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A group $H$ is called a retract of a group $G$ if there exist homomorphisms $f:H\longrightarrow G$ and $g:G\longrightarrow H$ such that $g\circ f=id_H$. By a trivial retract of $G$, I just mean the trivial group and $G$ itslef.

My question is that:

Is there a group admitting non-trivial retracts, and which is a retract of all its non-trivial retracts"?

Thanks in advance.

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    $\begingroup$ There are lots of groups without nontrivial retracts. $\endgroup$ – მამუკა ჯიბლაძე Sep 29 '17 at 5:22
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    $\begingroup$ So certainly the question should be "Is there a group admitting non-trivial retracts, and which is a retract of all its non-trivial retract"? $\endgroup$ – YCor Sep 29 '17 at 6:25
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    $\begingroup$ @AliTaghavi you'd be above 50 if you were not systematically spending your reputation for bounties. (Anyway, this is somewhat a bug of the system: it's absurd you can't comment because of bounties.) $\endgroup$ – YCor Sep 29 '17 at 11:57
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    $\begingroup$ @AliTaghavi yes. Let $A,B$ be infinite sets of the same cardinal with union $D$. For $C_k$ cyclic group of order $k$, let $C_2$ act nontrivially on $C_3$. Then $(C_3\rtimes C_2)^D=(C_3^D\rtimes C_2^D)=C_3^B\rtimes (C_3^A\rtimes C_2^D)$ and $C_3^A\rtimes C_2^D=(C_3^A\rtimes C_2^A)\times C_2^B$, so $(C_3\rtimes C_2)^D$ and $(C_3\rtimes C_2)^D\times C_2^D$ are isomorphic to retracts of each other. But the second one has nontrivial center unlike the first one, so they're not isomorphic. $\endgroup$ – YCor Sep 29 '17 at 12:07
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    $\begingroup$ @AliTaghavi According to answers to this Math Overflow question, there is an Abelian group $A$ such that $A\cong A^3$ but $A\not\cong A^2.$ Clearly $A$ is a retract of $A^2$ while $A^2$ is an retract of $A^3\cong A$. Therefore, $A$ and $A^2$ are non-isomorphic groups which are retracts of each other. $\endgroup$ – M.Ramana Sep 29 '17 at 18:05

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