In A. Nagy's Special Classes of Semirings, the first theorem is:
Theorem 1.1 A finitely generated semigroup is finite iff it is periodic and has the permutation property.
The definitions are as follows. Let $S$ be a semigroup.
Call $x \in S$ is periodic iff $\langle x \rangle$ is finite (really, this should be called "eventually periodic." Oh well.) Furthermore, say that $S$ is periodic iff every $x \in S$ is periodic.
$S$ has the permutation property iff there exists a natural number $n$ such that for every sequence of $n$ elements of $S$, there is a non-trivial way to permute that sequence such that the product of the original sequence equals the product of the new sequence.
Now the forward direction (finite implies periodic and permutation property) is pretty easy. However, the converse has me stumped. So, I've been toying with a simpler version, namely:
Proposition. Let $S$ denote a periodic semigroup with the permutation property, which is generated by $\{x,y\}.$ Then $S$ is finite.
Any ideas how to solve it? Mainly the problem is that the permutation property just seems trivially weak. For example, suppose that $6$ is the number of elements a sequence must have before we may conclude that a non-trivial permutation leaves the product invariant. So we might have a sequence like $xxxyxy.$ We may deduce that a non-trivial permutation leaves this product invariant. So? This is obvious; just swap any two $x$'s.
Presumably, then, the proposition can be proven using just the periodicity of $S.$
Ideas, anyone?
You can find the proof in this article
A. Restivo and C. Reutenauer, On the Burnside problem for semigroups. J. Algebra 89, 102–104 (1984)
The proof is short but relies on a deep result of Shirshov on combinatorics on words. For a complete discussion and extensions of this result (Theorem 3.4.1.), the bible is the book
Finiteness and Regularity in Semigroups and Formal Languages, by Aldo De Luca, Stefano Varricchio, Springer ISBN 978-3-642-64150-3.