# Necessary condition for left simple semigroup with an idempotent

Show that the following statements are equivalent;

1. $$S$$ is a left simple semigroup that contains an idempotent
2. $$S$$ is isomorphic to the direct product of a subgroup of $$S$$ and a left zero semigroup $$L$$

In one way this is kind of obvious. If $$S \cong G\times L$$, then the semigroup $$G\times L$$ is left simple and $$(1,l), l\in L$$ is an idempotent since $$(1,l)(1,l) = (1,ll) = (1,l)$$.

The other way seems a total catastrophe. Grillet, in his Semigroups V.4.4, suggests the Rees-Sushkevich theorem solves this problem. I don't understand how, though.

Hint 1: Let's show that $$S$$ is left simple iff $$S$$ is single $$\mathscr L$$-class.
If $$S$$ is a single $$\mathscr L$$-class, then for every $$a\in S$$, $$S = L_a$$. Let $$I\subseteq S$$ be a left ideal, $$x\in I$$ and $$S = L_x$$. Then for every $$b\in S$$, there is $$y\in S^1$$ s.t $$b = yx\in I$$ so $$I=S$$ and $$S$$ is left simple.
Conversly, let $$S$$ be left simple and $$L_x\subseteq S,x\in S$$. Since $$S$$ is left simple, then $$S^1x = S$$ and for every $$b\in S$$, $$b\in L_x$$ hence $$L_x=S$$.

My algebra textbook (not Grillet's) defines regularity:

$$a\in S$$ is regular if there exists an $$x\in S$$: $$a= axa$$.

In II.2.2 Grillet says that the following are equivalent

1. $$a\in S$$ is regular
2. L_a contains an idempotent
3. R_a contains an idempotent

So we have $$S = L_x$$ and every $$\mathscr R$$-class contains idempotent. Since the intersection of $$\mathscr L$$ and $$\mathscr R$$-classes produces $$\mathscr H$$-classes, I can say $$S$$ contains groups.

Hint 2: Since every $$\mathscr H$$-class contains an idempotent, then every $$\mathscr H$$-class is a subgroup. Idempotent in each $$\mathscr H$$-class must be the identity element (if $$g^2 = g$$ in a group, then $$g=1$$). So there's a one to one correspondence between the idempotents and the $$\mathscr H$$-classes? Since $$S$$ is left simple, then the $$\mathscr R$$ and $$\mathscr H$$-classes coincide?

Let $$e,f\in E(S)$$ be idempotents, then w.l.o.g $$e\in L_f$$ and $$e=sf, s\in S^1$$. Then $$ef = (sf)f = s(ff) = sf = e$$ and $$E(S)$$ is left zero semigroup.

All $$\mathscr H$$-classes have the same cardinality. So pick an $$\mathscr H$$-class $$H_a$$. Then $$H_a\times E(S)\cong S$$. (seems logical by the eggbox picture)

Define $$\varphi : S\to H_a\times E(S), s\mapsto (h,e)$$ where $$e$$ determines the $$\mathscr H$$-class where $$s$$ resides and $$h\in H_a$$ is the counterpart of $$s$$. (I hope this makes sense).

PROBLEM How to define the isomorphism? $$S\to H_x\times E(S), x\in E(S), s\mapsto (h,e)$$ runs into a problem. Namely, if $$s\mapsto (h,e)$$ and $$t\mapsto (h',f)$$, then $$st\mapsto (??, e)$$

To be continued..

Hint 1. Show that $S$ consists of a single regular $\mathcal{L}$-class. Now, each $\mathcal{R}$-class contains an idempotent.
Hint 2. Each $\mathcal{H}$-class is a group $G$ and the set of idempotents is a left zero semigroup.
• Further developments on hint 2. Is there an isomorphism $S\leftrightarrow H\times E(S)$ now? Apr 27, 2017 at 15:49