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A group $H$ is called a retract of a group $G$ if there exists homomorphisms $f:H\to G$ and $g:G\to H$ such that $gf=id_H$.

We know that a group $G$ is finite if and only if $G$ has finitely many subgroups.

Now my question is that a finitely generated group $G$ has finitely many finite retracts?

What I've tried: If $G$ is a finitely generated abelian group, then every retract of $G$ is a direct summand of $G$. So the number of finite retracts of $G$ is finite.

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    $\begingroup$ Do you mean "up to isomorphism" or do you assume that $H\subset G$? $\endgroup$
    – freakish
    Jul 4 at 15:05
  • $\begingroup$ @freakish I mean $H\subset G$. $\endgroup$
    – M.Ramana
    Jul 4 at 15:35
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    $\begingroup$ If instead of "up to isomorphism" you assume $H \subset G$, then your statement for finitely generated abelian groups is false. For a counterexample take $G = \mathbb Z \oplus \mathbb Z$. There are infinitely many subgroups $H \subset G$ which are retracts in that sense: given two relatively prime integers $a,b$ the infinite cyclic subgroup $H = \{(na,nb) \mid n \in \mathbb Z\}$ is a direct summand and therefore a retract. $\endgroup$
    – Lee Mosher
    Jul 4 at 15:39
  • $\begingroup$ @LeeMosher Thanks a lot for the comment. Here my question is about finite retracts but your example is about infinite retracts. $\endgroup$
    – M.Ramana
    Jul 4 at 15:43
  • $\begingroup$ I see, alright. $\endgroup$
    – Lee Mosher
    Jul 4 at 15:43

1 Answer 1

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Assuming, as you say in your comment, that you are interested in retracts as subgroups $H \subset G$, here is a counterexample.

Consider the infinite dihedral group $D_\infty = \langle a,b \mid a^2 = b^2 = \text{Id}\rangle $. It is a finitely generated subgroup with infinitely many cyclic subgroups of order 2, every one of which is a retract. Here are some details.

The group $D_\infty$ has an infinite cyclic normal subgroup $Z \subset D_\infty$ of index $2$. The quotient group $D_\infty / Z$ is the finite cyclic group of order $2$ (apologies for the link, but, it is after all one of the best things ever).

Every element of the unique non-identity coset $D_\infty - Z$ is conjugate to $a$ and has order $2$, these elements have the form $r_n = (ab)^na$, they are all different, and so we have infinitely many subgroups $H_n = \langle r_n \rangle$ each of which is cyclic of order 2. Each such subgroup $H_n$ is a retract: we have a quotient homomorphism $$g : D_\infty \mapsto D_\infty / Z \mapsto H_n $$ and an inclusion $$i : H_n \hookrightarrow D_\infty $$ and the composition $$H_n \xrightarrow{i} D_\infty \xrightarrow{g} H_n $$ is the identity.

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  • $\begingroup$ Thank you so much for your great answer. For me, it has some good points about retracts. I appreciate it. $\endgroup$
    – M.Ramana
    Jul 4 at 16:09
  • $\begingroup$ Can I ask another question? Is there finitely generated infinite groups with finitely many (finite) retracts? $\endgroup$
    – M.Ramana
    Jul 5 at 4:49
  • $\begingroup$ Asking a new question in the comments of an answered question means that your new question is invisible to everyone except you, me, and anyone who by happenstance has read this deeply into this particular question. Instead, post a new question (but make sure that you give it proper context if you desire for it not to be closed). $\endgroup$
    – Lee Mosher
    Jul 5 at 13:35
  • $\begingroup$ You are right. I got it. Consider it done. $\endgroup$
    – M.Ramana
    Jul 5 at 14:28

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