rectangular groups are completely simple and orthodox Let $S$ be a rectangular group. i.e $S$ is isomorphic to the direct product of a group and a rectangular band. 


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*Show that a semigroup is completely simple and orthodox if and only if it is a rectangular group.

*Show that $S$ is a rectangular group if and only if it is completely regular and satisfies the law
$$x^{-1}yy^{-1}x=x^{-1}x$$
I have now shown all of this besides the first implication in 1., that a completely simple and orthodox semigroup is a rectangular group. I believe I have to use the Rees Theorem, but I am not quite sure where I am supposed to use orthodox in my argument?
 A: Here is an answer for part 1:
Suppose $S$ is a rectangular group. Then $S=G\times R$ where $G$ is a group and $R=I\times \Lambda$ is a rectangular band. If $(g,r)\in S$ is an idempotent, then $(g,r)^2=(g,r)$ and so $g^2=g$ and $g$ is the identity of $G$. Hence the idempotents of $S$ are of the form $\{e\}\times R$, and this is clearly a subsemigroup. Hence $S$ is orthodox. 
The semigroup $S$ is also simple. If $(e,r), (e,s)\in S$ are idempotents and $(e, r)(e,s)=(e,s)(e,r)=(e,s)$, then $rs=sr=s$. If $r=(i,\lambda)$ and $s=(j,\mu)$, then $rs= (i,\mu)$ and $sr=(j,\lambda)$ and so $r=s$.  It follows that $(e,s)$ is primitive, and so $S$ is completely simple. 
For the converse, suppose that $S$ is completely simple and orthodox. By the Rees Theorem (Theorem 3.3.1 in Howie), every completely simple semigroup is isomorphic to a Rees matrix semigroup $\mathcal{M}[G; I, \Lambda; P]$ where $G$ is a group, $I$ and $\Lambda$ are some sets, and $P$ is a $|\Lambda|\times |I|$ matrix with entries in $G$. Since the set of elements of $S=\mathcal{M}[G; I, \Lambda; P]$ is $I\times G\times \Lambda$ it suffices to prove that the multiplication in $S$ is the same as the multiplication in direct product of the group $G$ with the rectangular band $I\times \Lambda$. For this it suffices to show that every entry of the matrix $P$ is the identity of $G$. By Theorem 3.4.2 in Howie, we can assume without loss of generality that the first row and first column of $P$ can be chosen to consist solely of the identity $1_G$ of the group $G$. 
In other words, we can assume that $1\in I\cap \Lambda$, and $p_{i, 1}=p_{1, \lambda}=1_G$ for all $i\in I$ and $\lambda\in\Lambda$.
Let $i\in I$ and $\lambda\in \Lambda$ be arbitrary. The element $(i,p_{\lambda, i}^{-1}, \lambda)$ is the only idempotent with first position $i$ and last position $\lambda$. Since $S$ is orthodox and $(i, p_{1,i}^{-1}, 1), (1, p_{1, \lambda}^{-1}, \lambda)$ are idempotents, it follows that 
$$(i, p_{1,i}^{-1}, 1), (1, p_{1, \lambda}^{-1}, \lambda)=(i, p_{1,i}^{-1}p_{1,1}p_{1,\lambda}^{-1}, \lambda)$$
is an idempotent too. But the only idempotent starting with $i$ and ending with $\lambda$ is 
$(i,p_{\lambda, i}^{-1}, \lambda)$ and so $p_{\lambda, i}^{-1}=1_G$, as required.
A: for part 2.
Suppose $S$ is a rectangular group so is isomorphic to $G$ x $B$ for a group $G$ and a rectangular band $B$. To show a semigroup is completely regular we have to show there is a unary operation from $a$ to $a^{-1}$ s.t;
i.) $(a^{-1})^{-1}=a$ 
ii.) $aa^{-1}a=a$   
iii.) $aa^{-1}=a^{-1}a$
In $S$ define this inverse as $(g^{-1},a)$ for the element $(g,a)$. By simple calculation using multiplication in  a direct product we can see all three conditions stated above are satisfied by this operation. So we can conclude that $S$ is completely regular.
It is also a very simple matter to show the equation $x^{-1}yy^{-1}x=x^{-1}x$ is satisfied by letting $x=(g,a), x^{-1}=(g^{-1},a), y=(h,b), y^{-1}=(h^{-1},b)$, remembering that $B$ is a rectangular band. 
Conversely, suppose $S$ is completely regular and satisfies $x^{-1}yy^{-1}x=x^{-1}x$. We will show $S$ is completely simple and orthodox and hence $S$ is a rectangular group by part 1.
To show $S$ is completely simple we first show it is simple, so we show $\mathcal{J}=S$ x $S$. Let $a,b \in{S}$. Need to find $x,y,u,v \in{S^1}$ such that $xay=b, ubv=a$. Let $x=bb^{-1}, y=a^{-1}b, u=aa^{-1}, v=b^{-1}a$. Subsitituing this into the above and using $x^{-1}yy^{-1}x=x^{-1}x$ to simplify we see $a$ $\mathcal{J}$ $b$ for all $a,b \in{S}$, so $\mathcal{J}=S$ x $S$ and hence $S$ is simple. Using Proposition 4.1.2 from Howie we can conclude $S$ is completely simple.
Now to show $S$ is orthodox we must show it is regular and the idempotents form a subsemigroup. Clearly $S$ is regular as its completely regular. Suppose $e,f$ are idempotents of $S$. As $S$ is completely regular we know i) ii) and iii) are all satisfied. so consider;
\begin{equation*}
e=ee^{-1}e=e^{-1}ee=e^{-1}e=(e^{-1}e)(e^{-1}e)=(e^{-1}e)(ee^{-1})=e^{-1}ee^{-1}=e^{-1}
\end{equation*}
Similarly $f=f^{-1}$. Consider $x^{-1}yy^{-1}x=x^{-1}x$. This holds for all $x,y \in{S}$, so in particular it holds for $e,f$, so;
\begin{equation*}
e^{-1}ff^{-1}e=e^{-1}e \: \Rightarrow \: effe=ee \: \Rightarrow \: efe=e
\end{equation*}
Hence $efef=ef$ so $ef$ is an idempotent so the idempotents form a subsemigroups, so $S$ is orthodox. Hence by applying 1. we get the deisred result.
