EPIC sets in group theory

Question

The following is well known and easy to prove.

• If $G$ is a group and for all $x,y$ in $G$ we have $(xy)^2=x^2y^2$, then $G$ is abelian.

The following is fairly well known and fairly easy to prove.

• If $G$ is a group and $k$ is an integer and for all $x,y$ in $G$ we have $(xy)^k=x^ky^k$ and $(xy)^{k+1}=x^{k+1}y^{k+1}$ and $(xy)^{k+2}=x^{k+2}y^{k+2}$, then $G$ is abelian.

Definition. Let $S$ be a set of integers such that the following is true.

• If $G$ is a group and for all $x,y$ in $G$, all $n$ in $S$ we have $(xy)^n=x^ny^n$, then $G$ is abelian.

Then $S$ is called an Equality-of-Powers-Implies-Commutativity set, or an EPIC set for short.

Problem. Determine all EPIC sets.

Examples.

• From above, $\{2\}$ is an EPIC set. It's also easy to show that $\{-1\}$ is an EPIC set.

• From above, any set consisting of three consecutive integers is an EPIC set.

• If $k,m$ are coprime then $\{k,k+1,m,m+1\}$ is an EPIC set. This is a generalisation of the previous example.

• It's clear that if $S$ is an EPIC set and $S\subseteq T$, then $T$ is an EPIC set. So we may as well just look for minimal EPIC sets.

Answer 1

The answer is given in the paper Abelian Forcing Sets, written by Joseph A. Gallian and Michael Reid. We have the following result:

Theorem: a set of integers $S$ is epic (called "abelian-forcing" in the paper) if and only if the gcd of the numbers $n(n-1)$ where $n$ runs over $S$ is equal to $2$.