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I am not familiar with much semigroup theory, but this question came up in my research and I've been unable to find much on it.

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What does "periodic" mean? –  Matthew Daws Aug 28 '10 at 20:44
    
@MD: I'm guessing it means that every element has finite order. –  Pete L. Clark Aug 28 '10 at 21:26
    
This guess seems to be confirmed by en.wikipedia.org/wiki/Semigroup#Structure_of_semigroups. –  Pete L. Clark Aug 28 '10 at 21:28
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2 Answers 2

up vote 4 down vote accepted

This question has been studied extensively more than 20 years ago. Most results are about semigroups satisfying identities. the first such result is by Morse and Hadlund: there exists an infinite 3-generated (infinite 2-generated) semigroup satisfying $x^2=0$ (resp. $x^3=0$). The latest results are in Sapir, M. V. Problems of Burnside type and the finite basis property in varieties of semigroups. (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 51 (1987), no. 2, 319--340, 447; translation in Math. USSR-Izv. 30 (1988), no. 2, 295--314. Here is a statement from that paper: All finitely generated periodic semigroups with identity $u=v$ in $n$ variables are finite if and only if all periodic finitely geneated groups with that identity are finite and the Zimin word $Z_{n+1}$ is not an isoterm for $u=v$ or $v=u$. Here Zimin words are defined by $Z_1=x_1, ..., Z_{n+1}=Z_nx_{n+1}Z_n$. A word $Z$ is an isoterm for $u=v$ if for every substitiution $\phi$, if $\phi(u)$ is a subword of $Z$, then $\phi(u)=\phi(v)$ (by a substitution I mean a map assigning words to variables). If a periodic semigroup has some algebraic structure properties, then its finiteness can be deduced sometimes from Shevrin's results (see, for example, Shevrin, L. N. Certain finiteness conditions in the theory of semigroups. (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 29 1965 553--566 or Shevrin, L. N. A general theorem on semigroups with certain finiteness conditions. (Russian) Mat. Zametki 15 (1974), 925--935. ). For example, if the semigroup is a union of groups (i.e. every element $x$ satisfies $x=x^p$ for some $p=p(x)$), then the semigroup is finite if and only if all subgroups are locally finite. There are many other results but I do not have time to list them here.

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This is known as the Burnside problem for semigroups. I believe there are many different properties that can ensure a finitely generated periodic semigroup to be finite, for example see here when all the finitely generated subgroups have finite order and every element is equal to a power of itself (this is stronger than periodicity, see the comment below), or here for when the semigroup has the permutation property.

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Finite order of subgroups is not enough. Se the example of Morse and Hadlund (their semigroup satisfies $x^2=0$ and all subgroups are trivial). –  Mark Sapir Oct 4 '10 at 5:44
    
Thanks, I believe it's fixed now. –  Gjergji Zaimi Oct 4 '10 at 6:31
    
The first reference reproves a result of Shevrin which by that time was 18 years old. The second reference is about proving that certain periodic semigroup is in fact infinite. –  Mark Sapir Oct 4 '10 at 12:15
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