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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!

J.G.131
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1 Answers1

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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.

Derek Holt
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  • You may want to add "$kG$-module with $k$, say, a prime field." – Moishe Kohan Jun 06 '24 at 19:25
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    @MoisheKohan I've done that, but it doesn't have to be a prime field. – Derek Holt Jun 06 '24 at 21:45
  • Then it would be wrong. If $k$ is a finite field with prime subfield $P$ and $k^n$ is an irreducible $kG$-module, then $P^n< k^n$ is still $G$-invariant and, hence, is a normal subgroup of the semidirect product different from $k^n$. – Moishe Kohan Jun 06 '24 at 23:33
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    @MoisheKohan As a very small example, let $k$ be the finite field of order $4$. Then for $G=C_3$, there is a $1$-dimensional $kG$-module with nontrivial action, and the corresponding semidirect product is isomorphic to $A_4$, which has $k$ as its unique nontrivial proper normal subgroup. – Derek Holt Jun 07 '24 at 07:32