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I am struggling with the rightward implication. Here is some of my working:

$(\Leftarrow)$

Every subgroup is contained within a maximal subgroup that is the $\mathcal{H}$ class of that subgroup's idempotent. As each $\mathcal{H}$ class is trivial by assumption then every subgroup is contained within a trivial maximal subgroup. Result.

$(\Rightarrow)$

Idea 1:

Something that involves:

Let every subgroup be trivial, take some element $s$ in the semigroup and suppose it has index $i$ and period $p>1$. Then we can form a subgroup $\{a^i,a^{i+1},\dots,a^{i+p-1}\}$ but this is a contradiction and hence every element of the semigroup has period $1$.

$\vdots$

Idea 2:

Since we're dealing with a finite semigroup we know it satisfies every (left/right) (ascending/descending) chain condition and hence $\mathcal{D}=\mathcal{J}$.

Maybe we could show now that every element is in the same $\mathcal{D}=\mathcal{J}$ class? Then by a corollary of Green's lemma we can use the fact that, since some subgroup $G$ we have $|G|=1$, every $\mathcal{H}$ class has size $1$?

$\vdots$

Idea 3:

Maybe I could fuse the above two approaches, somehow exploit the period $1$ property to show that every $\mathcal{J}=\mathcal{H}$ class has an idempotent in it. This would show that in every $\mathcal{D}$ class there is a trivial $\mathcal{H}$ class, and therefore by a corollary to Green's lemma we would have that every $\mathcal{H}$ class in every $\mathcal{D}$ class is trivial.

$\vdots$

I have also just tried to show that if $a\mathcal{H}b$ then $a=b$ but struggled with that.

Any hints are welcome.

Levent Michael Dasar
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1 Answers1

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Let $S$ be a finite semigroup. The following conditions are equivalent:

  1. there is an integer $n > 0$ such that, for all $x \in S$, $x^n = x^{n+1}$,

  2. $S$ is $\cal H$-trivial,

  3. the groups in $S$ are trivial.

Proof.

(1) implies (2). Suppose $a \mathrel{\cal H} b$. Then there exist $u, v, x, y \in S^1$ such that $ua = b$, $vb = a$, $ax = b$ and $by = a$, whence $uay = a$ and therefore $u^nay^n = a$. Since $u^n = u^{n+1}$, one has $a = u^nay^n = u^{n+1}ay^n = u(u^nay^n) = ua = b$. Therefore $S$ is $\cal H$-trivial.

(2) implies (3) follows from the fact that each group in $S$ is contained in some $\cal H$-class.

(3) implies (1). Since $S$ is finite, there exists an integer $n$ such that, for all $x \in S$, $x^n$ is idempotent. Let $x \in S$. Then the $\cal H$-class of $x^n$ is a group $G$, which is trivial by (3). Since $x^n$ and $x^{n+1}$ belong to $G$, one gets $x^n = x^{n+1}$.

Note. For algorithms checking this condition, see this question.

J.-E. Pin
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