Show $(\mathcal S\times\mathcal S,*)$ with $|\mathcal S|\ge 2$ & $(a,b)*(c,d) = (c,b)$ is a semigroup without a one & $(a,b)^2=(a,b)\forall(a,b)$

Question

Let $\mathcal S$ be a set with at least two elements. Consider the operation $\ast$ defined on $\mathcal S $$\times$$\mathcal S$ by

$$(a,b) \ast(c,d) = (c,b).$$

How can I show that $(\mathcal S\times\mathcal S,\ast)$ is a semigroup without identity in which all elements are idempotent?

(I'm preping for an exam, it's not either for homework neither I'm in the middle of an exam)

Thanks

Answer 1

What’s $(a,b)^2=(a,b)(a,b)$?

What’s the requirement for $(a,b)(x,y)=(a,b)$ and $(x,y)(a,b)=(a,b)$, for every $a,b\in S$?

Can you prove associativity?

Answer 2

Just expanding the definitions: you can check that (1) $*$ is associative [so $(S\times S,*)$ is a semigroup], (2) For any $(a,b)\in S\times S$, there is some $(c,d)\in S\times S$ such that $(a,b)*(c,d)\neq (c,d)$ [so there is no identity], and (3) $(a,b)*(a,b) = (a,b)$ for all $(a,b\in S\times S$ [so all elements are idempotent].

In the comments, you say you got stuck on (1). Let's check: $$ ((a,b)*(c,d))*(e,f) = (c,b)*(e,f) = (e,b) $$ and $$ (a,b)*((c,d)*(e,f)) = (a,b)*(e,d) = (e,b) $$ so the operation is associative.