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.

One quick observation: by compactness, every EPIC set contains a finite EPIC set, so we only need to analyze the finite sets. — Noah Schweber, Jan 19 '18 at 01:08
This answer, about Abelian Forcing Sets, seems to be relevant. The following posts are also related to this topic: https://math.stackexchange.com/questions/132807, https://math.stackexchange.com/questions/40996/. — Watson, Jan 30 '18 at 20:54

score 5 · Accepted Answer · answered Jan 30 '18 at 20:59

5

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 $2$.

answered Jan 30 '18 at 20:59

Watson

23,793

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Thanks for that. I think the proof of Gallian and Reid is essentially correct, though there is an inaccurate statement on page 3, lines 7-8: if "the gcd of the integers $n(n-1)$ as $n$ ranges over $S$ is 2... [then] $2\in S$". This is clearly false, for example, $S={3,4,5}$. What they mean is that if the gcd is $2$ then $(ab)^2=a^2b^2$ for all $a,b$, and I believe their proof of this is correct. – David Jan 31 '18 at 00:33
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I've also found that the same authors published a version of their paper here in which the above point is addressed and the argument simplified. – David Feb 01 '18 at 00:48

EPIC sets in group theory

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