Kourovka Notebook Problem 21.48

Definitions and mathematical statement

A quasimorphism on a group $G$ is a function $f:G\to\mathbb R$ with finite defect \[ D(f)=\sup_{g,h\in G}|f(gh)-f(g)-f(h)|<\infty. \] It is homogeneous if $f(g^n)=nf(g)$ for every $g\in G$ and every integer $n$. Equivalently, its restriction to each cyclic subgroup is a homomorphism into the additive group of real numbers. Homogeneity in particular gives $f(1)=0$ and $f(g^{-1})=-f(g)$.

Assume that $G$ admits an unbounded homogeneous quasimorphism which is not a homomorphism. Unbounded means $\sup_{g\in G}|f(g)|=\infty$; not being a homomorphism means that $f(gh)\ne f(g)+f(h)$ for at least one pair. Must $G$ contain a nonabelian free subgroup?

The desired conclusion is the existence of elements $a,b\in G$ such that every nonempty freely reduced word in $a^{\pm1},b^{\pm1}$ is nonidentity. Equivalently, the map from the free group on two generators sending its generators to $a,b$ is injective.

The group is arbitrary: finite generation, a topology, and a geometric action are not assumed. The defect has one finite bound over all pairs, not a bound depending on $g,h$. A real homomorphism alone, even an unbounded one, does not satisfy the hypothesis. Likewise a free quotient of $G$ is not a substitute for the explicitly required free subgroup without an accompanying argument producing its embedding.

Short English statement

Does a non-homomorphic homogeneous real quasimorphism force a group to contain a free subgroup of rank two?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2557 (KOU-21.48), 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 21.48. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.