Let $G$ be a group, $H, K \le G$. Under what conditions does $G/H \cong G/K \Rightarrow H \cong K$?

Question

Let $G$ be a group, $H, K \le G$. Under what conditions does $G/H \cong G/K \Rightarrow H \cong K$?

I believe there was a condition that could be simply stated under which it was true. Maybe having to do something with split exact sequences?

Answer 1

I'm not aware of any useful sufficient conditions other than extremely restrictive ones like that $G$ is finite elementary abelian (equivalently, a finite-dimensional $\mathbb{F}_p$-vector space) or cyclic (in which case a subgroup is determined by its index). Consider the following very simple counterexample: $G = C_2 \times C_4$ has two non-isomorphic subgroups $H = C_2 \times C_2, K = 1 \times C_4$ both of whom have quotient $C_2$. This is the smallest finite abelian group which is not elementary abelian or cyclic.

$G$ is also a counterexample to the analogous statement about whether $H \cong K$ implies $G/H \cong G/K$: this time if we take $H = C_2 \times 1$ and $K = 1 \times C_2$, the two subgroups are isomorphic but their quotients are $C_4$ and $C_2 \times C_2$ respectively. Actually this counterexample is Pontryagin dual to the previous one.