Does there exists a non-trivial free group $F$ and a finite group $L$ acting on $F$ such that the semidirect product $F\rtimes L$ is perfect?


Thanks @YCor for reformulating the question.



Perfect group that is split extension of a normal free subgroup of finite index
~~The question should be clarified.~~ *(done)*


Let me assume the question is as follows (I couldn't think of another nontrivial interpretation):



> 
> Does there exist a free group $F\neq 1$ and a finite group $L$ acting on $F$ such that the semidirect product $F\rtimes L$ is perfect?
> 
> 
> 


The answer is yes with $F=F\_9$.


Indeed, start from $H$ the derived subgroup of index 2 in the signed permutation group $C\_2\wr A\_5$. So $H$ has order $2^4.60$ and is perfect.


Let $F\_5$ be freely generated by $x\_1,\dots,x\_5$. Then $H$ naturally acts on $F\_5$: the alternating group permutes the generators, and the normal abelian subgroup acts by changing the sign of generators (because we restrict to $H$, we consider this changing the sign of two coordinates). For instance the element $x\_i\mapsto x\_i^{-1}$ for $i\le 2$ and $x\_i\mapsto x\_i$ for $i\ge 3$ is such an element.


Then the semidirect product is not perfect, but its subgroup of index $2$, $F\_9\rtimes H$ is perfect. Here $F\_9$ consists of elements of even length in $F\_5$.


