I was surprised by this question which provides an example of a finite group $G$ where normal isomorphic subgroups $H, K\triangleleft G$ do not imply isomorphic quotient groups, i.e. $G/H\not\cong G/K$.
Question: What condition must be met in order for the quotient groups to be isomorphic?
Trying to sort this out intuitively, I think like this:
- A finite group is fully described by its multiplication table and isomorphic groups have the same table.
- Still we evidently can't say "Let $J\triangleleft G$ be a subgroup with table $T$; form the quotient group $G/J$".
- It breaks down when trying to carry out coset multiplication in $G/J$.
- Because coset multiplication depends not only on multiplication in $J$ as specified by $T$ but also on multiplication of elements in $J$ with elements in the embedding group $G$.
So an attempt at an answer would be that $\forall g\in G, h\in H:gh=g\psi(h)$ where $\psi$ is an isomorphism between $H$ and $K$. I am trying to express that isomorphic elements of the subgroups interact similarly with elements of the embedding group. Is this the right conclusion?