Question about semigroups of permutations

Question

Consider a semigroup $A\subseteq Sym(X)$. For $a,b\in A$, say that $a\leq b$ iff $b=ac$ for some $c\in A$. Suppose that $\leq$ is a linear order on $A$. Does it follow that $A$ is abelian?

Answer 1

I think the answer is negative. Let $G$ be a non commutative totally bi-ordered group, for instance the free group with two generators. Let $P = \{x \in G \mid x \geqslant 1 \}$ be the set of positive elements of $G$. Then $x \leqslant y$ if and only if $x^{-1}y \in P$. In particular, there exists $z \in P$ such that $y = xz$. Conversely, if there exists $z \in P$ such that $y = xz$, then $z = x^{-1}y$ and hence $x^{-1}y \in P$. It follows that $P$ is a non commutative totally bi-ordered monoid in which $x \leqslant y$ if and only if there exists $z \in P$ such that $y = xz$.

Finally, by Cayley's theorem, $G$ is isomorphic to a subgroup of a permutation group and thus $P$ can imbedded into this permutation group.