Does Green's $\mathcal{J}$-relation define a total order on the equivalence classes?

Question

In a semigroup $S$ we define $a\le_\mathcal{J} b$ iff $a=xby$ for some $x,y\in S^1$. Defining $a\equiv_\mathcal{J} b$ by $a\le_\mathcal{J} b$ and $b\le_\mathcal{J} a$ gives a partial order on $S/\equiv_\mathcal{J}$. Is this order always total?

Answer 1

No it is not always a total order.

For example, in the semigroup $S$ with elements $a$, $b$, and $c$ and multiplication in the table below:

\begin{equation*} \begin{array}{c|ccc} &a&b&c\\\hline a&a&a&a\\ b&a&b&a\\ c&a&a&c \end{array} \end{equation*}

The $\mathscr{J}$-order on $S$ has $b$ and $c$ incomparable, $a\leq_{\mathscr{J}} b$ and $a\leq_{\mathscr{J}} c$.

More generally, every partially ordered set with a minimum element is order-isomorphic to the partial order of the $\mathscr{J}$-classes of some semigroup. A reference for this is Theorem 4(i) in the paper

C. J. Ash and T. E. Hall, Inverse semigroups on graphs, Semigroup Forum 11 (1975) 140–145.

Answer 2

It might be good to look at this order as the inclusion of principal two-sided ideals instead (as it's isomorphic). Perhaps the fact that you didn't look at it like this is why you didn't immediately answer the question yourself. By your nickname I assume you like rings. In rings, principal ideals in the ring-theoretic sense and principal ideals in the semigroup-theoretic sense are the exact same thing. In the commutative case, two-sided and one-sided principal ideals are the same thing too. Can you see examples of commutative rings you know in which the principal ideals don't form a chain?