# Examples of left reversible semigroups

I am looking for concrete examples of cancellative, left reversible semigroups. Left reversible semigroups are also called "Ore semigroups". See this wikipedia page for the definition of a left reversible semigroup. Of course, commutative semigroups are automatically left reversible, and I am looking for non-commutative examples.

Please also mention if these semigroups arise in an interesting setting.

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Thanks for the answers. I am quite satisfied now. –  Orr Shalit Nov 12 '09 at 2:16

Some examples and further references (and an interesting setting) are given in this paper by Laca:

http://arxiv.org/abs/math/9911135

See section 1.1, pages 2-3.

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Here is an example that I found in the book "Algebraic Theory of Semigroups, vol. I" by Clifford and Preston (the exercise on p. 36): The universal semigroup generated by two elements $a,b$ such that $ab = ba^k$. This semigroup can be concretely described as the set of of pairs $(i,j)$, with $i,j$ nonnegative integers, with multitplication $$(i,j)(m,n) = (i+m, k^m + n) .$$

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I had thought of this but wasn't sure about left reversibility. It might be worth observing, if Clifford & Preston don't, that the group completion of this guy is an example of a Baumslag-Solitar group. BS-groups are in general of interest to geometric group theorists, perhaps mined to exhaustion, perhaps not (I honestly don't know). –  Yemon Choi Nov 10 '09 at 22:54
Thanks, that's interesting. –  Orr Shalit Nov 11 '09 at 2:03

The semigroup $S= \langle a,b\mid a^2=b^2\rangle$ is two-sided reversible.

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I suppose that the non-zero elements of a left Ore domain would work --- presumably this is why they are sometimes called Ore semigroups.

To expand: a ring is a left Ore domain if it has no non-trivial zero-divisors and for every non-zero element s of the ring and every other element r of the ring one can find r' in the ring and s' non-zero and in the ring such that rs'=sr'.

Goldie's theorem says every left Noetherian ring without zero-divisors is a left Ore domain so the non-zero elements will form a left-reversible cancellative semigroup. In fact Goldie's theorem says a little more than this but I don't have time to check if the non-zero divisors will always give what you want in any left Goldie ring.

(It is possible I have my left and rights mixed-up here if so just swap them around!).

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I just did the check I said I didn't have time for, and yes, the non-zero divisors in a left Goldie ring (and therefore any left Noetherian ring) will be a left-reversible cancellative semigroup –  Simon Wadsley Nov 10 '09 at 13:33
I realised on the way home that my brain failed to move the statement of Goldie's theorem from its back to its front correctly. It only applies to semiprime (left) Goldie rings. i.e. there should be no nilpotent ideals. –  Simon Wadsley Nov 10 '09 at 18:54
Thanks for the answer. Naturally, I did not know of many concrete examples of Ore left domains, either. –  Orr Shalit Nov 12 '09 at 2:15
Do you know examples of left Noetherian rings? There are many. I can give some if you would like. –  Simon Wadsley Nov 12 '09 at 8:34

Handle with care - see comments: Imagine some system with states and transitions between them, e.g. an automaton as in computation theory or a Markov chain as in stochastics, for simplicity assume that you have an assigned initial state. Then you can label the transitions by letters and record your transition history, starting from the initial state, by forming words out of them. This gives you a semigroup. Now declare two words to be equal if they lead you from the initial state to the same state (careful - this is not the way you usually use automata in group theory). Then left reversibility means that you can, from two given states always go on to reach one same state, i.e. any two starts of your program can lead to the same outcome.

A non-deterministic terminating algorithm gives a meaningful example of such a thing. Non-deterministic means that there actually are several ways through your state diagram, terminating means that you always end in the final state.

Another (boring) example: The multiplicative semigroup of a ring - the zero always does the job. This is like an automaton which always has a one-step way from any state to the unique final state.

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These examples don't appear to be cancellative. –  Jonas Meyer Nov 10 '09 at 0:40
Absolutely true - I overread "cancellative"! The answer is bullshit - I will delete it tomorrow, if no one objects... –  Peter Arndt Nov 10 '09 at 2:18
Thanks. Please don't erase, that's an interesting example and explanation, and somebody might find it useful. –  Orr Shalit Nov 10 '09 at 20:39
Ok, I leave it, but it really has more flaws than just the cancellativeness issue. To make the semigroup multiplication globally defined (you have to be able to multiply with a given element, no matter in which state you are), you need a special kind of automaton... –  Peter Arndt Nov 11 '09 at 0:38