# Semidirect factors of free product of two groups

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?

• What is your definition of semidirect factor? You can write $Z*Z$ as $N \rtimes Z$, where $N$ is the normal closure of the first free factor, and $N$ is not even finitely generated ($Z = {\mathbb Z}$). May 12 '18 at 11:34
• @DerekHolt I wrote the definition of a semidirect factor before the question. May 12 '18 at 11:44
• I can say that when $G$ and $H$ are free groups with finite ranks, since $G*H$ is also free, every semidirect factor of $G*H$ has the form $K*L$ up to isomorphism, where $K$ and $L$ are semiderct factors of $G$ and $H$, respectively. May 12 '18 at 11:53
• As I said in previous comment, that is not true when $G=H={\mathbb Z}$, which a free group of rank $1$. Your statement in your post about the possible semidirect factors of ${\mathbb Z}*{\mathbb Z}$ is incorrect. The free group of countable infinite rank is also a semidirect factor of ${\mathbb Z}*{\mathbb Z}$. May 12 '18 at 12:37
• @DerekHolt There is a one-to-one correspondence between the set of all isomorphism classes of semidirect factors of $G$ and the set of all homotopy types of retracts of $K(G,1)$, where $K(G,1)$ denotes the Eilenberg-MacLane space of group $G$. Since $\mathbb{S}^1 \vee \mathbb{S}^1 =K(\mathbb{Z}*\mathbb{Z},1)$, so I think my statement in the post is true. May 12 '18 at 13:10

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 $G = N \rtimes H$.

• Thank you very much for explanation. It seems that your answer is completely true. But can you tell me why my answer is wrong in the below? May 14 '18 at 8:48
• Thank you in advance. May 14 '18 at 9:16

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 $\leq n$.

• Suppose that $G = N \rtimes H$ is a semidirect product. All of your arguments apply to $H$. They do not apply to $N$, and $N$ need not be finitely generated. According to your definition $N$ is also a semidirect factor of $G$. May 14 '18 at 9:28
• @DerekHolt Why don't my arguments apply to $N$? where does Normalness of $N$ apply here? I understood that when $H$ is an $r$-image of $G$, then $f(H)$ is not necessary normal. But I don't understand why the arguments don't apply to $N$. May 14 '18 at 10:14
• You have proved that $H$ is an $r$-image of $G$. You have not made any attempt to prove that $N$ is an $r$-image of $G$. May 14 '18 at 13:27
• @DerekHolt Excuse me. According to the above discussion, can we say that the number of r-images of a group $G$ is equal to the cardinality of the set of pairs $(N,H)$ of subgroups of $G$ such that $G$ is the semidirect product $N\rtimes H$? Jun 17 '18 at 11:36
• The question needs clarification on how you define equivalence. Is an $r$-image an isomorphism class of groups? When are two pairs $(N_1,H_2)$, $(N_2,H_2)$ regarded as being equivalent? But I think the answer will be no in any case because you could have decompositions $(N_1,H)$ and $(N_2,H)$ with $N_1 \not\cong N_2$. Jun 17 '18 at 13:52