According to answers to this Math Overflow question, there is an infinite rank abelian group $A$ such that $A\cong A^3$ but $A\not\cong A^2.$ Clearly $A$ is an 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.

Are there two finitely presented groups $G$ and $H$, such that $G\not \cong H$ which are retracts of each other?

Thank you in advance.

  • 1
    $\begingroup$ Just a comment on the abelian case. The question of whether there are non-isomorphic abelian groups that are summands of each other is Kaplansky's first "test problem" from 1954, and was answered by Sąsiada in 1961, before Corner showed in 1964 that there is an abelian group $A$ with $A\cong A^3$ but $A\not\cong A^2$. $\endgroup$ Oct 4 '17 at 14:43
  • $\begingroup$ @ Jeremy Rickard Thank you so much for your valuable comment. $\endgroup$
    – M.Ramana
    Oct 4 '17 at 16:19
  • 3
    $\begingroup$ Another comment: there exist two non-isomorphic finitely presented groups that are quotient of each other (this is a bit weaker than being retracts of each other). Indeed, take $G=BS(2,3)$ (Baumslag-Solitar): it's isomorphic to a quotient $G/N$, where $N$ is a nontrivial free group. Therefore, if $x$ belongs to a basis of $N$ and $M$ is the normal subgroup of $G$ generated by $x^2$, the quotient $G/M$ has an element of order 2 (the image of $x$), and $G$ and $G/M$ are isomorphic to quotients of each other. Since $G$ is torsion-free, it is not isomorphic to $G/M$. $\endgroup$
    – YCor
    Oct 4 '17 at 16:27
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    $\begingroup$ In a comment on mathoverflow.net/questions/256467, user მამუკა ჯიბლაძე says that they think that they have seen an example of the type that you ask for. $\endgroup$ Oct 4 '17 at 16:48
  • $\begingroup$ @ Ycor Thank you very much for your interesting comment. $\endgroup$
    – M.Ramana
    Oct 4 '17 at 16:58

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