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Let $\mathbb{Z}G$ be the group ring over a group $G$. It is a well-known fact that every finitely generated module over a commutative ring is Hopfian. Hence if $G$ is abelian, then every finitely generated $\mathbb{Z}G$-module is Hopfian. On the other hand, P. Hall proved that if $G$ is a polycyclic-by-finite group and $R$ is right noetherian ring with identity, then the group ring $RG$ is right noetherian. Hence if $G$ is polycyclic-by-finite, then every finitely generated $ZG$-module is noetherian (hence Hopfian).

The only groups which are known to have a Noetherian group ring are polycyclic-by-finite. But:

Are there other classes of groups $G$ for which every finitely generated $\mathbb{Z}G$-module is Hopfian?

M.Ramana
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    Probably not significant enough an extension of the examples you know to merit an answer, but if $G$ has a finite index subgroup with this property, then $G$ has the property. So abelian-by-finite groups work. – Jeremy Rickard Jun 18 '18 at 08:12
  • The claim that every finitely generated module over a commutative ring, doesn't it make a noetherianity assumption? if so this only yields finitely generated abelian groups (but otherwise it means that my previous comment is undue). – YCor Jun 18 '18 at 08:54
  • @YCor No, for commutative rings it doesn't need noetherianity: https://math.stackexchange.com/questions/145310/proving-that-surjective-endomorphisms-of-noetherian-modules-are-isomorphisms-and/145362#145362 – Jeremy Rickard Jun 18 '18 at 10:08
  • @JeremyRickard thanks for the link! it escaped by little google search... – YCor Jun 18 '18 at 10:15
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    It would be useful to also list groups for which some non-Hopfian finitely generated $ZG$-modules are known (if any!). – YCor Jun 18 '18 at 10:26
  • @YCor And if there are no known examples for free groups then there are no known examples for any group. M.Ramana: Do you know examples of non-Hopfian finitely generated modules for free groups? – Jeremy Rickard Jun 18 '18 at 12:37
  • @YCor Sorry, that question was addressed to the OP. – Jeremy Rickard Jun 18 '18 at 13:30
  • @JeremyRickard No, I don't. I was wondering if you could give me an example of this kind. – M.Ramana Jun 18 '18 at 16:31
  • For sofic groups $G$ it's known that finitely generated free $kG$-modules are Hopfian when $k$ is a field. – YCor Jun 19 '18 at 07:17
  • @JeremyRickard Could you tell me the reason of this claim: "if there are no known examples for free groups, then there are no known examples for any group"? Thanks in advance. – M.Ramana Jun 19 '18 at 10:42
  • @M.Ramana Every group $G$ is a quotient of a free group $F$ (possibly of infinite rank). So $\mathbb{Z}G$ is a quotient ring of $\mathbb{Z}F$. So a finitely generated non-Hopfian $\mathbb{Z}G$-module can be regarded as a $\mathbb{Z}F$-module, still finitely generated and non-Hopfian. – Jeremy Rickard Jun 19 '18 at 11:09
  • @JeremyRickard That's true. I understood. Thank you very much. – M.Ramana Jun 19 '18 at 11:43

1 Answers1

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I'll show that if $F$ is a free group of rank at least three, then $\mathbb{Z}F$ has a finitely generated non-Hopfian module. (So this is not an answer to the question, but it seems relevant and is too long for a comment.)

Suppose $R$ is any ring with a finitely generated non-Hopfian right module $M$.

Then the ring $M_2(R)$ of $2\times2$ matrices over $R$ is another example, because it's Morita equivalent to $R$ (or, more explicitly, the $M_2(R)$-module $(M\text{ }M)$ of row vectors with entries from $M$ is a finitely generated non-Hopfian $M_2(R)$-module).

Also $M_2(R)$ is generated as a ring by units: If $X=\{x_i\vert i\in I\}$ is a generating set for $R$ in the strong sense that there is not even a proper nonunital subring of $R$ containing $X$, then it is easy to check that $$Y=\left\{\pmatrix{x_i&1\\1&0}\middle\vert i\in I\right\}\cup\left\{\pmatrix{0&1\\1&0}\right\}$$ is a generating set for $M_2(R)$ consisting of units.

So if $F$ is the free group on the set $Y$, there is a natural surjective ring homomorphism $\mathbb{Z}F\to M_2(R)$, and so we can regard $(M\text{ }M)$ as a finitely generated non-Hopfian right $\mathbb{Z}F$-module.

If we take $R$ to be the ring $$R=\mathbb{Z}\left\langle x,y\middle\vert xy=1\right\rangle,$$ then the maps given by left multiplication by $y$ and $x$ exhibit the free right module $R$ as a proper direct summand of itself, and so $R$ is a non-Hopfian module for itself. Since $R$ is generated by two elements, $M_2(R)$ is generated by three units, and so is a quotient of $\mathbb{Z}F$ for $F$ free of rank three.

Jeremy Rickard
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  • Thank you so much for your example. So we conclude that not every finitely generated $\mathbb{Z}F$-modules is Hopfian. Is there any particular condition on a $\mathbb{Z}F$-module to make it Hopfian? – M.Ramana Jul 04 '22 at 15:56
  • @M.Ramana I don’t know, sorry. – Jeremy Rickard Jul 04 '22 at 17:18