Semigroup over a finite set that is nothing more than that

Question

Ok so a semigroup only has associativity.

Suppose i have a set $S = \{a,b,c,d\}$

Someone give me an example for the function $+:S \times S \rightarrow S$ (in table form) so that is only associative so that $(S, +)$ is just a semigroup and nothing more (meaning not commutative or anything else). Neutral element does not matter.

Answer 1

Two simple addition operators which would make the set $S$ a semigroup are $+_L:x+_Ly\mapsto x$ and $+_R:x+_R y\mapsto y$.

It's obvious that both of these operators are associative$$(a+_L b)+_Lc=a+_Lc=a$$$$a+_L(b+_Lc)=a+_Lb=a$$ $$(a+_Rb)+_Rc=b+_Rc=c$$$$a+_R(b+_Rc)=a+_Rc=c$$

But, neither of these operators have an identity, nor are commutative.

Answer 2

Actually 3 elements suffice to obtain a semigroup that is neither a monoid, nor a group, and is neither commutative nor idempotent. Take $S = \{a,b,0\}$ with $aa = a$, $ab=b$, $ba =0$ and $0a=a0=0b=b0$.

If you insist to have 4 elements, take $S =\{a,b,c,0\}$ with $aa = a$, $c = ab$, $ac = c$ and all other products equal to $0$. Technically speaking, this is the syntactic semigroup of the language $a^+b$.