I want to verify that the analogy of the Rees matrix theorem with completely simple semigroups holds. In particular, I want to prove: A completely simple semigroup $S$ is isomorphic to some semigroup $(T,\cdot)$ where $T=I\times G\times \Lambda$ with $P=(p_{\lambda_i})$ an $\Lambda\times I$ matrix with entries from $G$. Here, $G$ is a group.
We know by the original version of the Rees matrix theorem that there exists an isomorphism between $S^0=S\cup\{0\}$ and $T\cup\{0\}$, since $S^0$ will be completely $0$-simple. The operation $\cdot$ defined on $T\cup\{0\}$ is given by $(i, a, \lambda)(i', a', \lambda')=(i,ap_{\lambda_{i'}}a')$, if $p_{\lambda_{i'}}\neq 0$, and $=0$ if $p_{\lambda_{i'}}=0$.
Call one such isomorphism $\phi$, and by the proof of the Rees matrix theorem, we know $\phi$ looks like:
$\phi:T\cup\{0\}\longrightarrow S^0$, $\phi(i, a, \lambda)=r_iaq_{\lambda}$, for $(i, a, \lambda)\in T\cup\{0\}$, where $r_i$ and $q_{\lambda}$ are elements of the $\mathcal{R}$-classes and $\mathcal{L}$-classes of $S$, respectively.
What I want to know is, can we safely remove the adjoined zero from $S^0$ and $T\cup\{0\}$ in this isomorphism? The isomorphism itself doesn't look like it will change, even though the domain and codomain of the map itself changes.
Any hints or suggestions are much appreciated.