# What's an example of a finite, non-abelian, non-simple group that is *not* semidirectly reducible? [duplicate]

Say I want to classify all groups of a given order. The abelian case is completely understood by the structure theorem for finitely generated abelian groups.

Assume our group is non-abelian, and we somehow managed to find a normal subgroup $$N$$ (e.g., by considering the core of some Sylow subgroup). It would be pretty nice if we could go on to construct all semidirect products of $$N$$ with groups of the remaining order – but $$G$$ need not be semidirectly reducible*.

The only counterexample that comes to mind is $$0 \to C_2 \to C_4 \to C_2 \to 0$$, which does not admit an appropriate section, i.e. a subgroup complementing our normal subgroup with trivial intersection.

What are counterexamples in the non-abelian case?

## marked as duplicate by Community♦Nov 8 '18 at 10:15

What about $$\mathrm{Aut}(A_6)$$, where $$A_6$$ is the alternating group on six letters? $$A_6$$ is normal in $$\mathrm{Aut}(A_6)$$ and it has no complement. Or consider the Mathieu group $$M_{10}$$. It contains $$A_6$$ as a normal subgroup of index $$2$$ (indeed, $$M_{10}^{'} = A_6$$), but $$A_6$$ again has no complement in $$M_{10}$$. Although these groups are not simple, they are both almost simple.
If, instead, you are interested in soluble groups only, take an odd prime $$p$$ and consider the non-abelian group $$P$$ of order $$p^3$$ and exponent $$p$$. The centre of that group is $$\cong C_p$$ and it has no complement in $$P$$, so $$P$$ is a non-split extension of $$C_p$$ by $$C_p \times C_p$$. More generally, if $$G$$ is a non-abelian group and $$G/Z(G)$$ is abelian (we can guarantee this, for example, by requiring $$|G:Z(G)| = p^2$$ for some prime $$p$$) then $$Z(G)$$ can have no complement in $$G$$. (Do you see why?)
Of course, if we can find a normal subgroup $$N$$ of $$G$$ and $$\gcd(|N|,|G:N|)=1$$, then $$G$$ is a semi-direct product, because $$N$$ is guaranteed to have a complement in $$G$$ by the Schur-Zassenhaus theorem.
• Is there an obvious was to see that it has no complement? And do you mean the inclusion into the inner automorphisms, or something specific to $A_6$? – Luke Nov 7 '18 at 22:28