Forum Problem Lists Definitions FAQ

20.19 (2022)

Open

A subgroup $H$ of a group $G$ is called commensurated if for all $g \in G$, the index $|H : H \cap gHg^{-1}|$ is finite. Can a non-abelian free group (or a non-elementary hyperbolic group) contain two infinite commensurated subgroups $A$, $B$ with a trivial intersection? The answer is negative if $A$ or $B$ is normal.

Contributor: P.-E. Caprace

Comments

All comments are the responsibility of the user. Comments appearing on this page are not verified for correctness. Please keep posts mathematical and on topic. If you want to submit a proof (or a partial proof), please use the dedicated proof submission form rather than posting it in the comments.
Order by newest first or oldest first.

No comments yet. Be the first to comment.

Proof claims

Proof claims are the responsibility of the submitter. Appearance here does not mean the claim has been checked for mathematical correctness. Moderators only screen for spam, abuse, and obviously low-effort submissions.

No proof claims yet.