Rees matrix semigroup

Remarks

The product is associative: both ways of bracketing $(i, g, \lambda)(j, h, \mu)(k, f, \nu)$ give $(i, \, g p_{\lambda j} h p_{\mu k} f, \, \nu)$. So $\mathcal{M}(G; I, \Lambda; P)$ is a semigroup. It is simple: given $x = (i, g, \lambda)$ and $y = (j, h, \mu)$, one can solve $(j, a, \lambda) \, x \, (i, b, \mu) = y$ for $a, b \in G$. Rees's theorem: a semigroup is completely simple if and only if it is isomorphic to some $\mathcal{M}(G; I, \Lambda; P)$.

Example: if $I = \Lambda = \set{1}$ and $P = (e)$, then $(1, g, 1) \mapsto g$ is an isomorphism onto $G$. Example: if $G$ is trivial, $I = \set{1, 2}$ and $\Lambda = \set{1}$, the product is $(i, \lambda)(j, \mu) = (i, \mu) = (i, 1)$, so $xy = x$ for all $x, y$: a left zero semigroup.

Non-example: the monoid $(\N, +)$ is not isomorphic to any Rees matrix semigroup over a group. It has the proper ideal $\set{1, 2, 3, \dots}$, while a Rees matrix semigroup has no proper two-sided ideal.