Semidirect factors of free product of two groups

Question

Given a group $G$ with identity element 1, a subgroup $H$, and a normal subgroup $N$ of $G$; $G$ is called the semidirect product of $N$ and $H$, written $G = N\rtimes H$ , if $G = NH$ and $H\cap N=1$. Then $H$ is called a semidirect factor of $G$.

From the fact that every retract of $\mathbb{S}^1 \vee \mathbb{S}^1$ has the homotopy type of $*$, $\mathbb{S}^1$ or $\mathbb{S}^1 \vee \mathbb{S}^1$, one can conclude that ever semidirect factor of $\mathbb{Z}*\mathbb{Z}$ has the form $1$, $\mathbb{Z}$ or $\mathbb{Z}*\mathbb{Z}$ up to isomorphism.

Now let $G$ and $H$ be two groups. What is the form of semidirect factors of $G*H$? Or at least up to isomorphism?

Answer 1

Let $F =\langle x,y\rangle={\mathbb Z}*{\mathbb Z}$ be free of rank two.

Let $N$ be the normal closure of $\langle x \rangle$ in $G$. Then $N$ is free of infinite rank, with free generators $\{ y^{-k}xy^k : k \in {\mathbb Z} \}$.

Let $H = \langle y \rangle$. Then $N \unlhd G$, $NH=G$, and $N \cap H = 1$, so $G = N \rtimes H$.

Answer 2

First I show that there is a one to one correspondence between the set of all isomorphism classes of semidirect factors of $G$ and the set of all isomorphism classes of r-images of $G$. Recall that a group $H$ is called an $r$-image of $G$ if there exist homomorphisms $f:H\longrightarrow G$ and $g:G\longrightarrow H$ so that $g\circ f=id_H$.

I define the map $\Phi$ from the set of all isomorphism classes of semidirect factors of $G$ to the set of all isomorphism classes of r-images of $G$ by $\Phi ([H])=[f(H)]$, where $[.]$ denotes the isomorphism class. Clearly, $\Phi$ is injective. For surjection, let $G=N\rtimes H$. Then it is sufficient to define $g:G\longrightarrow H$ by $g(nh)=h$. If $f:H\longrightarrow G$ is the inclusion map, then $g\circ f=id_H$ and so $H$ is an $r$-image of $G$.

Now, let $F$ be a free group of finite rank $n$ and $H$ be a semidirect factor of $F$. Then by the above, there exist homomorphisms $f:H\longrightarrow F$ and $g:F\longrightarrow H$ so that $g\circ f=id_H$. One can define $\bar{f}:H/H' \longrightarrow G/G'$ and $\bar{g}:G/G' \longrightarrow H/H'$ by $\bar{f}(hH')=f(h)G'$ and $\bar{g}(xG')=g(x)H'$, respectively. Then $\bar{g}\circ \bar{f}=id$. This shows that $H/H'$ is an $r$-image of $F/F'$. By the above, $H/H'$ is a direct summand of $F/F'$ which implies that $H/H'$ is a free abelian group of finite rank $\leq n$. Since $H/H'$ and $H$ have the same rank, $H$ is free group of finite rank $\leq n$.