Forum Problem Lists Definitions FAQ

5.28 (1976)

Solved

Let $G$ be a group and $H$ a torsion-free subgroup of $G$ such that the augmentation ideal $I_G$ of the integral group-ring $\mathbb{Z}G$ can be decomposed as $I_G = I_H\mathbb{Z}G \oplus M$ for some $\mathbb{Z}G$-submodule $M$. Prove that $G$ is a free product of the form $G = H \ast K$.

Contributor: D. E. Cohen

Progress

This has been proved (W. Dicks, M. J. Dunwoody, Groups acting on graphs, Cambridge Univ. Press, Cambridge–New York, 1989).

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.