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Let $\mathcal{M}^0 = \mathcal{M}^0 ( G; I , \Lambda ;P)$ be a Rees matrix semigroup ($G$ a group, $I$, $\Lambda$ non-empty sets, $P=(p_{\lambda i})$ a $\Lambda \times I$ matrix over $G \cup \{0\}$ such that every row/column of $P$ contains at least one non-zero entry), where $P$ does not contain any $0$'s.

Show that the idempotents of $\mathcal{M}^0$ form a subsemigroup of $S \iff \forall \; i,j\in I$ and $\lambda , \mu \in \Lambda$ we have $p_{\mu i} p_{\lambda i}^{-1} p_{\lambda j} p_{\mu j}^{-1} = 1_G$.

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First write the general form of an idempotent of a Rees matrix semigroup – Boris Novikov Feb 28 '13 at 17:12

I suppose that $S = \mathcal{M}^0$ in your question. Further, if you don't mind, I will use the notation $J$ instead of $\Lambda$ and use only latin letters for the indexes, just to get a slightly more homogeneous notation.

Thus you have a Rees matrix semigroup $\mathcal{M}(G, I, J, P)$ (the zero is no longer needed in your case). Let $s = (i, g, j)$ be an element of $S$. Then $s^2 = (i, gp_{j,i}g, j)$ and hence $s$ is idempotent if and only if $g = p_{j,i}^{-1}$. Let now $e = (i, p_{j,i}^{-1}, j)$ and $f = (\ell, p_{m,\ell}^{-1}, m)$ be two idempotents. Then $ef = (i, p_{j,i}^{-1}p_{j,\ell} p_{m,\ell}^{-1}, m)$ is idempotent if and only if $p_{j,i}^{-1}p_{j,\ell} p_{m,\ell}^{-1} = p_{m,i}^{-1}$ or equivalently $p_{m,i}p_{j,i}^{-1}p_{j,\ell} p_{m,\ell}^{-1} = 1$, which is exactly your condition.

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IMHO your deleted answer with an example of a monoid having a lot of elements with one-sided inverses has salvageable material. Please consider editing and undeleting it. AFAICT the "pure" powers of a single generator have one-sided inverses. – Jyrki Lahtonen Aug 9 '13 at 17:16

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