Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question is related to and inspired by the question Why are groups more important than semigroups?.

I am curious why I don't see much studies on right groups.

On Pp.37 of Clifford& Preston's Algebraic Theory of Semigroups Vol. I : A semigroup is called a right group if it is right simple and left cancellative. It is equivalent to saying that for any element $a$ and $b$ of a semigroup $S$, there exists one and only one element $x$ of $S$ such that $ax = b$.

A right group is the direct product of a group and a right zero semigroup. A right group is the union of a set of disjoint groups. A periodic semigroup is a right group iff it is regular and left cancellative, etc.

A left group is the dual of a right group.

Most literature I found about right groups is very old (back in 50's and earlier). As far as I know, there is not much research on this subject in the past 30 years or so. Am I wrong about this? Or "right group" are common words so I do not get useful results when I google for it?

But I thought the right group concept could serve as a bridge between groups and semigroups. For example, a permutation group is a set of bijective mappings and a right zero transformation semigroup is a set of constant mappings, you have a right group if you take the direct product of a permutation group and a right zero transformation semigroup. What is this right group look like? Another example, an aperiodic semigroup contains no non-trivial groups, then what is a semigroup which contains no non-trivial right groups? What is a semigroup which contains a non-trivial right group? etc. etc.

share|improve this question
2  
It might be helpful if you defined what a right group actually is in the question. –  Tara B May 1 '12 at 8:29
    
@ymar, I am not too sure about that. I forgot to mention being a right group is equivalent to saying the semigroup is right simple and contains an idempotent element. And the idempotents are the identities of the disjoint groups. –  scaaahu May 1 '12 at 9:10
    
One more thing, every finite semigroup has an idempotent element. So, all finite right simple semigroups are right groups. –  scaaahu May 1 '12 at 9:22
    
What do you mean by "What [does] this right group look like?" It looks like a direct product of a group and a right semigroup. What more do you exactly want to know? –  Tara B May 2 '12 at 13:03
    
@Tara B, I meant the right group transformation semigroup which is isomorphic to the direct product of a proper subgroup of $S_n$ and a right zero transformation semigroup which is of cardinality less than n. The rank of the elements of the premutation group is n. the rank of the elements of the right zero transformation semigroup is 1. What is this right group transformation semigroup? This is the kind of questions I had in mind. I don't have answers. That's one of the reasons I ask this question. –  scaaahu May 3 '12 at 2:30
show 1 more comment

1 Answer 1

You mention in your question some theorems about right groups that can be found in Volume 1 of The Algebraic Theory of Semigroups. However, you don't distinguish between those given as theorems and those given as exercises, and the distinction is important here. The only one of them that's not an exercise is this:

Theorem 1.27. The following assertions concerning a semigroup $S$ are equivalent:

(i) $S$ is a right group.

(ii) $S$ is right simple, and contains an idempotent.

(iii) $S$ is the direct product $G\times E$ of a group $G$ and a right zero semigroup $E.$

(page 38.)

And indeed, this theorem, especially the equivalence between (i) and (iii), seems to explain pretty well what right groups are. It is a structure theorem -- it explains the structure of right groups in terms of "simpler" (or at least other) structures. For right zero semigroups, they're quite clearly very simple structures. Groups definitely aren't but it seems to be a general trend in semigroup theory to push problems towards groups and abandon them when groups are reached. I think P. A. Grillet explained it well on the first pages of his book on semigroups, but I seem to have lost the book.

I'm quite sure some interesting things can still be said about right groups but it's natural that they don't arouse very much interest in presence of such a theorem. Modulo group theory, it explains the structure of right groups pretty well.

share|improve this answer
    
I didn't distinguish between the theorem you cited and the exercises because the exercises are established results. I guess I should have changed the title of the question to "why right groups are not studied that much in semigroup theory?" I am more in favor of the semigroups than groups - as I said in the first line of the question. My real issue is, I need to know more about right groups, any significant studies have been done? If not, why? You may be right about the existence of the structure theorem is the reason. I guess I need to dig more. –  scaaahu May 1 '12 at 11:22
2  
From a semigroup theory point of view, we pretty much already know all about right groups. We understand direct products pretty well; right zero semigroups are straightforward; the only non-trivial part is the groups, but those are for group theorists to worry about. –  Tara B May 1 '12 at 14:19
1  
In other words, people who care about semigroups qua semigroups consider groups to be "known." If you can solve a semigroup problem in terms of groups, then you consider the problem 'solved', even if the group-theoretic problem is not really "solved". –  Arturo Magidin May 1 '12 at 19:00
    
@scaaahu May I ask why you need to no more about right groups? I'm curious where they came up. –  user23211 May 1 '12 at 19:56
    
@ymar,I am studying direct product. Every semigroup theorist knows about Krohn Rhodes decomposition which is about wreath product. If structure theorem suffices, we all can quit studying semigroups after K&R decomp was done. –  scaaahu May 2 '12 at 1:25
show 10 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.