# On the notion of torsion-freeness in semigroup theory

The following seems to be the "official" notion of torsion-freeness in the context of semigroups:

TF1. A (multiplicatively written) semigroup $\mathfrak A$ is torsion-free if there do not exist $a,b \in \mathfrak A$ and $n \in \mathbb N^+$ such that $a \ne b$ and $a^n = b^n$.

On another hand, I recently ended up with the following alternative idea:

TF2. A semigroup $\mathfrak A$ is torsion-free if, given $a \in \mathfrak A$, $a^m = a^n$ for some $m,n \in \mathbb N^+$ with $m \ne n$ only if $a$ is idempotent.

Both of these generalize the usual notion of torsion-freeness for groups. Also, it is not difficult to check that TF1 implies TF2, but not viceversa. So, my questions are:

Q1. What about existing literature concerning torsion-free semigroups in the sense of the second definition? Q2. Could you point out some reasons why the former definition should be preferred to the latter?

For the record, this is somehow related to question 105851.

• I do not think TF1 is ever used. "Torsion-free" means "no torsion". "Torsion" means elements of finite order. Element $x$ has finite order if it satisfies $x^m=x^n$ for $m\ne n$, $m,n\ge 0$. So "torsion-free" means TF2. – user6976 Sep 3 '12 at 0:36
• People in logarithmic geometry usually uses monoids and not only semi groups, but for them the second definition is the right one. The first is somehow referred to as "integral" making reference to integral domains. – Filippo Alberto Edoardo Sep 3 '12 at 0:40
• I think Clifford and Preston use the first version in their book. For commutative it is equivalent to the Grothendieck group being torsion-free. – Benjamin Steinberg Sep 3 '12 at 0:54
• @Ben: where do they use it? If the semigroup is a monoid, one also assumes that $x\ne 1$ (as for groups). – user6976 Sep 3 '12 at 1:08
• For commutative monoids, TF1 is used in the case of cancelation only, as far as I know. Then TF1 means that the monoid embeds into torsion-free (in the usual sense) Abelian group. That is of course not a very reasonable terminology since it confuses two things, but it is indeed used sometimes. – user6976 Sep 3 '12 at 1:15

I believe your first definition is primarily used by people in commutative semigroup theory. Most people in non-commutative semigroup theory that I know prefer the second. An important example is the free profinite monoid. Every periodic element is an idempotent so it is torsion-free in the second sense. This is a non-trivial theorem due to Rhodes and myself and relies on projectivity of maximal subgroups and torsion-freeness of projective profinite groups.

I think the same maybe true for Stone-Čech compactifications of free monoids, but perhaps they have some nilpotent cyclic subsemigroups. I believe their maximal subgroups are torsion-free.

• Thank you, Benjamin. Let me just add a link to the arXiv preprint of your paper: arxiv.org/pdf/math/0611896v1.pdf. I put it on the top of the pile of must-read things. – Salvo Tringali Sep 2 '12 at 19:28

It has been a few years since this question was posed, but I hope the following answer helps. In my own work on semigroups, the definition of torsion-free that we've used is the following:

Definition: A semigroup $$S$$ is torsion-free when for any two commuting elements $$s,t\in S$$, if $$s^n=t^n$$ for some positive integer $$n$$, then $$s=t$$.

This generalizes the notion of periodic that is encapsulated by TF2, and simultaneously generalizes TF1 to the case of noncommutative semigroups.

TF2 is fine for periodicity, but it generally doesn't capture important notions related to torsion-freeness. For instance, it is well known that an abelian group is orderable if and only if it is torsion-free. A commutative semigroup is similarly orderable if and only if it is cancelative and TF1.

A noncommutative semigroup that is orderable is still cancelative, and is torsion-free is we use the definition I gave about. Other results similarly generalize. Moreover, the definition I've given actually works for nonabelian groups (whereas TF1 without the commuting hypothesis does not).

• What I find a bit surprising with this definition is that it doesn't only rely on the subsemigroup generated by individual elements (in particular it doesn't really rely on a notion of torsion element). I'm wondering if this notion has a name in group theory. – YCor Nov 23 '20 at 21:57
• @YCor Actually, it does only depend on the (commutative) subsemigroup generated by $s$ and $t$. (Perhaps you replaced "commuting" with "central"?) – Pace Nielsen Nov 23 '20 at 22:39