Question. (Is this statement true?) Given non-simple, centerless group $G$ such that there exists exactly one non-trivial normal subgroup $N\triangleleft G$, then $G/N$ must be isomorphic to some subgroup $<G$.
Motivation. See this post. (My question is an extension of the one posed in the linked post.) In short, I'm looking for a non-simple, centerless group which lacks any semidirect product decomposition. I've tried looked at (i) Symmetry groups $S_n$, (ii) Dihedral groups $D_{2n}$ where $n$ is odd, (iii) quotient groups of the Quaternion group $Q$, (iv) etc., but none of them provide the desired contradiction. Does anybody have any ideas on how to proceed? Thank you!
Best Answer
No, any nonsplit extension of a nontrivial irreducible module $kG$-module with $k$ a finite field by a simple group $G$, such as ${2^3} \cdot L_3(2)$ (or $q^3 \cdot L_2(q)$ for an odd prime power $q>3$) is a counterexample.