# Classification of minimal sets of properties proving a group is Abelian

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.

• See Joseph A. Gallian and Michael Reid, Abelian Forcing Sets, Amer. Math. Monthly, Vol. 100, No. 6 (Jun. - Jul., 1993), pp. 580-582. They show that a set $S$ of integers is abelian forcing iff the greatest common divisor of the numbers $n(n-1)$, for $n\in S$ is equal to $2$. – James Jul 10 at 1:00
• @James: you should write this up as an answer rather than as a comment. – IJL Jul 10 at 10:27
• @IJL Sure. I've done so now. – James Jul 10 at 22:24

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 $$S$$.

See the Monthly article: Joseph A. Gallian and Michael Reid, Abelian Forcing Sets, Amer. Math. Monthly, Vol. 100, No. 6 (Jun. - Jul., 1993), pp. 580-582.

Gallian and Reid show that a set $$S$$ of integers is abelian forcing if, and only if, the greatest common divisor of the numbers $$n(n−1)$$, for $$n\in S$$ is equal to $$2$$.

They show how this makes it particularly easy to see that standard exercise examples, such as $$S = \{2\}$$, $$S = \{-1\}$$ and $$S = \{ k, k+1, k+2 \},$$ (for $$k$$ an integer) are all abelian forcing sets, and note that those first two are the only singletons $$S$$ that are abelian forcing.

• +1 for the definitive reference about what constitutes an af set. I don't see an easy way to go from all af sets to all maf sets, other than by saying "and for all proper subsets $T \subset S$, the gcd of $n(n-1)$ for $n\in T$ is not equal to $2$. [I realize that I should have broken this into two questions, because the answer by @Gabe Conant nails the second issue and I can't mark two answers as "the" answer. ] – Mark Fischler Jul 11 at 16:24