# Is there a notion of “sequential” idempotence?

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?

• If you want your property to hold whatever is the product order, then it means that the semigrouo generated by your functions is regular. – C.P. Nov 20 at 22:17
• @C.P. A regular semigroup is not necessarily idempotent. – J.-E. Pin Nov 21 at 8:10
• 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" :). – C.P. Nov 21 at 8:44

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.