Kourovka Notebook Problem 21.7

Definitions and mathematical statement

A finite left brace is a finite set $B$ with operations $+$ and $\circ$ such that $(B,+)$ is an abelian group, $(B,\circ)$ is a group, and \[ a\circ(b+c)=(a\circ b)-a+(a\circ c) \qquad(a,b,c\in B), \] where $-a$ is the inverse in the additive group. The additive identity is also the multiplicative identity. No right distributive law is assumed. A finite group is called an IYB-group if it is isomorphic to the multiplicative group $(B,\circ)$ of some finite left brace. This brace definition fixes the equivalent formulation used in the source.

For a finite group $G$, a Sylow $p$-subgroup is a subgroup of order the largest power of $p$ dividing $|G|$. The group $G$ is soluble if its derived series, formed by repeatedly taking the subgroup generated by commutators, terminates at $1$.

Suppose $G$ is a finite soluble group and every Sylow subgroup of $G$ is an IYB-group. Must $G$ be an IYB-group? Equivalently, can one put an abelian addition on the underlying set of $G$ so that its given group multiplication and that addition satisfy the brace identity?

The brace structures assumed to exist on the Sylow subgroups may be different and need not agree on intersections or be compatible with conjugation in $G$. The conclusion asks for one brace structure on the whole group, not merely separate structures on its prime-power subgroups.

Short English statement

If every Sylow subgroup of a finite soluble group is the multiplicative group of a left brace, must the whole group admit such a structure?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2516 (KOU-21.7), 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 21.7. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.