About Groups and Isomorphic Quotients

Question

So I was studying some group theory and I came up with this question: Say you have a group $G$ and two isomorphic normal subgroups $A,B$, is it true then that $$\dfrac{G}{A}\cong\dfrac{G}{B}$$ Turn out, it is not, take $G=\mathbb{Z}_4\times\mathbb{Z}_2$ and $A=\mathbb{Z}_2\times1$ and $B=1\times\mathbb{Z}_2$, so $G/A\cong\mathbb{Z}_2\times\mathbb{Z}_2$ and $G/B\cong\mathbb{Z}_4$ which are not isomorphic to each other. Now, my question is, is there any desirable property that relates isomorphic subgroups and their respective quotients? Perhaps if the groups are not finite, or different than direct product of groups (?). Thanks a lot!

Answer 1

I am surprised someone hasn't said to look at group extensions or group cohomology. One does actively consider when a group $G$ is built as an extension of a normal subgroup $N$ by a group $Q$ and how many ways that can be done. The "how many" has possible ambiguity.

You could ask first, how many ways does $Q$ act on $N$, that is, $N$ is normal so pulling back to $G$, $Q$ will conjugate $N$ producing autormophisms. It gets complicated if $N$ is not abelian as then you have to keep track of what $N$'s own inner automorphisms are doing. So most group cohomology assumes $N$ is abelian. So ingredient 1 is $\theta:Q\to \mathrm{Aut}(N)$.

Then for fixed $\theta$ you can ask how many ways you can build a group $G$ such that it has a normal $N$ with a quotient $Q$ such that the action of $Q$ on $N$ is $\theta$. For that you just right down a projection $\pi:G\to Q$ and a section $\ell:Q\to G$ and see what is the obstruction $f(x,y)$ to the homomorphsim property $$\ell(xy)f(x,y)=\ell(x)\ell(y).$$ You get what is called factor set or 2-cocyclic $f:Q\times Q\to N$. There is a natural equivalence class to put on these -- do they differ by just conjugation -- so you get co-boundaries, or rather cohomology $H^2(N,Q,\theta)$. This capture all extensions.

3rd, just because two extensions might be different, the groups may yet be isomorphic without the extensions being the same. (You can build the same group two different ways). So you then have to consider the action of $\mathrm{Aut}(N)\times\mathrm{Aut}(Q)$ on $H^2(N,Q,\theta)$ dividing up into orbits.

All this is mostly impractical except with small groups but it does in fact give a complete answer in theory. You can read about this in most group theory texts under the section titled "cohomology". If you text doesn't have such a section, consider that a somewhat bad group theory reference.