## Groups are precisely those semigroups which are both left and right simple

Prove that a semigroup $S$ is a group if and only if it is both left and right simple. Exhibit a left simple semigroup which is not right simple.

We begin with a lemma.

Lemma: Let $S$ be a semigroup. Then $S$ is left simple (right simple) [simple] if and only if $Sa = S$ ($aS = S$) [$SaS = S$] for all $a \in S$. Proof: Suppose $S$ is left simple, and let $a \in S$. Certainly $SSa \subseteq Sa$, so that $Sa$ is a left ideal. Thus $Sa = S$ for all $a \in S$. Conversely, suppose $Sa = S$ for all $a \in S$, and let $L \subseteq S$ be a left ideal. Now $L$ is nonempty by definition; say $a \in L$. Then $Sa \subseteq L$, and so $S \subseteq L$. Thus $L = S$, and in fact $S$ is the only left ideal of $S$. So $S$ is left simple. The proofs for right simple and simple are similar. $\square$

Now let $S$ be a group and let $a \in S$. If $x \in S$, then we have $x = xe = xa^{-1}a$; in particular, $x \in Sa$. So $S = Sa$ for all $a$. By the lemma, $S$ is left simple. Similarly, $S$ is right simple.

Now suppose the semigroup $S$ is both left and right simple. Let $a \in S$. Since $Sa = S$, there exists an element $e \in S$ such that $ea = a$. Now let $b \in S$ be arbitrary. Since $aS = S$, there exists $c \in S$ such that $b = ac$. Now $b = ac = eac = eb$, so in fact $e$ is a left identity element of $S$. Similarly, there is a right identity element $f$, and we have $e = ef = f$, so that $e$ is a two-sided identity.

Now let $a \in S$ be arbitrary. Since $Sa = aS = S$, there exist elements $b,c \in S$ such that $ba = ac = e$. Now $b = be = bac = ec = c$, and we have $b = c$. Thus $a$ has a two sided inverse with respect to $e$. Since $a$ was arbitrary, every element has an inverse, and so $S$ is a group.

Now consider the semigroup $S = \{a,b\}$ with $xy = x$ for all $x,y \in S$. (That is, $S$ is the left zero semigroup on two elements.) Suppose $T \subseteq S$ is a left ideal. Now for all $x \in S$ and $y \in T$, we have $xy = x \in T$. Thus $T = S$, and so $S$ is left simple. However, $S$ is not a group, and so (by the previous argument) cannot be right simple.