From J. M. Howie's Fundamentals of Semigroup Theory, Exercise 9, page 139: A semigroup $S$ is called orthodox if it is regular and the idempotents form a subsemigroup. How would I show that a completely regular semigroup $S$ is orthodox if and only if $\left(\forall x,y \in S\right) \space xy=xyy^{-1}x^{-1}xy$?
Things I know:
A semigroup $S$ is completely simple and orthodox $\iff$ it is a rectangular group (that is, $S \cong G \times B$, where $G$ is a group and $B$ is a rectangular band.
$S$ is a rectangular group $\iff$ it is completely regular and satisfies $x^{-1}yy^{-1}x=x^{-1}x.$
A Clifford semigroup is defined as a completely regular semigroup $\left(S, \mu, \space^{-1}\right)$ in which, $\forall\space x,y \in S,$ $$\left(xx^{-1}\right)\left(yy^{-1}\right)=\left(yy^{-1}\right)\left(xx^{-1}\right).$$
Any help you can give would be much appreciated.
Hints. (1) First prove that if $e$ is idempotent, then $e^{-1} = e$.
(2) Suppose that $S$ is completely regular and satisfies the identity $$ (*) \qquad xy=xyy^{-1}x^{-1}xy. $$To prove that the idempotents form a semigroup, apply $(*)$ with $x$ and $y$ idempotent.
(3) Suppose now that $S$ is orthodox. To prove $(*)$, observe that $xyy^{-1}x^{-1}xy = (xx^{-1}x)yy^{-1}x^{-1}x(yy^{-1}y) = x(x^{-1}xyy^{-1})^2y$