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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.

M.Ramana
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    There are lots of groups without nontrivial retracts. – მამუკა ჯიბლაძე Sep 29 '17 at 05:22
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    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"? – YCor Sep 29 '17 at 06:25
  • Yes. I mean that the group $G$ must contains non-trivial retracts. – M.Ramana Sep 29 '17 at 07:14
  • This blog post discusses some related results: https://berstein2015.wordpress.com/2015/04/18/a-group-with-a-quotient-isomorphic-to-the-direct-square/ . – HJRW Sep 29 '17 at 08:51
  • My apology for this comment in the answer part. I can not give a comment since I am under 50 reputation. Are there two non isomorphic groups which are retract of each others? – Ali Taghavi Sep 29 '17 at 08:10
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    @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.) – YCor Sep 29 '17 at 11:57
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    @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. – YCor Sep 29 '17 at 12:07
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    @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. – M.Ramana Sep 29 '17 at 18:05
  • @YCor l honestly thank you very much for your two previous comments: Your interesting example and your comment on my reputation. – Ali Taghavi Oct 01 '17 at 19:26
  • @M.Ramana thank you very much for the link and your interesting counter example. – Ali Taghavi Oct 01 '17 at 19:28
  • @YCor why not make that an answer? – Joshua Grochow Apr 15 '20 at 18:03
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    @JoshuaGrochow my example addressed Ali Taghavi's question in a comment, but not the OP. For instance, $C_2$ is a nontrivial retract of the group $G$ given in my comment, but $G$ is not isomorphic to any retract of $C_2$. – YCor Apr 15 '20 at 18:09

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