Perfect group that is split extension of a normal free subgroup of finite index

Question

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.

Answer 1

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 $F_5$.

Answer 2

An alternative construction of such a group: let $G$ be the free product of two copies of $A_5$, i.e., $G=A_5*A_5$. This group $G$ is perfect and it is also virtually free, so that any torsion-free subgroup is free. Consider the kernel of any group homomorphism from $G$ to $A_5$ that is an isomorphism when restricted to each of the two copies of $A_5$. This kernel is free of rank 59, and since $G$ does contain copies of $A_5$ that map isomorphically to the quotient $A_5$, $G$ is isomorphic to $F_{59}\rtimes A_5$.

The calculation that the rank of the kernel is 59 is as follows: the Bass-Serre tree for $G=A_5*A_5$ is a tree with $G$-action with one free orbit of edges, and two orbits of vertices, each with stabilizer a copy of $A_5$. Since the kernel of the map $G\rightarrow A_5$ has trivial intersection with the vertex stabilizers, it acts freely on the tree. It thus has two free orbits of vertices and 60 free orbits of edges. So the quotient graph is two vertices joined by 60 edges, which has fundamental group $F_{59}$.

Answer 3

This is a follow-up to IJL's answer. Let me describe it with an explicit action.

Let $G$ be a group acting on a set $X$ and $x_0\in X$. Write $X'=X\smallsetminus\{x_0\}$.

Then $G$ acts on the free group $F_{X'}$ (with basis $(u_x)_{x\in X'}$ as follows: $$g\cdot u_x=u_{gx}u_{gx_0}^{-1}\qquad (\text{with } u_{x_0}:=1)$$

We can then consider the semidirect product $G\ltimes F_{X'}$.

If $X'=G$ (left action), one can check that this is (*) indeed isomorphic to $G\ast G$. More generally if $X=G/H$, this is the double $G\ast_H G$. In particular, if $G$ is perfect and $X$ is $G$-transitive, then this is a perfect group (it is a bit messy to check directly).

(*) To check this, start from $G\ast G$. Let $F$ be the kernel of the canonical homomorphism onto $G$ that is identity on both factors. Let $i_1,i_2$ be the two inclusions of $G$. The kernel is generated by the $j(g)=i_1(g)i_2(g^{-1})$ for $g$ ranging over $G\smallsetminus\{1\}$ (straightforward). The argument in IJL's answer, at least for $G$ finite, shows that this kernel is free of rank $|G|-1$. Since we have a generating family of the right rank, it has to be a free generating subset. Computing the action of $G\sim i_1(G)$ on the generators of $F$ exactly yields the desired formula above.

In the amalgam case, observe that $j(g)$ only depends on the class of $g$ in $G/H$ and $j(1)=1$, and the remainder is similar.