I am trying to tackle the following semigroup question. I can't see why my answer is wrong but I haven't used the fact the semigroup is COMPLETELY simple anywhere so I think there must be an error somewhere in my proof. Any advice?
Question
Consider a completely simple semigroup S, represented as a Rees matrix semigroup $M[G; I, \Lambda; P]$ over a group $G$. Prove that
$(i, g, l) \mathrel{\mathcal{L}} (j, h, m) \iff l = m$
$(i, g, l) \mathrel{\mathcal{R}} (j, h, m) \iff i = j$
$(i, g, l) \mathrel{\mathcal{H}} (j, h, m) \iff i = j \text{ and } l = m$
$\mathcal{D} = \mathcal{J} = S \times S$.
Attempted Answer
Let $S$ be completely simple, recall this means $S$ has no proper two-sided ideals but has minimal left and right ideals.
(i,g,l)$\mathcal{L}$(j,h,m)$\iff\exists s=(a.b.c),t=(d,e,f) \in S$ such that $(a.b.c)(i,g, l)=(j,h,m)$ and $(d,e,f)(j,h,m)=(i,g, l)$
$\iff (a, bP_{ci}g, l)=(j,h,u)$ and $(d,eP_{fj}h,m)=(i, g, l)$
$\iff a=j, d=i, l=m, bP_{ci}g=h, eP_{fj}h=g$
Now $a,b,c,d,e,f $ are arbitrary we can select these values to satisfy the above equations thus the only necessary and sufficient condition is that $l= m$.
A similar proof then works for the $\mathcal{R}$ relation, and we use the fact that $\mathcal{H}$ related $\iff$ both $\mathcal{L}$ and $\mathcal{R}$ are related to do the third part.
Next onto $D$, we recall that $D= \mathcal{L} \circ R=R \circ \mathcal{L}$ and thus $(\alpha,\beta) \in J \iff (\alpha,\beta) \in \mathcal{L} \circ R \iff \exists u \in S\ $ such that $\alpha \mathrel{\mathcal{L}} u \mathrel{\mathcal{R}} \beta\ $.
Now letting $\alpha=(a,b,c) , u=(d,e,f), \beta=(g,h,i)$ we observe that we need $(a,b,c) \mathrel{\mathcal{L}} (d,e,f) \mathrel{\mathcal{R}} (g,h,i) \iff c=f, d=g\ $ but $u$ is an arbitrary element so we can always make this happen thus all elements are $\mathcal{D}$-related.
Lastly onto $\mathcal{J}$ relation. $(a,b) \in \mathcal{J} \iff$ they generate the same two-sided ideal. However $S$ is simple so the only two-sided ideal is $S$ itself as there are no proper two-sided ideals, thus all elements generate the same two-sided ideal so all elements are $\mathcal{J}$ related.
Your proof seems to be correct, so the problem has another nature: by Rees’ characterization theorem, a semigroup is completely simple if and only if it is isomorphic to a Rees matrix semigroup over a group.