To determine whether every finitely generated group has finitely many retracts up to isomorphism, we start by recalling that a retract of a group ( G ) is a subgroup ( H ) for which there exists a homomorphism ( r: G \to H ) (a retraction) with ( r(h) = h ) for all ( h \in H ). Retracts are finitely generated because they are images of homomorphisms from a finitely generated group.

...

No.

There are finitely generated groups that have infinitely many pairwise nonisomorphic retracts.

One convenient way to see this is to start from a countable free product of the groups you want to appear as retracts, and then use a standard embedding theorem (via small–cancellation/HNN techniques) that places that countable group as a retract of a 2–generated group.

Here is a concrete scheme. ...

A definite answer is not known at present.

What is asked is whether, for every finitely generated group G, the set

  { H ≤ G | H is a retract of G } / ≅

is finite. (Here two retracts are identified when they are isomorphic as abstract groups.)

• Many standard families – finitely generated abelian groups, free groups, surface groups, right–angled Artin groups, poly-cyclic groups, one–relator...

<think> We are given that there are finitely generated groups with infinitely many retracts (subgroups that are images of retractions, i.e., idempotent endomorphisms). However, in the examples provided (infinite dihedral group, free abelian group of rank 2, and free groups), while there are infinitely many retracts, they fall into only finitely many isomorphism classes. The question is whether thi...

Answering the question: Does every finitely generated group have finitely many retracts up to isomorphism?

The answer is no. There exist finitely generated groups that have infinitely many retracts which are pairwise non-isomorphic.

The examples you provided (, , ) are excellent starting points, and your observation that they have finitely many ...

First, let us restate the problem in equivalent “algebraic” terms. A subgroup H≤G is a retract if there is a homomorphism
r:G→H
with
r|H=Id_H.
Equivalently G splits as a semidirect product
G = Ker r ⋊ H.
We want to know:

“If G is finitely generated, can it ever have infinitely many non-isomorphic retracts?”

All of the standard examples of f.g. groups with infinit...

This is a great question about the structure of retracts in finitely generated groups. Let me analyze this carefully.

The answer is no - there exist finitely generated groups with infinitely many retracts up to isomorphism.

Here's a counterexample:

Consider the free group on two generators. This group has infinitely many non-isomorphic retracts.

**Construction ...

Does every finitely generated group have finitely many retracts up to isomorphism?

To answer this question, I need to carefully analyze what it means for a subgroup to be a retract and examine whether finitely generated groups can have infinitely many non-isomorphic retracts.

Understanding Retracts

A retract of a group G is a subgroup H for which there exists a homomorphism r: G → H (calle...

At present nobody knows the answer – the problem is open.

What is known

1. For a number of large classes of finitely generated groups one can show that there are only finitely many retracts up to isomorphism. • Finitely generated abelian groups
   (every retract is a direct summand, hence its isomorphism type is determined by the rank, which is bounded by the rank of ...