Terminology for semigroups where every element is a product

Solution 1:

I don't know of a specific name for these semigroups (although there might be one), but here is a characterization of the finite semigroups satisfying this property.

First, a brief remainder about Green's relations. If $S$ is a semigroup, let $S^1$ be the monoid obtained from $S$ by adjoining an identity element if needed. Next define a preorder $\leq_{\cal J}$ on $S$ by setting $s \leqslant_{\cal J} t$ if there exist $a, b \in S^1$ such that $s = atb$. Then $s \mathrel{\cal J} t$ if and only if $s \leqslant_{\cal J} t$ and $t \leqslant_{\cal J} s$.

Theorem. A finite semigroup $S$ satisfies $S^2 = S$ if and only if its maximal $\cal J$-classes for the order $\leqslant_{\cal J}$ are regular.

Proof. Let $E$ be the set of idempotents of $S$. By [1, Proposition III.9.2, p. 81], one has $S^n = SES$ for all $n \geqslant 0$. If $S^2 = S$, then $S = SES$. In particular, for each element $s \in S$, there is an idempotent $e$ such that $s \leqslant_{\cal J} e$. It follows that every maximal $\cal J$-class contains an idempotent and hence is regular.

Conversely, suppose that every maximal $\cal J$-class contains an idempotent. Let $s \in S$. Then there is an idempotent $e$ such that $s \leqslant_{\cal J} e$ and there exist $a, b \in S^1$ such that $s = aeb$. If $a \in S$ or $b \in S$, then $s \in S^2$. Otherwise, $a = b = 1$ and $s = e =e^2 \in S^2$. Thus $S = S^2$.

[1] S. Eilenberg, Automata, languages, and machines. Vol. B. Pure and Applied Mathematics, Vol. 59. Academic Press, New York-London, 1976.

Solution 2:

I'm not aware of a standard term, but in one paper, I and my co-authors called such semigroups 'decomposable' (because every element can be decomposed as a product of two other elements). I'm not sure it's a very good term, since there are a lot of possibilities one could imagine for the definition of 'decomposable semigroup'. (Indeed, it has at least one other definition already: https://www.sciencedirect.com/science/article/pii/S0747717113000722; we should probably have checked for this!)