Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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

Consider for example the semigroup consisting of all constant functions on a set $X$ [acting on the right], 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'.]

share|cite|improve this answer
Thanks a lot, that's an interesting example. – Artem Pelenitsyn Sep 16 '12 at 16:47
How is this f a left identity? If you compose it with a constant function, that's only value is y\neq x, then you get another constant function. – JSG May 24 '13 at 16:38
@user4514: Ah, I should have specified which side the functions are acting on, as obviously it makes a difference. I'll fix this now. – Tara B May 26 '13 at 7:20

Take the semigroup $S = \{a, b, 0\}$ with $a^2 = a$, $ab = b$ and every other product equal to $0$. Then $a$ is a left identity (since $ax = x$ for all $x \in S$) but it is not an identity (since $ba = 0$). You can define this semigroup in three other equivalent ways:

(1) As a transformation semigroup on $\{1, 2, 0\}$. Just take the semigroup generated by $a = [1, 0, 0]$ and $b = [2, 0, 0]$.

(2) As a semigroup of matrices. Just take $a = \pmatrix{1 & 0\\ 0 & 0}$, $b = \pmatrix{0 & 1\\ 0 & 0}$ and $0 = \pmatrix{0 & 0\\ 0 & 0}$.

(3) As the syntactic semigroup of the regular language $a^*b$ [or as the transformation semigroup of its minimal automaton, which is just (1)].

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


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.