Honestly, I have been looking for an a finite Quasicommutative semigroup by surfing the web, but I could't. May I ask here to give me an example for such this kind of semigroup. I tried to built one of them by using GAP, but it fails. Thank you so much. I am new to it.

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I don't understand the problem. A semigroup is called quasicommutative if for all elements $a,b$ there is some $r \geq 1$ such that $ab=b^r a$. In particular every commutative semigroup is also quasicommutative, and there are lots of examples. Or are you looking for a finite quasicommutative semigroup which is not commutative? – Martin Brandenburg Feb 4 '13 at 11:19
Dear Martin, I know that and I have found out that the Clifford (1961) made a methd to constuct them, but I ask for a sample. I though maybe someone gave me an example. In fact, my question is not a problem, but it is asking for a finite quasicommutative semigroup. I have none of them in my hand to work. Sorry if I get the time of all users. Thanks. – Babak S. Feb 4 '13 at 12:24
I want it non-commutative. – Babak S. Feb 4 '13 at 12:25
I've worked in semigroups for 15 years and never saw this definition. One can easily construct artificial examples but I don't know offhand a natural example that is noncommutative. – Benjamin Steinberg Feb 4 '13 at 16:45
@Martin and @Sorouh: Is it right the definition written above for quasicommutativity? I have found some online references where it is required $ab=(ba)^r$ instead of $ab=b^r a$. – boumol Feb 4 '13 at 18:09

The simple examples are Hamiltonian groups. Then you can construct Clifford semigroups from them.

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Thanks Boris for the time. I think, I should focus to Clifford's paper again, however it is very old one. – Babak S. Feb 5 '13 at 18:20
Babak, in your semigroups all idempotents are central. There are some articles on semigrous with commuting idempotents, e.g. C.Ash, Finite semigroups with commuting idempotents, J. of the Australian Math. Society (1987). Maybe they will be useful for you. – Boris Novikov Feb 5 '13 at 18:51

The following is (for me) a bit much for a series of comments. EDIT: It is also not appropriate as an answer. The fun begins when I alternate quantifiers and turn the intended property into a semigroup identity. In an actual quasicommutative semigroup, $ab=b^3a$ would not say anything about b having torsion. Even in the finite semigroup case, there is no justification for saying that all nontrivial powers are rth powers. I leave the comedy of inferences for those who like to see how easy and wrong it can be to deduce something from a single conflation of (for all)(exists) and (exists)(for all). END EDIT. It might interest one who wants to construct an example.

The propety as described by Martin Brandenburg is a semigroup identity that implies $b^2=b^{r+1}$. Further, one finds that for $i>1$, $b^i=c^r$ for some $c$ and that $r$th powers commute with everything in the semigroup. An attempt at a reduced form for words on k letters in such a semigroup (this part needs checking before taking seriously) is something like wp where w is a squarefree word (so w does not have abcabc or similar subwords) on some subset S of the k letters and p is possibly the empty word and otherwise is an $r$th power involving only powers of the letters not in S. The subsemigroup of square or higher powers will be commutative.

If the above is correct, I see finite noncommutative examples looking like a small set of extra elements adjoined to a commutative semigroup, where the square of any extra element lies in the commutative semigroup. Not being a semigroup theorist, I sympathize with the original poster and his/her plight.