# Pair of short exact sequences of groups

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.

• 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). Jul 3 at 1:05

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 $$N_1N_2, so that $$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.

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)

• 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. Jul 3 at 2:28
• @IanAgol: Oh, that's a nice observation! I'll rewrite the answer to include that idea. Jul 3 at 23:21