$H$ and $K$ are subgroups of $G$ and $H$ normalizes $K$. Is $H/K$ a group?

Question

I don't see any problem defining a group in this manner. $(aK)(bK):=(ab)K$. Let $ak_1$ and $bk_2$ be elements from $aK$ and $bK$. Then $ak_1bk_2=abk_3k_2$ since $H$ normalizes $K$. So the operation is well-defined. That it is a group follows from the fact that $H$ is a group.

Answer 1

If $H$ normalizes $K$, that means $H \subseteq N_G(K)$, which does not imply $K \subseteq H$. So $H/K$ has no meaning in general. You can construct a counterexample finding a group $G$ with $H=Z(G)$ non-trivial and $K \nsubseteq Z(G)$ for some subgroup $K$. $D_4$, the dihedral of $8$ elements, would work here...