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Let N denote the free semigroup of rank 1. Say that a semigroup T is a small extension of N if N embeds in T and |T - N| is finite. Is there some kind of classification of small extensions of N? What do these semigroups look like?

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David: I'm afraid I just don't understand what is meant by a classification. The paradigms I'm most familiar with are: classification of finite simple groups up to isomorphism, by listing them; classification of (as many as possible) nuclear C*-algebras up to isomorphism in terms of K-theoretic data; classification of essentially normal operators, in terms of essential spectra and index data (BDF). Could you please edit your question to explain what would be a "classification" in the sense you desire? – Yemon Choi Apr 18 '10 at 19:47

David, I don't know if you still care about this. The people at St. Andrews say that a semigroup S has finite Rees index in a semigroup T if S is a subsemigroup of T and |T-S| is finite. They prove that this forces S and T to have a lot in common. For example, if memory serves one is finitely presented iff the other is. Probably the same is true for things like residual finiteness. I don't know if their results help for what you want.

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This is probably not the kind of answer you are looking for, but we can take any finite semigroup $R$, and define $T=N\sqcup R \sqcup \{\theta\}$ where $\theta$ is a formal symbol; we then make $T$ into a semigroup by declaring the product of $\theta$ with anything to be $\theta$ itself, declaring that $nr=\theta=rn$ for all $n\in N$ and all $r\in R$, and then keeping the same products on $N$ and $R$ as before. Then $T$ is a semigroup, $N$ embeds as a subsemigroup, and $N$ is a cofinite subset of $T$; but I find it hard to see how we can say anything useful about $T$.

Basically, I think you need to come up with a refined version of your question, perhaps by restricting $T$ to lie in some class of semigroups which you care about.

Update: In the comments, David asks if a finite extension of N (in the sense defined in the question) must contain a left or right zero for example. It is not too difficult to see that this need not be the case; take a finite semigroup S (which necessarily has a zero element) and stick it at the front of N, so that the product of $s\in S$ and $n\in N$ is just $n$.

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"restricting T to lie in some class of semigroups which you care about." I want to prove something about small extensions of N, so I can't restrict myself to a smaller class. In your comment, you point out that adjoining a zero to the set N union R and forcing products nr and rn to be zero gives a small extension of N. I agree, but I want some kind of classifying statement. Not all small extensions of N have a zero, but do all small extensions of N have a left or right zero (for example)? – david mccune Apr 18 '10 at 6:53
What Yemon demonstrated is that the problem of classifying small extensions is at least as hard as the problem of classifying finite semigroups, which in turn is at least as hard as the problem of classifying finite groups... you probably want to ask a more specific question. – Qiaochu Yuan Apr 18 '10 at 7:07
@Qiaochu: Classifying finite semigroups is MUCH worse than classifying finite groups! David: I really feel you need to think harder about what you want to prove; the lesson of semigroup theory (cf. Howie's book for instance) is that we really have to restrict ourselves to certain classes in order to get anywhere. The answer to your last question is "no" and it is instructive to try and think of the simples, most trivial example. – Yemon Choi Apr 18 '10 at 8:21
[I deleted an intemperate and inaccurate comment, after further reflection. I still maintain that a more restricted/precise question seems likely to admit a more fruitful answer.] – Yemon Choi Apr 18 '10 at 8:40
If the answer to my original question is "no" (my original question being is there some kind of classification of small extensions of N), then you can just say so. I'm trying to prove that small extensions of N don't show up in a certain class of semigroups (and I have strong experimental evidence to suggest that they don't), so I don't think I need to "think harder about what I need to prove." I have proved my result for subclasses of small extensions of N, so I have already restricted myself down. – david mccune Apr 18 '10 at 18:25

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