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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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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. – ndkrempel Mar 22 '11 at 19:57
@ 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 – Poove Mar 22 '11 at 21:03
@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. – ndkrempel Mar 22 '11 at 23:07
@ ndkrempel: G has no normal subgroups isomorphic to a free abelian group of finite rank. (dont want to assume G is torsion free. – Poove Mar 23 '11 at 10:07

An example exists in this paper. In fact they also construct a Hopfian group $H$ with finite (cyclic) normal subgroup $F$ and $H/F$ non-Hopfian.

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@Mark, could you tell on what page the group is constructed? – Igor Belegradek Mar 28 '11 at 19:06
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. – Mark Sapir Mar 28 '11 at 21:20
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. – Mark Sapir Mar 29 '11 at 4:05
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. – Andreas Thom Mar 30 '11 at 14:24

Edit: this post 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 provides a negative answer to the question. Thus the group $G$ below refers to 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 I restrict to $d_{ii}\in\langle t\rangle$ but it's not important. [end edit]

I haven't checked but here are some guidelines to show the group 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 naturally isomorphic to $F_p[t,1/t]$. Our group is $H=G/M$, where $M\subset U^3$ is generated by $F_p[t]$ and $t^{-2}$. Let $f$ be a surjective endomorphism of $H$.

1) 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 $G/U^3$.

2) 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).

3) 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 $w\cdot t^a$ ($w\in F_p*$) and the coefficient $24$ is multiplied by $vt^b$. So, taking a commutator (that should kill the "modulo something"), we obtain that in the action of $f$ on $H$, the coefficient $14$ should be multiplied by a monomial. This multiplication should stabilize $M$ so this is multiplication by a scalar in $F_p*$, which implies that f is actually an automorphism.

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I just edited tex by inserting dollar signs etc. – Igor Belegradek Apr 2 '11 at 23:11
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)? – BS. Jul 22 '12 at 16:58
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. – BS. Jul 22 '12 at 17:41

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