# Are there any interesting and natural examples of semigroups that

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
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
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.)
karton

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 $$a=e$$.