TL;DR: I have a definition, and I'm wondering if it already has a name or has been studied.

Suppose we have a sequence of operations (or if we want to be mathematical, functions whose domains and ranges are all the same) $f_1, f_2, ..., f_n$.

I'll call the sequence idempotent if the composition $f_1f_2...f_n$ is idempotent.

But I call the sequence strongly idempotent if any initial subset of the operations $f_1...f_k$ is idempotent for all $1 \leq k \leq n$.

For example, if we take the domain of integers, set $f_1(x) = x + 1$ and $f_2(x) = 0$, then the sequence $(f_1, f_2)$ is idempotent but not strongly idempotent.

My question is, does this definition/idea already have a name? Are there any known useful properties of strong idempotency, or is it known to be useful in any particular areas?

  • $\begingroup$ If you want your property to hold whatever is the product order, then it means that the semigrouo generated by your functions is regular. $\endgroup$
    – C.P.
    Nov 20, 2019 at 22:17
  • $\begingroup$ @C.P. A regular semigroup is not necessarily idempotent. $\endgroup$
    – J.-E. Pin
    Nov 21, 2019 at 8:10
  • $\begingroup$ A finite aperiodic regular semigroup is, though. I don't know for infinite. Do you? I was imprecise indeed. Thank you for the: "pan sur le bec" :). $\endgroup$
    – C.P.
    Nov 21, 2019 at 8:44

1 Answer 1


Your question would probably fit better on MathsStackExchange, but here is an answer.

I don't think there is any specific name for your definition. Since any semigroup is isomorphic to a transformation semigroup, your question is about finitely generated semigroups that are quotients of the semigroup $S_n$ with presentation $$ \langle a_1, \ldots, a_n \mid a_1^2 = a_1, (a_1a_2)^2 = a_1a_2, \ldots, (a_1 \dotsm a_n)^2 = a_1 \dotsm a_n \rangle $$ It turns out that $S_n$ is infinite for $n > 1$, but I don't know whether it has been studied in the literature.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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