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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)))

Chad Brewbaker
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    For programmers out there, finite semigroups are a collection of functions on a finite set. Finite groups are collections of permutation functions. Classifying sets of permutation functions took decades, https://en.wikipedia.org/wiki/Classification_of_finite_simple_groups – Chad Brewbaker Jan 27 '17 at 19:14
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    A quick google search came up with this: http://www.arsmathematica.net/2012/04/25/classification-of-finite-simple-semigroups-and-moufang-loops/ – David Hill Jan 27 '17 at 19:24
  • @davidHill I saw that, going to take a deeper look at the underlying paper. I have a bunch of semigroup equations in OEIS. I should see which ones are Moufang. http://oeis.org/A239750 – Chad Brewbaker Jan 27 '17 at 19:30
  • @DavidHill In my group work, decomposition into prime order permutations has been most useful. In Semigroup land you can factor by self composition. Treelike outside with a groupish center and idempotent in the middle. – Chad Brewbaker Jan 27 '17 at 19:39

2 Answers2

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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.

J.-E. Pin
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    If you set edges from $a$ to $a^{i}$ for all elements of the semigroup, you get a set of connected components surrounding every idempotent. In the middle they have a group-like structure that is all cycles. In the outside you have trees that lead into the center. The 3 element monoid corresponds to the tree-like elements on the outside? – Chad Brewbaker Feb 01 '17 at 15:16
  • I am sorry, but I don't understand your sentence "In the outside you have trees that lead into the center". First of all, what does "outside" mean in this context? Anyway, in the 3-element monoid of my answer, all elements are idempotent, and thus your graph consists of three loops $1 \to 1$, $a \to a$ and $b \to b$. – J.-E. Pin Feb 02 '17 at 10:13
  • Ah ok. Under iteration, $a^i$ -> $a^{i+1}$, you get forgetful/reluctant functions which then enter a cycle. The paths before the cycle form paths on a tree if you zoom out. – Chad Brewbaker Feb 02 '17 at 21:31
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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:

  1. a two-element semigroup
  2. a finite simple group
  3. 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.

Martin Sleziak
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arsmath
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