Let $ \pi,V $ be a representation of a perfect group $ G $.
I'm interested in sufficient conditions for a semi direct product like $ V \rtimes_\pi G $ to be perfect.
Requiring that $ \pi $ is faithful and or irreducible isn't enough because of trivial counterexamples using abelian groups.
I'm especially interested in the condition that $ \pi $ is fixed point free (the only vector $ v $ fixed by all of $ G $ is the zero vector). Is that sufficient for $ V \rtimes_\pi G $ to be perfect?
Motivation: Let $ V $ be the $ n-1 $ dimensional vector space of length $ n$ binary vectors whose entries sum to 0. There is a natural permutation representation $ \pi $ of the alternating group $ A_n $ on $ V $. This representation is fixed point free, in fact it acts transitively on the nonzero vectors. For $ n \geq 5 $ the semdirecrt product $ V \rtimes A_n $ is perfect.
Consider the Weil representation of $ Sp_{2n}(p) $ on $ V=\mathbb{F}_p^{2n} $ for $ p $ an odd prime. This representation is fixed point free, in fact it acts transitively on the nonzero vectors. $ V \rtimes Sp_{2n}(p) $ is perfect (except when $n=1,p=3 $ in which case the symplectic group is not perfect).
Is it good enough for $ \pi $ to be fixed point free? Or do we need that $ G $ is transitive on the nonzero elements of $ V $?
If you have some totally different sufficient condition for a semi direct product to be perfect feel free to share, these are just some thoughts.
And more generally I'm interested in any condition on a (possibly non split) extension of a perfect group $ G $ $$ 1 \to V \to E \to G \to 1 $$ Which is sufficient to imply that $ E $ is perfect.
EDIT: Really nice sufficient condition from Derek Holt: Let $$ 1 \to V \to E \to G \to 1 $$ be a SES with $ V $ abelian and $ G $ perfect. Then if $ V $ is a (nontrivial) irreducible $ G $ module then $ E $ must be perfect. The statement for semi direct products is just the special case when the sequence splits.
Requiring nontriviality of the module just avoids the case that $ E $ is a direct product of $ G $ with a cyclic group of order $ p $, which is technically an irreducible $ G $ module.
Necessary and sufficient conditions for the semidirect product to be perfect are that $G$ is perfect (which you are assuming) and that the action of $G$ on $M/{\rm Rad}(M)$ is fixed point free. The second condition is equivalent to $M$ having no nontrivial homomorphisms onto the trivial module, or that the action of $G$ on ${\rm Dual}(M)$ is fixed point free.
It is not sufficient for the action of $G$ on $M$ to be fixed point free. For example, $A_5$ has a $5$-dimensional fixed point free module $M$ over ${\mathbb F}_2$, with a $4$-dimensional irreducible submodule $N$, and $M/N$ is the trivial module, so $M \rtimes G$ is not perfect.