Are there any interesting and natural examples of semigroups that

petrusrexcs 2021-12-30 Answered
Are there any interesting and natural examples of semigroups that are not monoids (that is, they don't have an identity element)?

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Kayla Kline
Answered 2021-12-31 Author has 3186 answers
Step 1
Consider the following definition of a group:
Definition A semigroup S is said to be a group if the following hold:
1. There is an \(\displaystyle{e}\in{S}\) such that \(\displaystyle{e}{a}={a}\) for all \(\displaystyle{a}\in{S}\)
2. For each \(\displaystyle{a}\in{S}\) there is an element \(\displaystyle{a}^{{-{1}}}\in{S}\) with \(\displaystyle{a}^{{-{1}}}{a}={e}\)
At one point in my life, it seemed natural to ask what happens if we replace axiom 2 with the very similar axiom
2'. For each \(\displaystyle{a}\in{S}\) there is an element \(\displaystyle{a}^{{-{1}}}\in{S}\) with \(\displaystyle{a}{a}^{{-{1}}}={e}\).
It is a fun exercise to work out some of the consequences that result from this. Here are a few facts about a semigroup S which satisfies 1 and 2':
If e is the unique element of S satisfying axiom 1, then S is a group
If S has an identity (in the usual sense) then S is a group
The principal left ideal \(\displaystyle{S}{a}={\left\lbrace{s}{a}{\mid}{s}\in{S}\right\rbrace}\) is a group for all \(\displaystyle{a}\in{S}\), and in fact all principal left ideals of S are isomorphic as groups.
It is not difficult to find examples of such semigroups that are not groups. For example, consider the following set of \(\displaystyle{2}\times{2}\) matrices (with matrix multiplication as the operation):
\[\left\{(\begin{array}{c}a & b \\ 0 & 0 \end{array})| a,\ b \in \mathbb{R},\ a\ne0\right\}\]
Or, an example that appears as exercise 30 in section 4 of Fraleigh's abstract algebra text: the nonzero real numbers under the operation \(\displaystyle\times\) defined by \(\displaystyle{a}\times{b}={\left|{a}\right|}{b}\).
Certainly it is debatable whether or not semigroups satisfying axioms 1 and 2′ are "interesting" or "natural". But I guess I think they are. And, I am not the only one (or the first one, by a long shot!) to think this. See Mann, On certain systems which are almost groups (MR).
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lovagwb
Answered 2022-01-01 Author has 4566 answers
Step 1
Let \(\displaystyle{\left({M},\ \cdot,\ {e}\right)}\) be a monoid and let \(\displaystyle{M}^{{\circ}}\) be the set of finite words constructed from M. For two words let's define operation \(\displaystyle\times\) as point-wise application of \(\displaystyle\cdot\), truncating according to the shorter word:
\(\displaystyle{\left({u}_{{{1}}},\cdots,{u}_{{{m}}}\right)}\times{\left({v}_{{{1}}},\cdots,{v}_{{{n}}}\right)}={\left({u}_{{{1}}}\cdot{v}_{{{1}}},\cdots,{u}_{{\min{\left({m},\ {n}\right)}}}\cdot{v}_{{\min{\left({m},\ {n}\right)}}}\right)}\)
Then \(\displaystyle{\left({M}^{{\circ}},\ \times\right)}\) is a semigroup, but it's not a monoid. Any candidate y for the unit element would have only a finite length, so for any x that is longer we'd have \(\displaystyle{y}\times{x}\ne{x}\). The unit element would have to be an infinite sequence \(\displaystyle{\left({e},\ {e},\cdots\right)}\), but \(\displaystyle{M}^{{\circ}}\)
This example is not arbitrary, it is closely related to zipping lists and convolution.
Step 2
Update: A very simple example of a semigroup that is not a monoid is \(\displaystyle{\left({\mathbb{{{Z}}}},\ \min\right)}\). While min is clearly associative, there is no single element in \(\displaystyle{\mathbb{{{Z}}}}\) that would serve as the identity. (It is actually a homomorphic image of the previous example, mapping words to their length.)
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karton
Answered 2022-01-09 Author has 9103 answers

I don't know if this counts as interesting, but a simple example is what C programmers know as the comma operator: Ignore the first argument and return the second.
Or written as multiplication rule: \(ab=b\) for all \(a,\ b\in S.\)
Or written as multiplication rule: \(ab=b\) for all \(a,\ b\in S.\). This is easily shown to be a semigroup, but as long as there are at least two elements in S, this is not a monoid, as with neutral element e you'd have for all \(a\in S\) the identity \(a=e\).

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