Does there exist a pair of finite groups $G$ and $H$ satisfying both of the short exact sequences $1 \rightarrow G \rightarrow H \rightarrow A_4 \rightarrow 1$ and $1 \rightarrow G \rightarrow H \rightarrow D_6 \rightarrow 1$? Of course the homomorphisms $G \to H$ in these short exact sequences are not the same.

$\begingroup$ I am not sure how to continue this, but if we had a functor $F: \textrm{Grp} \to A$, where $A$ is an abelian category, we could infer that $[F(A_4)] = [F(D_6)]$ in the Grothendieck ring of $A$ (which are known in some cases) and see if there are obstructions. I haven't found a satisfying option though. My attempts: homotopy groups of classifying spaces and character rings (which could work, but I am not smart enough to work this out). $\endgroup$– Andrea MarinoJul 3, 2022 at 1:05
2 Answers
Call two finite groups $Q_1$ and $Q_2$ compatible if there exists a finite group $G$ with two isomorphic normal subgroups $N_1$ and $N_2$ such that $G/N_i\cong Q_i$.
One can show the following:
Proposition: If two groups are compatible, then they have subnormal series of the same length with the same factor groups appearing in the same order.
Proof: Let $Q_1$ and $Q_2$ be compatible with $G$ a witness of minimal order, and $N_1$ and $N_2$ the two corresponding isomorphic normal subgroups and let $\alpha$ be an isomorphism from $N_1$ to $N_2$. Let $M=N_1\cap N_2$. Note that $M$ and $\alpha(M)$ are isomorphic and normal in $N_2$, so $N_2/M$ and $N_2/\alpha(M)$ are compatible, with $N_2$ as a witness.
But $N_2/M\cong N_1N_2/N_1$ while $N_2/\alpha(M)\cong N_1/M\cong N_1N_2/N_2$. Minimality of $G$ implies that $N_1N_2<G$, so that $Q_1$ and $Q_2$ have $G/N_1N_2$ as a nontrivial common quotient, but moreover the corresponding normal subgroups are compatible, so the result follows by induction. $\square$
I've read somewhere that the above argument (which is in some sense a generalisation of the one by Robert) is due to Sims, but I'm not sure the argument itself is actually written anywhere.
In particular, it shows that $A_4$ and $D_6$ are not compatible, because they don't have such subnormal series. (Any series for $A_4$ has a $C_3$ "on top", and in $D_6$, a $C_2$ "on top".)
I've been interested in the question of determining which groups are compatible for a while. I think it's an interesting question and the answer is not known. See
Giudici, Glasby, Li, Verret, Arctransitive digraphs with quasiprimitive local actions, Journal of Pure and Applied Algebra 223 (2019) 12171226
for some motivation and further results.
The answer is no. To help keep our notation straight, assume that there is a finite group $H$ and normal subgroups $G_1$ and $G_2$ of $H$ such that $G_1 \cong G_2$ and such that we have short exact sequences $$1 \longrightarrow G_1 \longrightarrow H \stackrel{f}{\longrightarrow} A_4 \longrightarrow 1$$ and $$1 \longrightarrow G_2 \longrightarrow H \stackrel{g}{\longrightarrow} D_6 \longrightarrow 1.$$ Choose this $H$ such that its cardinality is as small as possible.
Define $$\overline{G}_1 = G_1 / G_1 \cap G_2 \quad \text{and} \quad \overline{G}_2 = G_2 / G_1 \cap G_2.$$ The homomorphism $f$ induces an isomorphism between $\overline{G}_2$ and a nontrivial normal subgroup of $A_4$, and the homomorphism $g$ induces an isomorphism between $\overline{G}_1$ and a nontrivial normal subgroup of $D_6$. What is more, since $G_1$ is isomorphic to $G_2$ the groups $\overline{G}_1$ and $\overline{G}_2$ have the same cardinality. Examining the nontrivial normal subgroups of $A_4$ and $D_6$, we see that the only possibility is that these subgroups are actually the entire groups, i.e. that $$\overline{G}_1 \cong D_6 \quad \text{and} \quad \overline{G}_2 \cong A_4.$$ This implies that we have short exact sequences $$1 \longrightarrow G_1 \cap G_2 \longrightarrow G_1 \longrightarrow D_6 \longrightarrow 1$$ and $$1 \longrightarrow G_1 \cap G_2 \longrightarrow G_2 \longrightarrow A_4 \longrightarrow 1.$$ We conclude that $G = G_1 = G_2$ is a group that of cardinality strictly smaller than $H$ that fits into the desired exact sequences, contradicting the minimality of the cardinality of $H$.
(the original post only contained a bunch of observations about the problem, but in the comments Ian Agol pointed out that one could use them as above to give a negative answer)

3$\begingroup$ Assume $H$ was minimal with respect to this property, then your argument gives a contradiction (a sort of “descent” argument). Replace $H$ with $G$ and $G$ with $G_1\cap G_2$. Then this is a smaller pair of groups with the same property by your observation. $\endgroup$– Ian AgolJul 3, 2022 at 2:28

$\begingroup$ @IanAgol: Oh, that's a nice observation! I'll rewrite the answer to include that idea. $\endgroup$– RobertJul 3, 2022 at 23:21