Forum Problem Lists Definitions FAQ

15.40 (2002)

Open

Let $N$ be a nilpotent subgroup of a finite simple group $G$. Is it true that there exists a subgroup $N_1$ conjugate to $N$ such that $N \cap N_1 = 1$?

Contributor: E. P. Vdovin

Progress

The answer is known to be affirmative if $N$ is a $p$-group.

Comment of 2013: an affirmative answer for alternating groups is obtained in (R. K. Kurmazov, Siberian Math. J., 54, no. 1 (2013), 73–77).

*Yes it is true; moreover, for any two nilpotent subgroups $H, K$ of a (non-abelian) finite simple group $G$ there is $g \in G$ such that $H \cap K^g = 1$ (T. C. Burness, H. Y. Huang, Preprint, 2025, https://arxiv.org/abs/2508.03479).

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.