A group $G$ is Hopfian iff every epimorphism $f:G\to G$ is an automorphism. A group $G$ is weakly Hopfian iff every r-homomorphism $r:G\to H\cong G$ is an isomorphism. In other words, $G=K\rtimes H$ and $H\cong G$ imply $K=1$.
Trivially, every Hopfian group is weakly Hopfian. I was wondering if someone could give me an example of a weakly Hopfian group which is not Hopfian. I particularly interested in finitely presented examples. Thanks in advance.
An example is the Prüfer group $G = \mathbf Z[1/p] / \mathbf Z$. The multiplication-by-$p$ map gives a nontrivial epimorphism. On the other hand every proper subgroup of $G$ is finite, so $G$ does not have the form $K \times H$ (or equivalently $K \rtimes H$ since $G$ abelian) for nontrivial $K,H \le G$.
Edit: Sorry, I missed the requirement that $G$ should be finitely presented. Maybe $\mathrm{BS}(2,3)$ is an example.
Yes, a classical construction of Abels does the job. Fix a prime $p$, define the group $G_0$ 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 $u_{ij}\in\mathbf{Z}[1/p]$, and $d_{ii}\in p^\mathbf{Z}$. Let $e_{14}$ denote the map $x\mapsto\begin{pmatrix} 1&0&0&x\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$, and $G=G_0/e_{14}(\mathbf{Z})$.
Abels observed that $G$ is not Hopfian (a surjective endomorphism with kernel of order $p$ being induced by conjugation by the diagonal matrix $(p,1,1,1)$), and proved that $G_0$ and hence $G$ is finitely presented.
But $G$ is weakly Hopfian. Indeed, let $f$ be a surjective endomorphism of $G$. Since $G_0/e_{14}(\mathbf{Z}[1/p])$ is residually finite, the kernel of $f$ is contained in $e_{14}(\mathbf{Z}[1/p])/e_{14}(\mathbf{Z})$. If this were split, since it is central it would be a direct factor, but this is absurd because $e_{14}(\mathbf{Z}[1/p])\subseteq [G_0,G_0]$.
