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..


1 Answer 1


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.

  • $\begingroup$ I think I have managed the Hint. But I can't see it, yet :/ $\endgroup$
    – AlvinL
    Apr 26, 2017 at 9:43
  • $\begingroup$ Further developments on hint 2. Is there an isomorphism $S\leftrightarrow H\times E(S)$ now? $\endgroup$
    – AlvinL
    Apr 27, 2017 at 15:49
  • $\begingroup$ The answer is Yes. $\endgroup$
    – J.-E. Pin
    Apr 28, 2017 at 8:38
  • $\begingroup$ I have problems defining the isomorphism. The thing I suggested, doesn't work. It's a bijection, but not necessarely a homomorphism. $\endgroup$
    – AlvinL
    Apr 29, 2017 at 14:48

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