6
$\begingroup$

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.

$\endgroup$
  • 5
    $\begingroup$ 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 $2$. $\endgroup$ – James Jul 10 at 1:00
  • $\begingroup$ @James: you should write this up as an answer rather than as a comment. $\endgroup$ – IJL Jul 10 at 10:27
  • $\begingroup$ @IJL Sure. I've done so now. $\endgroup$ – James Jul 10 at 22:24
8
$\begingroup$

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 $S$.

$\endgroup$
10
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ +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. ] $\endgroup$ – Mark Fischler Jul 11 at 16:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.