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Are there two non-isomorphic groups $G_1$ and $G_2$ so that $G_1$ is an r-image of $G_2$ and $G_2$ is an r-image of $G_1$?

Recall that a group $H$ is called an r-image of a group $G$ if there are homomorphisms $f$ and $g$ from $H$ to $G$ and from $G$ to $H$, respectively, so that $gf=id_H$.

I know that there are two non-isomorphic groups which are isomorphic to a subgroup of each other, like two free groups of rank 2 and 3.

Thanks in advance.

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1 Answer 1

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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 an $r$-image of $A^2$ while $A^2$ is an $r$-image of $A^3\cong A$. Therefore, $A$ and $A^2$ are non-isomorphic groups which are $r$-images of each other.

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  • $\begingroup$ I don't immediately see how the isomorphism between $A$ and $A^3$ must factor through $A^2$. Sure, there are homomorphisms going both ways (inclusion of subspaces, projections), but those do not compose to an isomorphism. Or is the $r$-image constructed some other way? $\endgroup$
    – Arthur
    Aug 30, 2017 at 8:23
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    $\begingroup$ @Arthur I think it is fine. The point is that $A\hookrightarrow A^2\twoheadrightarrow A$ (and this splits), so $A$ is an $r$-image of $A^2$, and that $A^2\hookrightarrow A^3\twoheadrightarrow A^2$ (and this splits), so $A^2$ is an $r$-image of $A^3$. As $A\cong A^3$, this answers the question. $\endgroup$
    – user1729
    Aug 30, 2017 at 9:29
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    $\begingroup$ @user1729 Of course, how silly of me. We don't have the maps go 1-2-3 or 3-2-1 (and 2-3-4 or 4-3-2), but rather 1-2-1 and 2-3-2. $\endgroup$
    – Arthur
    Aug 30, 2017 at 9:32
  • $\begingroup$ @Arthur That's perfect. Thank you very much. $\endgroup$
    – M.Ramana
    Aug 30, 2017 at 11:02
  • $\begingroup$ @Arthur I have a question about your answer. I know that if one of $G_1$ or $G_2$ (in the question) is hopfian group, then $G_1$ and $G_2$ are isomorphic. So the torsion free abelian group $A$ (mentioned in the above answer) can not be finitely generated. Is it true that $A$ has finite rank? $\endgroup$
    – M.Ramana
    Aug 30, 2017 at 11:54

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