Proving a semigroup properties are equivalent

Question

I want to prove that the two clauses for a semigroup being a group are equivalent but am not sure how to go about it?

Prove the following are equivalent $$1)\quad\forall a\in S, aS=S \quad \text{and} \quad Sa=S$$ $$2) \quad \exists e\in S,\forall a\in S \quad ea=a\\ \forall a\in S, \exists a^{-1}\in S \quad a^{-1}a=e$$

I know that the first statement is equivalent to $\forall a,b\in S,\exists x,y\in S \quad ax=b \text{ and }ya=b$ so don't know whether this should be the way to go to firstly prove $(1)\implies(2)$ before $(2)\implies(1)$?

Answer 1

You must assume that $S$ is nonempty (some folks allow the empty semigroup); otherwise, (1) clearly holds for the empty set, but (2) does not. So assume $S$ is nonempty.

(1)$\implies$(2). Assume that for all $a\in S$, $aS=S=Sa$.

For each $a$ there exists $e_a\in S$, which may depend on $a$, such that $e_aa=a$. If $b\in S$, then there exists $x\in S$ such that $b=ax$, and therefore $e_ab = e_a(ax) = (e_aa)x = ax = b$. Thus, we can pick the same $e$ for all elements of $S$, proving that there exists $e\in S$ such that $ex=x$ for all $x\in S$.

Now fix that $e$. Then from $Sa=S$, we conclude the existence of an element $a^{-1}$ with the desired properties.

(2)$\implies$(1): Existence of a one-sided identity and inverses on the same side imply that we have a group, from which (1) easily follows