I was introduced to semigroups today and had a question. So all the examples of semigroups I was given were either monoids or groups. So I was curious, does there exist a semi-group which is not abelian and does not contain identity?
I tried to construct an example, but every example I tried to construct had an identity element.
Thanks!
If $S$ is a set with at least two elements, and multiplication is defined by $x\cdot y=y$, then $(S,\cdot)$ is a noncommutative semigroup with no right identity. (Of course every element is a left identity.)
Let $A^{\ast}$ be the free monoid on a nonempty set $A$; this is just the set of all words using $A$ as an alphabet. Let $e$ be the identity. Since no element of $A^{\ast}$ has an inverse other than $e$, $A^{\ast}-\{e\}$ is a semigroup without identity, and it is nonabelian provided $A$ has more than one element.
Take the set of all finite, non-empty sequences of elements in some non-empty set $S$ with operation given by concatenation (commonly called the free semigroup on $S$). That is, take $S^+=\bigcup_{n\geq 1}{S^n}$ where for $(s_1,\ldots, s_n),(t_1,\ldots, t_m)\in S^+$ we have their concatenation given by
$$(s_1,\ldots, s_n)(t_1,\ldots, t_m)=(s_1,\ldots, s_n,t_1,\ldots, t_m).$$
This is non-commutative so long as $S$ contains at least two elements; if $x,y$ are such distinct elements, then $(x)(y)=(x,y)\neq (y,x)=(y)(x)$.
A simple example would be the set of all non-empty strings (over some alphabet) with the concatenation operator. The empty string would be the identity, but we excluded it from the domain, so there is no identity.
It might be a little unsatisfying to create a semigroup with no identity simply by removing the identity, but if we take any semigroup without an identity and define a new identity element $e$ such that $e\cdot x = x\cdot e = x$ for all $x$ in the semigroup, then we always create a semigroup with an identity – so any example we can possibly give is just a semigroup with its identity removed (and which remains closed under its operation when we removed the identity).
It depends on exactly what you mean by “does not contain identity”. What about one sided identities? What if there is more than one distinct identity? The existence of a single unique identity element is a property of groups. When you are not talking about groups, you have to be more specific about what you are excluding. In this example, every element is a left identity, but there is no right identity or two-sided identity. I copied this example straight from Wikipedia (see the last example in the list), but in my defense I built an identical example myself before I ran across it.
only two elements {e, f} ∗ defined by e ∗ e = f ∗ e = e and f ∗ f = e ∗ f = f both e and f are left identities, but there is no right identity and no two-sided identity
The operation is associative, which you can see if you list out all combinations of two operations on three elements, e.g.
(e * f) * e = f * e = e
e * (f * e) = e * e = e
(e * f) * f = f * f = f
e * (f * f) = e * f = f
…
Yes, and you can do it with just two elements: Consider the set $\{a, b\}$ with the operation that all products are equal to $a$. This is clearly associative, and neither element is an identity.
You may recall from group theory that every group embeds into a symmetric group.
There is an analagous result that every semigroup embeds into the semigroup of functions on some set. So an easy way to construct examples of semigroups is to take a set $S$, and then consider some subset of the functions $S \to S$ that is closed under composition.
For your question in particular, we could take
The set of functions $f: \mathbb{N} \to \mathbb{N}$ such that the image of $f$ is finite, under composition. (Or replace $\mathbb{N}$ with any infinite set.)
The set of functions $f: \{0,1,2\} \to \{0,1\}$ under composition.
The set of bounded, continuous $f: \mathbb{R} \to \mathbb{R}$ under composition.
In all these examples, the idea is to take the set of functions on some set under composition (function composition is always associative), but to further restrict the set in some way that excludes the identity function. Function composition is generally nonabelian, so we should get a nonabelian semigroup without identity.
The Cayley table
$$\begin{array}{c|ccccccc}
\ast & 0 & 1 & 2 & 3 & 4 & 5 & 6\\\hline
0 & 4 & 2 & 6 & 0 & 4 & 5 & 6\\
1 & 3 & 5 & 1 & 6 & 4 & 5 & 6\\
2 & 0 & 5 & 2 & 6 & 4 & 5 & 6\\
3 & 4 & 1 & 6 & 3 & 4 & 5 & 6\\
4 & 4 & 6 & 6 & 4 & 4 & 5 & 6\\
5 & 6 & 5 & 5 & 6 & 4 & 5 & 6\\
6 & 4 & 5 & 6 & 6 & 4 & 5 & 6\\
\end{array}$$
defines a non commutative semigroup with no (left nor right) identity, over the finite set $\{0,1,2,3,4,5,6\}$ (reference).
Furthermore, as pointed out in the answer by @bof, finite magmas like the one in this answer are associative, non commutative (the Cayley table is not symmetric), have no left identity, hence have no identity.