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We know that there exists a group $G$ with normal subgroups $N_1,N_2$ where $N_1\cong N_2$ but ${G\over N_1}\not\cong {G\over N_2}$.

Since this is the case it is natural to wonder about which pairs of groups can each be represented by quotients of some group by isomorphic subgroups. In particular, for which pairs of groups $H_1,H_2$ do there exist a group $G$ and normal subgroups $N_1,N_2$ of $G$ such that $N_1\cong N_2$, $H_1\cong {G\over N_1}$, and $H_2\cong {G\over N_2}$?

Are there any nice conditions on such pairs guaranteeing the existence of these quotients? Or is it perhaps even possible to find such a $G,N_1,N_2$ whenever $H_1$ and $H_2$ are of the same order?

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    $\begingroup$ Re: your last sentence, note that as long as we allow $G$ to be infinite even having different order isn't an obstacle. $\endgroup$ Nov 3, 2021 at 5:01
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    $\begingroup$ And if you allow infinite groups you can always take $G=H_1^\mathbb{N}\times H_2^\mathbb{N}$. $\endgroup$ Nov 3, 2021 at 5:32
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    $\begingroup$ If you restrict to finite groups, though, a necessary condition is not just that $H_1$ and $H_2$ have the same order but also that they have the same composition factors. $\endgroup$ Nov 3, 2021 at 5:34

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This is a question in which I've been interested for almost ten years! I don't want to get too much into detail about my motivation, but you can get an idea if you check:

M. Giudici, S .P. Glasby, C. H. Li and G. Verret, Arc-transitive digraphs with quasiprimitive local actions, J. Pure Appl. Algebra 223 (2019) 1217–1226.

The question there is a bit more general, your question is the restriction when both permutation groups are regular (and so we can just think of them as abstract groups). This restriction is considered at the end of Section 3, see Corollary 3.5 for example. I think this is a very natural question, and I know a little bit about it, but I wish I knew more!

To be explicit, I am assuming that (finite) $H_1$ and $H_2$ are given, and we want to know if there exists a finite $G$ having isomorphic normal subgroups $N_1$ and $N_2$, such that $G/N_i\cong H_i$. I will say that $H_1$ and $H_2$ are compatible in this case.

As Eric Wofsey pointed out, a necessary condition is that $H_1$ and $H_2$ have the same composition factors. One can improve this and prove that $H_1$ and $H_2$ must have a composition series with the same factors appearing in the same order. (The proof is quite nice and short, as a hint, consider a minimal witness $G$.)

So, for example, $A_4$, the alternating group of degree $4$, is not compatible with the dihedral group of order $12$, even though they have the same multiset of composition factors.

Note that compatibility is not an equivalence relation. For example, $\mathrm{SL}(2,5)$ and $S_5$ are not compatible (for the reason above) but they are both compatible to $A_5\times C_2$ (see Example 3.6 in the paper above).

I know many constructions to show that certain groups are compatible. But I am not able to show that the above necessary condition is not sufficient, although I do believe it is not sufficient.

As a final small taste, the smallest pair of which I cannot determine the status is $A_4$ and $C_{12}$. I believe they are not compatible, but cannot prove it. (I asked about it once on the Group Pub Forum, which did not resolve the issue.)

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