MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm learning about periodic languages, and I'm confused over the vocabulary used to describe the periodicity of (syntactic) monoids.

If I understand correctly, a monoid M is periodic if : $$(\forall m \in M)(\exists i \neq j)[m^i = m^j],$$ and it is aperiodic if : $$(\forall m \in M)(\exists k)[m^k = m^{k+1}],$$ and then an aperiodic monoid is periodic. Where does that bizarre vocabulary come from?

And in the same vein, what would be the book you'd recommend on monoids and semigroups?

Thank you.

share|cite|improve this question
Nomenclature is seldom optimal! We also have irreducible but non-reduced schemes, and other niceties. – Mariano Suárez-Alvarez Jul 17 '10 at 3:00
In the definition of aperiodicity, you had (I corrected it) inverted the quantifiers. This was harmless for finite monoids but not for infinite ones. – Duchamp Gérard H. E. Sep 18 '15 at 13:28

The notion "periodic monoid" comes from the notion "periodic element", that is an element $a$, such that the sequence $a,a^2,a^3,\dots $ is eventually periodic. In particular every finite monoid is periodic. The notion "aperiodic monoid" (usually in this case we are talking about finite monoids) means "all subgroups are trivial", that is in the Crohn-Rhodes decomposition all "cycles" are trivial. One of the best books about (finite) monoids and their applications to automata and languages is Eilenberg, Automata, languages, and machines. Vol. A., B. For abstract semigroups, there is still the classic book by Clifford and Preston, "The algebraic theory of semigroups", and more recent books of Howie, "Fundamentals of semigroup theory" and Lawson, "Inverse semigroups. The theory of partial symmetries" among many others.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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