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.

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