Does a free group $F$ of finite rank $n$ have finitely many retracts (as a subgroup)?

Question

A subgroup $H$ of a group $G$ is called a retract of $G$ if there exists an epimorphism $r:G\to H$ such that $r(h)=h$ for all $h\in H$.

Does a free group $F$ of finite rank $n$ have finitely many retracts (as a subgroup)? Does a retract of a free group $F$ of finite rank have a specific form?

What I've tried: Let $G=\mathbb{Z}$. Let $H\neq 1$ be a retract of $G$. Then there is an epimorphism $r:G\to H$ such that $r(h)=h$ for all $h\in H$. So $G=H\ltimes \mathrm{ker}(r)$. Since $\mathrm{ker}(r)\lhd G$, if $\mathrm{ker}(r)\neq 1$, then we have $\mathrm{ker}(r)=n\mathbb{Z}$, hence $H\cong \mathbb{Z}_n$ which is a contradiction to the fact that $H$ is a subgroup of free group $\mathbb{Z}$. Thus $\mathrm{ker}(r)=1$, and then $H=\mathbb{Z}$. We conclude that either $H=1$ or $H=G$.

Infinitely many unless $n\le 1$: Each free factor of $F_n$ is a retract and there are infinitely many free factors. — Moishe Kohan, Aug 11 '22 at 15:28
An interesting question is whether every retract is a free factor. If so, there are only finitely many retracts up to the action of $Aut(F_n)$. — Moishe Kohan, Aug 13 '22 at 14:37
@MoisheKohan That's a really question. I'll consider it and try to prove it. Thanks very much for the comment. — M.Ramana, Aug 13 '22 at 14:43
My question has a negative answer: https://mathoverflow.net/questions/25944/does-every-retraction-of-free-groups-arise-from-projection-to-a-subset-of-a-free — Moishe Kohan, Aug 13 '22 at 16:11

score 1 · Accepted Answer · answered Aug 13 '22 at 14:27

I close this question by giving the answer of @MoisheKohan in the comments: For $n\geq 2$, the statement is false because each free factor of $F_n$ is a retract and there are infinitely many free factors. Thanks to @MoisheKohan for pointing this out.

Does a free group $F$ of finite rank $n$ have finitely many retracts (as a subgroup)?

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