In a semigroup, the product of two subsets is always a subsemigroup

Question

In a semigroup, the product of two subsets is always a subsemigroup. Let $A$,$B$ be two subset of semigroup G. $$AB=\{ab:a\in A,b \in B\}$$

I try to prove this result but i haven't got any idea. So please give me a hint. Please don't give me any answers.

Answer 1

We have an XY problem. The OP asked about proving something that's obviously false. In a comment it turns out what the OP really wants is this:

If $G$ is a finite group, $A\subset G$ and $|A|>|G|/2$ then $A^2=G$.

Of course $A^2$ might mean $\{a^2:a\in A\}$, but then the result is clearly false, since $|A^2|\le|A|$. Assuming instead that $A^2=\{ab:a,b\in A\}$ it's true by a simple counting argument:

Fix $x\in G$ annd define $$B=\{a^{-1}x:a\in A\}.$$Since $G$ is a group, $$|A|+|B|=2|A|>|G|.$$So $A\cap B\ne\emptyset$, and if $b=a^{-1}x\in B\cap A$ then $x=ab$.

Answer 2

No, the product of two subsets has no reason to be a subsemigroup. What is true however, is that $\mathcal{P}(S)$, equipped with the product you defined, is a semigroup.