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Mar 21, 2020 at 16:21 history edited YCor CC BY-SA 4.0
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Dec 3, 2016 at 20:27 comment added YCor @AndreasThom you're right. The modification by "BS." of my argument (see my answer) seems to fix the issue (in any case, $H$ will not be residually finite, because a finite-by-(residually finite) finitely generated group is always Hopfian).
Mar 30, 2011 at 14:24 comment added Andreas Thom That is a very nice example. But how does one show that killing $t^{-2}$ gives a Hopfian group? I doubt that this quotient is residually finite.
Mar 29, 2011 at 4:05 comment added user6976 In fact, as I was told by Yves de Cornulier, the group is not a quotient of Abels' group $A_n$ by its center but the quotient of the analog $B_n$ of group $A_n$ over $\mathbb{F}_p[t,t^{-1}]$ over a central copy of $\mathbb{F}_p[t]$ (see 5.10 in the paper). Then the factor-group is not Hopfian, its factor by the (finite) subgroup generated by $t^{-2}$ is Hopfian. If one then kills $t^{-1}$ as well, one get a non-Hopfian group again. I hope he himself will explain here the Hopfian property of these groups.
Mar 28, 2011 at 21:20 comment added user6976 Take the Abels' group $A_n$ (pages 19,20). Then $A_n/Z$ is not Hopfian while there is a central cyclic subgroup there F such that $(A_n/Z)/F$ is Hopfian.
Mar 28, 2011 at 19:06 comment added Igor Belegradek @Mark, could you tell on what page the group is constructed?
Mar 28, 2011 at 13:36 history edited user6976 CC BY-SA 2.5
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Mar 27, 2011 at 10:53 history answered user6976 CC BY-SA 2.5