Let $S$ be a non-empty, possibly infinite, set of integers, all of which are greater than $1$. For a given group $G$, let $S[G]$ denote the collection of statements
$$
\forall (n \in S, a \in G, b\in g) \,\,(ab)^n=a^nb^n
$$
For some sets $S$, $S[G]$ is sufficient to prove $G$ is Abelian. For example, the set $\{2\}$ is sufficient, since $(ab)^2 = a^2b^2\implies ba = ab$.


Define a minimal abelian forcing (maf) set as any $S$ such that $S[G]$ proves $G$ is Abelian, but for no proper subset $T \subset S$ does $T[G]$ prove $G$ is Abelian.



> 
> Has anybody studied the question of classifying all minimal abelian forcing sets? And has anybody addressed the question of whether there exists an infinite maf?
> 
> 
> 


For example, some mafs are:


* $S = \{2\}$
* $S = \{3,5\}$
* $S = \{n+3,n+4,n+5\}, n\in \Bbb Z^+$


The line of attack I have tried is to exploit the fact that $(ab)^n = a^nb^n$ implies that $n$-th powers commute with $(n-1)$-th powers and that $(ab)^n = a^nb^n \implies (ab)^{n-1} = b^{n-1} a^{n-1} $. This easily confirms the examples listed above, but it is hard to get a systematic test of whether a given set $S$ is maf.




---


I realize that using the word "forcing" has jargon implications. But I felt it was better than minimal abelian *proving* since the abbreviation of that would be map and that would be more confusing.



Classification of minimal sets of properties proving a group is Abelian
Here is an answer to the second question. Call $S$ an *abelian forcing set* if $S[G]$ implies that $G$ is abelian. 


**Proposition.** If $S$ is an abelian forcing set then some finite subset of $S$ is an abelian forcing set. So there is no infinite minimal abelian forcing set. 


**Proof.** Given $n>0$, let $\phi\_n$ be the statement


$$
\forall x\forall y\, (xy)^n=x^ny^n.
$$


This is a first order statement in the language of groups. 


Now suppose $S$ is an abelian forcing set. Let $\Sigma$ be the set consisting of $\phi\_n$ for all $n\in S$, together with the axioms for groups. Let $\psi$ be the statement
$$
\forall x\forall y\, xy=yx.
$$
Then our assumption is that $\psi$ is a logical consequence of $\Sigma$. By the Compactness Theorem for first order logic, there is some finite subset $\Sigma\_0$ of $\Sigma$ such that $\psi$ is a logical consequence of $\Sigma\_0$. So if $S\_0$ is the set of $n$ such that $\phi\_n$ is in $\Sigma\_0$, then $S\_0$ is an abelian forcing set and a finite subset of $S$. 


