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Dear All! I guess this should be well-known, but I cannot find it in literature:

Is it true that if $H$ is a subgroup of finite index in $G$, and $H$ is hopfian, then $G$ is also hopfian?

In case when the groups are finitely generated, the answer is "yes" by R. Hirshon.

(And the other way around question has answer no, as was kindly answered by Kate Juschenko, even for finitely generated case -- which is quite bizarre!!!)

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  • $\begingroup$ Could you give the reference for the Hirshon result? $\endgroup$
    – HJRW
    Commented Mar 5, 2011 at 17:43
  • $\begingroup$ Here it is: \texttt{ams.org/tran/1969-141-00/S0002-9947-1969-0258939-3/…} $\endgroup$
    – Victor
    Commented Mar 5, 2011 at 18:07
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    $\begingroup$ I think, you should repost this as separate question. the post loses structure, when you change the question after the answer is posted. $\endgroup$
    – Kate Juschenko
    Commented Mar 6, 2011 at 8:50
  • $\begingroup$ Well, I thought about reposting the question, but then it would mean for people to read about the same story once more again and... it's sort of saving their time. And of course that I read the original paper of Baumslag and Solitar and didn't see that what they wrote is true about NORMAL subgroup says something only about my illiteracy or what knows what. I was thinking perhaps for 3 years (!!!!) that there is a (logical) mistake in their paper. Anyway, what is really missing is whether hopficity passes upwards without the finite generation assumption. $\endgroup$
    – Victor
    Commented Mar 6, 2011 at 9:27
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    $\begingroup$ (Not least because, as the person who correctly answered your question, Kate deserves to have her answer accepted.) $\endgroup$
    – HJRW
    Commented Mar 6, 2011 at 11:16

1 Answer 1

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Baumslag-Solitar group BS(12,18) is Hopfian, but it contains normal subgroup of finite index which is not Hopfian.

it is proved in the following paper:

Baumslag; Solitar: Some two-generator one-relator non-Hopfian groups.

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  • $\begingroup$ It's what I thought for some time, but I think there is a mistake in their paper!!!!!!!!!! Ok, it is true that $BS(k,n)=\langle a,b:a^{-1}b^ka=b^n>$ is hopfian if and only if $k$ and $n$ are meshed by I don't see why $BS(2,3)$ is a subgroup of FINITE INDEX in $B(12,18)$ They perhaps meant to take a subgroup generated by $a$ and $b^6$ -- but they generate indeed a subgroup of infinite index -- just from normal forms for HNN-etxensions. $\endgroup$
    – Victor
    Commented Mar 5, 2011 at 14:38
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    $\begingroup$ the normal closure of $<a,b^6>$ in $B(12,18)$ is not hopfian and of index $6$. $\endgroup$
    – Kate Juschenko
    Commented Mar 5, 2011 at 17:04

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