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). $\endgroup$ Jul 3, 2022 at 1:05

2 Answers 2


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 non-trivial 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, Arc-transitive digraphs with quasiprimitive local actions, Journal of Pure and Applied Algebra 223 (2019) 1217-1226

for some motivation and further results.

See also https://math.stackexchange.com/questions/4295186/which-pairs-of-groups-are-quotients-of-some-group-by-isomorphic-subgroups/4296206#4296206


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 Agol
    Jul 3, 2022 at 2:28
  • $\begingroup$ @IanAgol: Oh, that's a nice observation! I'll rewrite the answer to include that idea. $\endgroup$
    – Robert
    Jul 3, 2022 at 23:21

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