Are the quotient groups in a composition sequence necessarily subgroups?

Question

Does there exist a finite group G and a normal subgroup N of G so that G/N is a simple group and G/N is not isomorphic to any subgroup of G ?

Answer 1

Yes, the perfect group $\operatorname{SL}(2,5)$ of order 120 is one. By Cauchy's theorem you won't find an example with $G/N$ abelian, so this is more or less the typical example. You want a non-split extension, and you probably want $N$ to be small to avoid coincidentally containing $G/N$. A good way to do this are the non-simple quasi-simple groups.

If $G/Z(G)$ is non-abelian simple, and $H \leq G$ is such that $H \cong G/Z(G)$, then $H \cap Z(G) \leq Z(H) = 1$, so we get a semi-direct product of $H$ and $Z(G)$, but such a semi-direct product is direct since $Z(G)$ is central, hence $G \cong G/Z(G) \times Z(G)$.

All we need then is for $Z(G) \leq [G,G]$, that is, for $G$ to be perfect, so that $G$ is quasi-simple.

There are examples that are not quasi-simple. It'd be nice to see a proof that they all have a (non-simple) quasi-simple subquotient, so that this example really is “typical.”