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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
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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
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@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
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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
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@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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3 Answers

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)].

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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'.]

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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. –  user4514 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
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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
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