Kourovka Notebook Problem 21.118

Definitions and mathematical statement

A group $R$ is residually finite if every $r\ne1$ has nonidentity image under some homomorphism $R\to Q$ to a finite group. Equivalently, its finite-index normal subgroups have trivial intersection.

A discrete group $N$ is amenable if it has a left-invariant mean on bounded real functions: a linear functional $m:\ell^\infty(N,\mathbb R)\to\mathbb R$ with $m(1)=1$, $m(f)\ge0$ whenever $f\ge0$, and $m(f\circ L_a)=m(f)$ for every $a\in N$, where $L_a(x)=ax$.

Does there exist a group $G$ for which no short exact sequence \[ 1\longrightarrow N\longrightarrow R\longrightarrow G\longrightarrow1 \] can have both $R$ residually finite and $N$ amenable? Exactness means that the first map is injective, the last is surjective, and the image of $N$ is exactly its kernel. Thus the equivalent quotient formulation is $G\cong R/N$ with $N\trianglelefteq R$ amenable.

A negative answer would assert that every group admits at least one such presentation as a quotient; the witnessing group $R$ and kernel $N$ could vary with $G$. A positive answer requires one group for which all possible choices fail. There are no finite-generation, countability, or finite-presentation restrictions on the groups in this formulation, and amenability is imposed on the kernel rather than on the quotient.

Short English statement

Is some group impossible to obtain from a residually finite group by quotienting out an amenable normal subgroup?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2627 (KOU-21.118), 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, version 47 (30 September 2026), new problems, Problem 21.118. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.