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Let $G$ be a group. Recall that a group $H$ is called a retract of $G$ if there exist homomorphisms $g:G\longrightarrow{H}$ and $f:H\longrightarrow G$ so that $g\circ f=id_H$. The homomorphism $g$ is called a retraction.

My question is that:

Is there an infinite sequence $\{ H_i \}_{i\in \mathbb{N}}$ of groups so that the following conditions hold?

1- $H_1$ is a finitely presented group.

2- $H_{i+1}$ is proper retract of $H_i$ with the retraction $g_{i}:H_i \longrightarrow H_{i+1}$, for all $i\geq 1$ (equivalently, $H_i$ admits a semidirect product decomposition $L_i\ltimes N_i$ with $N_i\neq 1$ and $L_i\simeq H_{i+1}$);

Thanks in advance.

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    $\begingroup$ At least there exists a sequence of finitely generated (fg) groups with this property. Indeed there exists a fg group $G\neq 1$ isomorphic to $G\times G$ (whether such $G$ exists finitely presented is unknown), so one gets a sequence of nontrivial retract homomorphisms $G\to G\to G\to\dots$ $\endgroup$ – YCor Oct 12 '17 at 16:25
  • $\begingroup$ @YCor Thank you for your comment. I understood from your comment that we can consider the sequence $\cdots \longrightarrow G\times G\times G \longrightarrow G\times G \longrightarrow G$ with projections maps as retractions. Did I understand your mean right? $\endgroup$ – M.Ramana Oct 13 '17 at 12:35
  • $\begingroup$ Of course not, your sequence exists for any group and is unrelated to your question. Let $G$ be isomorphic to $G\times H$ and let $f$ be an isomorphism $G\to G\times H$, $H\neq 1$, $p$ the canonical projection $G\times H\to G$ and $q=p\circ f$. I mean the sequence $G\to G\to G\dots$ with the (retract) homomorphism $q$ everywhere. $\endgroup$ – YCor Oct 14 '17 at 7:10
  • $\begingroup$ It's possibly an open question, and is definitely an excellent question; I'm surprised by the small number of upvotes. $\endgroup$ – YCor Oct 15 '17 at 22:00
  • $\begingroup$ By Lemma 1.3 in the paper "Finiteness Conditions for CW-complexes" by C.T.C.Wall, if $G$ is a finitely presented group and $H$ is a retract of $G$, then $H$ is also a finitely presented group. In fact, in the proof of this lemma, the generators of $H$ and $G$ are the same, but the relations of $H$ contain the relations of $G$ and other relations. $\endgroup$ – M.Ramana Dec 26 '17 at 8:49

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