If $S$ is a right group, then every $\mathcal L$-class is a $\mathcal H$-class and also a subgroup.

Question

A semigroup $S$ is called right simple if it contains no proper right ideal. A semigroup that is right simple and left cancellative is called a right group. This is equivalent to saying that, for any elements $a$ and $b$ of $S$, there exists one and only one element $x$ of $S$ such that $ax = b$.

For a semigroup the following relations (Green's relations) are defined: $a \mathcal L b :\Leftrightarrow S^1 a = S^1 b$, $a \mathcal R b :\Leftrightarrow a S^1 = b S^1$ and $a \mathcal H b :\Leftrightarrow a \mathcal L b \land a \mathcal R b$.

Now I want to show, that each $\mathcal L$-class of a right group $S$ is also an $\mathcal H$-class, and is a subgroup.

The following observation is useful. $$ a \mathcal R b \mbox{ iff } \exists s,t \in S^1 : a = bs, b = at. $$ because let $a \mathcal R b$, then $aS^1 = bS^1$ and so choosing $a\cdot 1$ there must exist some $s \in S^1$ with $a = a\cdot 1 = b \cdot s$, and analogously $b = at$. Now let $a = bs, b = at$ for some $s,t \in S^1$ and let $s' \in S^1$, then $a\cdot s' = (bs)s' = b(ss')$ and so $a \cdot s'$ which implies $aS^1 \subseteq bS^1$, likewise $bS^1 \subseteq aS^1$. A dual statement holds for the $\mathcal L$-relation.

Let $L_a$ be an $\mathcal L$-class (an equivalence class with representative $a$), I have to show that every two elements form $L_a$ are also $\mathcal R$-related, but this is equivalent to showing that for $u,v \in L_a$ there exists $s,t$ with $u = vs$ and $v = ut$. Using that in a right group $ax = b$ has a unique solution I got unique $s,t$. By the way, with the same reasoning it follows that for all elements $a,b \in S$ it follows that $a \mathcal R b$, so that it consists of just one $\mathcal R$-class, namely $S$ itself.

To show that $L_a$ is a subgroup it is enough to show that there exists a right identity in $L_a$ and that every element of $L_a$ has a right-inverse in $L_a$. For the first claim pick any element, for example $a \in L_a$ and then let $e$ be the unique solution of the equation $ae = a$. Let $b \in L_a$, then $b = sa$ because they are $\mathcal L$-related, so $be = (sa)e = s(ae) = sa = b$ and $e$ is a right-identity for each $b \in L_a$. Now using again the unique solvability of $ax = b$ in a right group in $L_a$ we got that for every $b \in L_a$ there exists a unique $b'$ with $bb' = e$.

But I still have to show that i) $e \in L_a$ and ii) $b' \in L_a$, but have no idea how to show this, any ideas?

Answer 1

Hint: I would show that:

• there is only one $\mathscr{R}$-class in $S$ (which implies that every $\mathscr{L}$-class is an $\mathscr{H}$-class);
• use the observation that if $H$ is an $\mathscr{H}$-class of a semigroup, then either $H^2\cap H=\varnothing$ or $H$ is a subgroup of $S$.

Alternatively, you could do Exercise 2.6.6 in Howie's Fundamental's of Semigroup Theory, where you will prove that a right group is isomorphic to the direct product of a group ($\mathscr{H}$-class of $S$) with a right zero semigroup $E$ (the idempotents of $S$).