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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 $gf=id_H$.

Given a group $G$ with identity element 1, a subgroup $H$, and a normal subgroup $N$ of $G$; $G$ is called the semidirect product of $N$ and $H$, written $G = N\rtimes H$ , if $G = NH$ and $H\cap N=1$. Then $H$ is called a semidirect factor of $G$.

We can easily see that a group $H$ is a retract of $G$ iff $Im(f)$ is a semidirect factor of $G$ (In fact, by the above notation, we have $G=ker(g)\rtimes Im(f)$).

Of course, finite groups have finitely many retracts. On the other hand, by Does finitely generated groups have finitely many finite retracts? , not every finitely generated group has finitely many (finite) retracts. My questions is that is there any finitely presented group with only finitely many (finite) retracts?

M.Ramana
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1 Answers1

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How about $\mathbb{Z}$, which can be written as a semidirect product only in the trivial way?

$\mathbb{Z}\oplus\mathbb{Z}_2$ also works.

Chris Sanders
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  • Thank you very much for your answer. Do finitely generated abelian groups with finitely many retracts have a specific form? For example, do they have the form $\mathbb{Z}\oplus G$, where $G$ is a finite abelian (or nonabelian) group? – M.Ramana Jul 05 '22 at 06:59
  • Could you please tell me why every FGAG with FMR is of the form $\mathbb{Z}\oplus G$, where $G$ is finite abelian? – M.Ramana Jul 05 '22 at 08:23
  • By https://math.stackexchange.com/questions/2671807/the-number-of-internal-direct-summands-of-a-finitely-generated-abelian-group , $\mathbb{Z}^2$ has infinitely many direct summands, as you said. – M.Ramana Jul 05 '22 at 08:24