A semigroup with three or four elements without identity

Question

Does there exist a semigroup with three or four (or finite) elements, without identity?

I tried to construct such an example, but every example I tried to construct had an identity element.

Answer 1

Here is another example. With three elements take $aa=ab=ac=a$ and $ba=bb=bc=b$ and $ca=cb=cc=c$

Then any product e.g. $vwxyz$, however bracketed, is equal to the first element appearing in it i.e. $v$. So this is associative. But there is no identity.

This can be extended to provide an associative multiplication without identity on sets of various sizes.

Answer 2

A semigroup is closed and associative under the operation (which we will call multiplication) but need have no other properties.

Consider a set $S$ with three elementsw $\{a,b,c\}$ with multiplication defined by $$ \forall x,y \in S : xy = a$$ It is trivial to see that this is colsed, and easy to see that associativity holds (the second more than one element is involved, the answer becomes $a$ and stays that way).

Yet there is no identity. In particular, there is no element $e$ such that $eb=b$.

Answer 3

$$\begin{array}{c|ccc} \ast & 1 & 2 & 3\\\hline 1 & 1 & 2 & 2\\ 2 & 2 & 2 & 2\\ 3 & 2 & 2 & 2\\ \end{array} \qquad\text{and}\qquad \begin{array}{c|cccc} \bullet & 1 & 2 & 3 & 4\\\hline 1 & 1 & 2 & 2 & 4\\ 2 & 2 & 2 & 2 & 4\\ 3 & 2 & 2 & 2 & 4\\ 4 & 4 & 4 & 4 & 4\\ \end{array}$$ are two examples of finite, associative magmas with no identity element, of order $3$ and $4$ respectively.

Also: Does a non-abelian semigroup without identity exist?