petrusrexcs

## Answered question

2021-12-30

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

### Answer & Explanation

Kayla Kline

Beginner2021-12-31Added 37 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 $e\in S$ such that $ea=a$ for all $a\in S$
2. For each $a\in S$ there is an element ${a}^{-1}\in S$ with ${a}^{-1}a=e$
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 $a\in S$ there is an element ${a}^{-1}\in S$ with $a{a}^{-1}=e$.
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 $Sa=\left\{sa\mid s\in S\right\}$ is a group for all $a\in S$, 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 $2×2$ matrices (with matrix multiplication as the operation):

Or, an example that appears as exercise 30 in section 4 of Fraleigh's abstract algebra text: the nonzero real numbers under the operation $×$ defined by $a×b=|a|b$.
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).

lovagwb

Beginner2022-01-01Added 50 answers

Step 1
Let be a monoid and let ${M}^{\circ }$ be the set of finite words constructed from M. For two words let's define operation $×$ as point-wise application of $\cdot$, truncating according to the shorter word:

Then 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 $y×x\ne x$. The unit element would have to be an infinite sequence , but ${M}^{\circ }$
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 . While min is clearly associative, there is no single element in $\mathbb{Z}$ that would serve as the identity. (It is actually a homomorphic image of the previous example, mapping words to their length.)

karton

Expert2022-01-09Added 613 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
Or written as multiplication rule: $ab=b$ for all . 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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