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Are there (infinite) non-isomorphic groups $G, H$ such that there are surjective group homomorphisms $f: G\to H$ and $g: H\to G$?

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Yes, an example is $\mathbb{Z}/2 \times \mathbb{Z}^{\mathbb{N}}$ and $\mathbb{Z}^{\mathbb{N}}$. This even works in the category of rings.

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    $\begingroup$ I hope that the word "even" is not confusing: The rings also surject onto each other, hence also the underlying abelian groups, but that they are not isomorphic as rings does not necessarily imply that they are not isomorphic as abelian groups. $\endgroup$
    – Martin Brandenburg
    Commented Oct 9, 2014 at 18:32
  • $\begingroup$ Bonus question: Is there an example with finitely generated $k$-algebras? $\endgroup$
    – Martin Brandenburg
    Commented Oct 9, 2014 at 18:37
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The first example that comes to mind, $G=\bigoplus_{i=1}^\infty\mathbb{Q}$ and $H=\mathbb{Q}/\mathbb{Z}\oplus\bigoplus_{i=1}^\infty\mathbb{Q}$, seems to work.

FYI, a related example $G=\bigoplus_{i=1}^\infty\mathbb{Q}$ and $H=\mathbb{Z}\oplus\bigoplus_{i=1}^\infty\mathbb{Q}$ handles your question with "surjective" replaced by "injective".

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    $\begingroup$ For the question with "injective" instead of "surjective", it sufficies to remember that the free group with $2$ generators contains the free group with $3$ generators as a subgroup. $\endgroup$
    – Francesco Polizzi
    Commented Oct 9, 2014 at 15:01
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    $\begingroup$ @FrancescoPolizzi: sure, that's another good example. I just like the fact that this pair of questions allows for almost identical examples. $\endgroup$
    – Vladimir Dotsenko
    Commented Oct 9, 2014 at 15:11

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