Kourovka Notebook Problem 21.22

Definitions and mathematical statement

A group is Hopfian if every surjective endomorphism of it is injective, hence an automorphism. It is finitely generated if some finite set generates it using multiplication and inverses. Let $G$ and $H$ be finitely generated Hopfian groups.

Define the restricted base group \[ B=\bigoplus_{h\in H}G =\{f:H\to G:\{h:f(h)\ne1\}\text{ is finite}\}, \] with pointwise multiplication. Distinct copies of $G$ commute inside this direct sum, even when $G$ itself is nonabelian. The group $H$ acts on $B$ by left translation of the indices: $(h\cdot f)(x)=f(h^{-1}x)$. The standard restricted wreath product is $G\wr H=B\rtimes H$, with multiplication \[ (f,h)(g,k)=(f\,(h\cdot g),hk). \] The question is whether $G\wr H$ must be Hopfian for every such pair. Thus every surjective homomorphism $\Phi:G\wr H\to G\wr H$ would have to have trivial kernel.

The map $\Phi$ is arbitrary; it is not assumed to preserve the displayed base group, the embedded acting group, or the projection onto $H$. The restricted support condition is essential: replacing $B$ by all functions $H\to G$ gives a different wreath product. No abelianity, residual finiteness, torsion condition, or finite-presentation hypothesis is added to the factors. A counterexample requires both factors to be Hopfian even though their restricted wreath product is not.

Short English statement

Is the standard restricted wreath product of two finitely generated Hopfian groups necessarily Hopfian?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2531 (KOU-21.22), Kourovka Notebook - New Problems, Issue 21. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 The Kourovka Notebook, 21st edition, new problems, Problem 21.22. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.