# Semigroup with exactly one left(right) identity?

Are there any examples of a semigroup (which is not a group) with exactly one left(right) identity (which is not the two-sided identity)? Are there any “real-world” examples of these (semigroups of some more or less well-known mathematical objects) or they could only be “manually constructed” from abstract symbols (a, b, c…) subject to operation given by a Caley table?

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Certainly, yes. – Artem Pelenitsyn Sep 15 '12 at 9:48
How can we tell whether a given manually constructed examples does not show up in any real-world context? – Rasmus Sep 15 '12 at 9:54
@tomasz: If you rename $a$ to $1$ and $b$ to $0$, you'll notice that you've described the multiplication of the two-element field. – celtschk Sep 15 '12 at 10:54
And therefore it is not an example of what Artem is asking for, since $a$ is a two-sided identity. – Tara B Sep 15 '12 at 11:00
@celtschk: Touche. Change it to $\{a,b,c\}$ with $a\cdot x=x$, $y\cdot x=b$ if $y\neq a$. :) That's what I had thought of in the first place, but I wanted to make it minimal and I overdid it. :) – tomasz Sep 15 '12 at 11:01
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## 2 Answers

Consider for example the semigroup consisting of all constant functions on a set $X$, together with one non-constant idempotent function $f$ (for example, let $f$ fix some point $x\in X$ and send every other point to some $y\neq x$). Then $f$ is a unique left identity, and $f$ is not a right identity.

In general I think it's probably helpful to think about this question in terms of transformation semigroups.

EDIT: Since this question has been sitting around with no accepted answer for a while, I'll state my last sentence a bit more strongly: You can determine exactly which transformation semigroups have a single left (or right) identity, and since every semigroup is isomorphic to a transformation semigroup, doing this will give you all examples. [Although I just noticed that the OP hasn't been on this site for about a month, so I guess the question might remain 'unanswered'.]

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 Thanks a lot, that's an interesting example. – Artem Pelenitsyn Sep 16 '12 at 16:47

Take a finite semigroup $S$. Then $S$ has an idempotent element $e$ since $S$ is finite.

Let $T = \{se : s \in S\}$. Then $T$ is a subsemigroup of $S$. We have $e \in T$ because $e = ee$. And $e$ is a right identity of $T$ since $(se)e = s(ee) = se$ for all $s \in S$.

My problem with this example is that I don't think $e$ is the only right identity of $T$ for every such $T$ and this is probably still not real world enough.

I believe if we give more conditions when we construct $T$, we might be able to get a semigroup with only one right identity. But, it may not be a real example for OP.

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 You need to make sure that $r$ is not a left identity. Also, I'm not sure how any of those qualifies as real-world. ;) – tomasz Sep 16 '12 at 10:46 How do you 'adjoin a right identity'? You need to define how to multiply by it on the left. It's not a straightforward thing like adjoining a two-sided identity or a zero. – Tara B Sep 16 '12 at 10:51 @TaraB I believe "adjoin" is doable if $S$ has more than one element. I didn't like this idea because it's not natural. – scaaahu Sep 16 '12 at 11:18 You guys are too worry about “real-world'ness”… @scaaahu show an interesting approach, but as it is already mentioned there is now guarantee that the e is the only one identity in T. – Artem Pelenitsyn Sep 16 '12 at 16:52