Character determines primitivity conjecture

This article is about a conjecture in the following area in/related to group theory: permutation groups. View all conjectures and open problems

This conjecture is believed to be true

History

This conjecture was made by: Wielandt

This conjecture was made by Wielandt at the Sixth All-Unio Symposium on Group Theory, held in Cherkassy in 1978.

Statement

In terms of permutation representations

Given two permutation representations (viz group actions) of a finite group, both having the same character, one of them is primitive if and only if the other is. In other words, the character of a permutation representation determines whether or not it is primitive.

In terms of subgroups of the symmetric group

Let $P$ and $Q$ be two isomorphic subgroups of the symmetric group $Sym(n)$ on $n$ letters, such that for every conjugacy class $C$ of $Sym(n)$, $|P\cap C|=|Q\cap C|$. Then, $P$ is a maximal subgroup of $Sym(n)$ if and only if $Q$ is.

Progress towards the conjecture

Solvable groups

The conjecture has been settled for the case of solvable groups. In other words, if $G$ is a solvable group, and there are two permutation representations of $G$ with the same character, one of them is primitive if and only if the other is.

References

• Kourovka notebook: unsolved problems in group theory

External links

• Google Print link for the preview of the Kourovka notebook