Call two finite groups $Q_1$ and $Q_2$ $\textit{compatible}$ 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:

$\textbf{Proposition:}$ If two groups are compatible, then they have subnormal series of the same length with the same factor groups appearing in the same order.

$\textit{Proof:}$ 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

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

Call two finite groups $Q_1$ and $Q_2$ $\textit{compatible}$ 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:

$\textbf{Proposition:}$ If two groups are compatible, then they have subnormal series of the same length with the same factor groups appearing in the same order.

$\textit{Proof:}$ 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.

Call two finite groups $Q_1$ and $Q_2$ $\textit{compatible}$ 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:

$\textbf{Proposition:}$ If two groups are compatible, then they have subnormal series of the same length with the same factor groups appearing in the same order.

$\textit{Proof:}$ 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