Hopfian property preserved by extensions with finite kernel? 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?  
 A: 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.  
A: 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 $Q$.
