Question:
Show that $A$ is a subsemigroup of $S$ if and only if $A^{2}\subset A$. The subset here may not necessarily be proper.
My approach,
Suppose $A$ is a subsemigroup of $S$, then for all $x,x\in A, x^{2}\in A$. Does this mean that, $A^{2}\subset A$ since $(x,x)\in A^{2}$?. I am not sure!
Conversely, suppose $A^{2}\subset A$, then for any $(x,x)\in A^{2}$, $x^2\in A$ since $A^{2}\subset A$. This implies $A$ is a subsemigroup of $S$.
I am trying to convince myself with this proof, but it seems to me that what I did is not a good approach. Can somebody help me?
First, note that here $A^2$ is not the set of pairs of elements of $A$. Instead, it represents the set of all products of two elements of $A$. (Recall that in general, if $R$ and $T$ are subsets of a semigroup $S$, then $RT = \{rt\mid r\in R, t \in T\}$; here we have $R=T=A$).
So the fact that $x\in A$ implies $x^2\in A$ is one part, but not all that is needed to show that $A^2 = \{aa'\mid a,a'\in A\}$ is contained in $A$ when $A$ is a subsemigroup.
Conversely, if you assume that $A^2\subseteq A$, you are assuming that for all $a,a'\in A$, you have $aa'\in A^2\subseteq A$. This should easily lead to $A$ being a subsemigroup.