Let $G$ be a group which is Hopfian and given a short exact sequence $1\to F \to H \to G \to 1$ with $F$ a finite normal subgroup of $H$. Is $H$ Hopfian?
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1$\begingroup$ I suspect you will need stronger assumptions on $G$, for example that any quotient of $G$ by one of its finite normal subgroups is Hopfian. $\endgroup$– ndkrempelCommented Mar 22, 2011 at 19:57
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$\begingroup$ @ ndkrempel: why the math symbols are not readable here? did you mean the group G or H on which i would need stronger assumptions? sorry for disturbing $\endgroup$– PooveCommented Mar 22, 2011 at 21:03
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4$\begingroup$ @Poove: Must be a problem at your end. I meant conditions on G. Another possible condition would be to require G to be torsion-free, I think that's enough for the result to hold... It would be helpful if you gave more details about your particular G, if possible. $\endgroup$– ndkrempelCommented Mar 22, 2011 at 23:07
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$\begingroup$ @ ndkrempel: G has no normal subgroups isomorphic to a free abelian group of finite rank. (dont want to assume G is torsion free. $\endgroup$– PooveCommented Mar 23, 2011 at 10:07
2 Answers
An example exists in this paper, p19-20. In fact they also construct a Hopfian group $H$ with finite (cyclic) normal subgroup $F$ and $H/F$ non-Hopfian.
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$\begingroup$ @Mark, could you tell on what page the group is constructed? $\endgroup$ Commented Mar 28, 2011 at 19:06
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1$\begingroup$ 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. $\endgroup$– user6976Commented Mar 28, 2011 at 21:20
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$\begingroup$ 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. $\endgroup$– user6976Commented Mar 29, 2011 at 4:05
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$\begingroup$ 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. $\endgroup$ Commented Mar 30, 2011 at 14:24
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$\begingroup$ @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). $\endgroup$– YCorCommented Dec 3, 2016 at 20:27
It seems that the answer is no: there exists an exact sequence $$1\to F\to H\to Q\to 1$$ with $F$ finite (central), $H$ non-Hopfian, and $Q$ Hopfian (with in addition, $H$ finitely generated solvable).
Mark Sapir's answer refers to a group constructed here (see 5.10), which is Abels' group over the ring $\mathbf{F}_p[t,1/t]$, and which probably be used to provide a negative answer to the question. Define the group $G$ as the group of matrices
$$\left(\begin{array}{rrrr} 1 & u_{12} & u_{13} & u_{14}\newline 0 & d_{22} & u_{23} & u_{24}\newline 0 & 0 & d_{33} & u_{34}\newline 0 & 0 & 0 & 1\newline \end{array}\right) $$ where $u_{ij}\in\mathbf{F}_p[t,1/t]$, and $d_{ii}\in\mathbf{F}_p[t,1/t]^\times=\langle t\rangle\mathbf{F}_p^\times$. (Actually one can restrict to $d_{ii}\in\langle t\rangle$ but it's not important.) Let $e_{14}$ denote the map $x\mapsto\begin{pmatrix} 1&0&0&x\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$.
As suggested by user "BS.", let $N$ be the central subgroup $e_{14}(\mathbf{F}_p[t^2])$ and $M$ the larger central subgroup $e_{14}(\mathbf{F}_p[t^2]\oplus \mathbf{F}_pt^{-1})$. (In a the initial post, $M$ and $N$ were chosen as other central subgroups but unfortunately $G/M$ failed to be Hopfian). So we have the central exact sequence with finite kernel $$1\to M/N\to G/N\to G/M\to 1.$$ Conjugation by the diagonal matrix $(t^2,1,1,1)$ is an automorphism of $G$, which maps $N=e_{14}(\mathbf{F}_p[t^2])$ strictly into itself and hence $H=G/N$ is non-Hopfian.
I haven't completely checked but here are some guidelines to show the group $Q=G/M$ is Hopfian.
Write the original group (given by $4\times 4$ triangular matrices) as $G=D\ltimes U$ with $D=\mathbf{Z}^2$ and $U$ its unipotent part. Set $U^2=[U,U]$ and $U^3=[U,U^2]$, which is central and equal to $e_{14}(\mathbf{F}_p[t,1/t])$. Let $f$ be a surjective endomorphism of $Q$.
check that the center of $G$ is precisely $U^3$. It follows that $f$ induces a surjective endomorphism of $G/U^3$. Since this group is linear, it is Hopfian so this is an automorphism of $G/U^3$.
Describe the group of automorphisms of $G/U^2 = \mathbf{Z}^2\ltimes F_p[t,1/t]^3$. (It should be reasonably easy to describe).
Deduce a description of the group of automorphisms of $G/U^3$, or at least describe how these automorphisms act on $U^2/U^3$, showing that modulo something, the coefficient $12$ is multiplied by a monomial $w\cdot t^a$ ($w\in \mathbf{F}_p^*$) and the coefficient $24$ is multiplied by $vt^b$. So, taking a commutator (which should kill the "modulo something"), we obtain that in the "action of $f$ on $G$", the coefficient $14$ should be multiplied by a nonzero monomial. This multiplication should stabilize $M$ so this is multiplication by a scalar in $\mathbf{F}_p^*$, which implies that f actually induces an automorphism of $Q$.
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$\begingroup$ I just edited tex by inserting dollar signs etc. $\endgroup$ Commented Apr 2, 2011 at 23:11
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$\begingroup$ Sorry if I missed something, but doesn't conjugation by $diag(t^2,1,1,1)$ define an automorphism of $G$ that sends $M$ strictly inside itself (implying $G/M$ non-Hopfian)? $\endgroup$– BS.Commented Jul 22, 2012 at 16:58
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1$\begingroup$ On the other hand, a similar idea might work : take $N=F_p[t^2]$ and $M=<N,t^{-1}$. Then $G/N$ is non-Hopfian, but since no non-trivial translation of $\mathbb{Z}$ sends $E={-1,0,2,4,...,2k,...}$ (strictly) inside itself, the conjugation counterexample breaks down and your argument might work. $\endgroup$– BS.Commented Jul 22, 2012 at 17:41
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$\begingroup$ @BS. thanks for the correction; this seems to fix the argument. $\endgroup$– YCorCommented Dec 3, 2016 at 20:24