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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?

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  • $\begingroup$ 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 ($Z = {\mathbb Z}$). $\endgroup$
    – Derek Holt
    May 12 '18 at 11:34
  • $\begingroup$ @DerekHolt I wrote the definition of a semidirect factor before the question. $\endgroup$
    – M.Ramana
    May 12 '18 at 11:44
  • $\begingroup$ 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. $\endgroup$
    – M.Ramana
    May 12 '18 at 11:53
  • $\begingroup$ 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 ${\mathbb Z}*{\mathbb Z}$. $\endgroup$
    – Derek Holt
    May 12 '18 at 12:37
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    $\begingroup$ @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. $\endgroup$
    – M.Ramana
    May 12 '18 at 13:10
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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 $G = N \rtimes H$.

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  • $\begingroup$ 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? $\endgroup$
    – M.Ramana
    May 14 '18 at 8:48
  • $\begingroup$ Thank you in advance. $\endgroup$
    – M.Ramana
    May 14 '18 at 9:16
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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 $\leq n$.

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  • $\begingroup$ 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 $G$. $\endgroup$
    – Derek Holt
    May 14 '18 at 9:28
  • $\begingroup$ @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 $N$. $\endgroup$
    – M.Ramana
    May 14 '18 at 10:14
  • $\begingroup$ 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 $G$. $\endgroup$
    – Derek Holt
    May 14 '18 at 13:27
  • $\begingroup$ @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 $N\rtimes H$? $\endgroup$
    – M.Ramana
    Jun 17 '18 at 11:36
  • $\begingroup$ 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 $N_1 \not\cong N_2$. $\endgroup$
    – Derek Holt
    Jun 17 '18 at 13:52

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