# Does finitely generated groups have finitely many finite retracts?

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.

• Do you mean "up to isomorphism" or do you assume that $H\subset G$? Jul 4 at 15:05
• @freakish I mean $H\subset G$. Jul 4 at 15:35
• 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. Jul 4 at 15:39
• @LeeMosher Thanks a lot for the comment. Here my question is about finite retracts but your example is about infinite retracts. Jul 4 at 15:43
• I see, alright. Jul 4 at 15:43

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.