of $G$, hence normal (in fact, characteristic). But $G/K$ is isomorphic to the dihedral group of order $8$, which is nonabelian.

More generally, you can take let $G$ any nilpotent group of class at least $3$, and let $H=[G,G]$ and $K=[H,G]$. Then $G/H$ is abelian, but $G/K$ is not abelian because its commutator subgroup is $H/K$, nontrivial.

Or take any group for which $H=[G,G]$ is a proper nontrivial subgroup, and $[H,H]$ is proper and nontrivial ($H$ nonabelian). For instance, any solvable group of solvability length three or more. For instance, take $G=S_4$, $H=A_4$, and $K$ the subgroup formed by the identity and the three involutions $(12)(34)$, $(13)(24)$, and $(14)(23)$.

Answer 2

I don't have a specific example, but a nice characterization of abelian quotients is the following:

Let $G$ be a group and consider its derived subgroup $G':=\langle[a,b]\mid a,b\in G\rangle$, where $[a,b]=a^{-1}b^{-1}ab$. If $N\trianglelefteq G$ then $G/N$ is abelian if and only if $N\geq G'$.

This means that you shall look for a group $G$ whose derived subgroup admits a non trivial subgroup, which is normal on $G$.

Edit: take the special linear group $SL(2,5)$: it's perfect, which means that $G'=G$, but it's not simple, therefore it has a non trivial subgroup of its derived subgroup, which means that it's a good counterexample thanks to the statement above.