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?

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


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. 

