Semigroup analogue to the classification of finite simple groups? Is there a semigroup analogue to the classification of finite simple groups?
If so what are some of the major results?
--Edit, reference links---
Classification of Finite Simple Groups
Special Classes of Semigroups 
One of my semigroup equation sequences in OEIS, g(f(x)) = f(f(f(x)))
 A: In group theory, the classification of finite simple groups reduces the classification of all finite groups to the so-called extension problem. Roughly speaking, the extension problem consists in describing a group in terms of a particular normal subgroup and quotient group.
There is no semigroup analogue to this theory, but a weaker classification scheme exists. A semigroup $S$ divides a semigroup $T$ if $S$ is a homomorphic image of a subsemigroup of $T$. The Krohn–Rhodes theorem states that every finite semigroup $S$ divides a wreath product of finite simple groups, each dividing $S$, and copies of the 3-element monoid $\{1, a, b\}$ in which $aa = ba = a$ and $ab = bb = b$.
The best reference on this theory is the (advanced) book
[1] J. Rhodes, B. Steinberg. The $q$-theory of finite semigroups. Springer Verlag (2008). ISBN 978-0-387-09780-0.
A: In semigroup theory, the term "simple" has a specialized meaning.  If you interpret it to mean that the semigroup has no quotients other than itself and the trivial semigroup, then there is a complete classification.  In semigroup theory, these are called "congruence-free semigroups".  (In universal algebra, "simple" means no non-trivial quotients.)
There is a complete classification of finite congruence-free semigroups.  They are either:

*

*a two-element semigroup

*a finite simple group

*a completely 0-simple semigroup.  Most of these are not congruence-free, but the ones that are have a complete description.  They are basically paramatrized by 0-1 matrices.

The proof that a congruence-free semigroup that is not covered by cases 1 or 2 must have a zero is sketched in this Math Overflow answer.  (A zero is just an element $0$ such that $0a = a0 = 0$ for all $a$.)
I don't know of a great online source for the third case.  This preprint states the theorem, with reference, as Theorem 3.1, but it's not the main subject of the paper.
