It's a part of an exercise on completely simple semigroups from Clifford and Preston, Algebraic theory of Semigroups.
Here $\Box$ denotes the empty set.
Let $S$ be a simple finite semigroup, and $T\subseteq S$ it's subsemigroup. Show that $T$ is a simple semigroup.
In fact, if $H_{ij} = R_i\cap L_j$, $i\in I$ and $j\in J$ where $R_i$, $i\in I$ are all minimal right ideals of $S$ and $L_j$, $j\in J$ are all minimal left ideals of $S$, then since $S$ is completely simple (because it's finite and simple), $S = \bigcup_{i\in I,\ j\in J} H_{ij}$, a rectangular band of groups, i. e. each $H_{ij}$ is a group and $H_{ij}H_{i'j'} = H_{ij'}$.
Then there are $I'\subseteq I$ and $J'\subseteq J$ such that $H_{ij}' = T\cap H_{ij}\neq \Box$ if and only if $i\in I'$ and $j\in J'$ and $T$ is a sum of groups $H_{ij}'$ with $i\in I'$, $j\in J'$.
I could show easily that there exists such sets $I'$ and $J'$ by setting $I' = \{ i\in I: \exists_{j\in J} H_{ij}' \neq \Box \} $ and $J' = \{ j\in J: \exists_{i\in I} H_{ij}' \neq \Box \} $, and using the fact that $(T\cap H_{ij})(T\cap H_{i'j'})\subseteq T\cap (H_{ij}H_{i'j'})$.
Showing that $H_{ij}'$ are groups for $i\in I'$ and $j\in J'$ was also fairly easy, since it's a finite subsemigroup, for any $a\in H_{ij}'$ there exists $n$ so that $a^n$ is idempotent, but there's only one idempotent, namely the identity $e$, so $a^n = e$ for some $n$. So it's a group.
But something seems to escape me, it doesn't seem like enough to claim that $T$ is simple. Can I prove that $(T\cap H_{ij})(T\cap H_{i'j'})= T\cap (H_{ij}H_{i'j'})$ ? If yes, I could use an exercise which says that rectangular band of groups is completely simple.
Best Answer
This is a not so trivial result in the structure theory of semigroups, and it can actually be established in a slightly more general setting, as follows:
Before we proceed with the proof, let us prepare a few preliminary notions and notations:
In order to efficiently produce a proof let us arrange a couple of auxiliary results beforehand:
Proof: Fix a certain $a \in S$ as $S$ is nonempty. According to the general theory of monogenous semigroups, under our hypothesis of torsion the subsemigroup $[a]_{\mathrm{Sg}}=\{a^n\}_{n \in \mathbb{N}^*}$ contains an idempotent; however, the only idempotent a group contains is its unit, so we must have $1_G \in [a]_{\mathrm{Sg}} \subseteq S$ and therefore $S \leqslant_{\mathrm{Mon}} G$; consider now an arbitrary $x \in S$ and by the same reasoning conclude that $1_G \in [x]_{\mathrm{Sg}}$, in other words that there must exist $k \in \mathbb{N}^*$ such that $x^k=1_G$; this means that $x^{-1}=x^{k-1}$ so either $k \geqslant 2$ and thus $x^{k-1} \in [x]_{\mathrm{Sg}} \subseteq S$ or $k=1$ in which case $x=x^{-1}=1_G \in S$ as we have already seen; in either case, $x^{-1} \in S$, hence $S \leqslant_{\mathrm{Sg}}G$. $\Box$
Proof: Let us introduce the canonical projections $p: A \times B \to A, q: A \times B \to B$ which are of course semigroup morphisms; if we denote $p(P)=M, q(P)=N$ it is immediate that $P \subseteq M \times N$. In order to prove the reverse inclusion, consider an arbitrary $x \in M$ and $y \in N$; by definition, there must exist $u, v \in P$ such that $p(u)=x, q(v)=y$ so we infer that $p(uv)=p(u)p(v)=p(u)=x$ and $q(uv)=q(u)q(v)=q(v)=y$ which in other words means that $(x, y)=uv \in P$, establishing the desired in/conclusion. $\Box$
Proof: As $G$ is a partition of $S$, there exists a unique map $\rho: S \to \Lambda \times M$ with the property that $\rho(x)=\tau \Leftrightarrow x \in G_{\tau}$ for all $x \in S, \tau \in \Lambda \times M$; furthermore, we introduce the canonical projections $\pi: \Lambda \times M \to \Lambda, \pi': \Lambda \times M \to M$ as well the compositions $p=\pi \circ \rho, q=\pi' \circ \rho$; we thus have $$(p(x)=\lambda \land q(x)=\mu) \Leftrightarrow x \in G_{\lambda \mu}$$ for all $x \in S, \lambda \in \Lambda, \mu \in M$. We remark straight away that owing to property 2) we have that $p \in \mathrm{Hom}_{\mathrm{Sg}}(S,\ _{\mathrm{s}} \Lambda), q \in \mathrm{Hom}_{\mathrm{Sg}}(S, M_{\mathrm{d}})$.
Let us also abbreviate $1_{G_{\lambda \mu}}=e_{\lambda \mu}$ and notice that since $S \neq \varnothing$ we must have $\Lambda, M \neq \varnothing$ which allows us to fix some arbitrary $\alpha \in \Lambda, \beta \in M$.
As $G_{\lambda \mu}$ is a subgroup, it is immediate that $G_{\lambda \mu} \subseteq \Gamma_{\bot}<e_{\lambda \mu}>$; hence, by introducing $J_{\lambda \mu}=(G_{\lambda \mu})_{\mathrm{d}}$ we infer that $J_{\lambda \mu}=(x)_{\mathrm{d}}$ for any $x \in G_{\lambda \mu}$. By property 2) we have that $e_{\lambda \mu}e_{\lambda \mu'} \in G_{\lambda \mu'}$ and since obviously $e_{\lambda \mu}e_{\lambda \mu'} \leqslant_{\mathrm{d}} e_{\lambda \mu}$ we infer that $J_{\lambda \mu'} \subseteq J_{\lambda \mu}$; the reverse inclusion is established analogously, by interchanging $\mu$ and $\mu'$. Thus $J_{\lambda \mu}=J_{\lambda \mu'}$ for any $\mu, \mu' \in M$ and we conclude that for any $x \in G_{\lambda \mu}, y \in G_{\lambda \mu'}$ we have $x \Gamma_{\mathrm{d}}y$; hence we obtain that $$p^{-1}(\{\lambda\})=\bigcup_{\mu \in M} G_{\lambda \mu} \subseteq \Gamma_{\mathrm{d}}<e_{\lambda \beta}>$$
The reverse inclusion is immediate: since $\{\lambda\}$ is a right ideal of $_{\mathrm{s}}\Lambda$, thus $p^{-1}(\{\lambda\})$ is easily seen to be a right ideal of $S$ and $e_{\lambda \beta} \in p^{-1}(\{\lambda\})$, hence $\Gamma_{\mathrm{d}}<e_{\lambda \beta}> \subseteq (e_{\lambda \beta})_{\mathrm{d}} \subseteq p^{-1}(\{\lambda\})$. We have thus exhibited for every $\lambda \in \Lambda$ a right ideal which is at the same time a right-Green class, hence a minimal right ideal. In particular $p^{-1}(\{\alpha\})$ is a minimal right ideal. Furthermore let us remark that this result actually establishes the fact that $$\Gamma_{\mathrm{d}}=\mathrm{Eq}(p)$$
since the right-Green classes are none other than the fibres of $p$.
By reasoning dually we infer that $q^{-1}(\{\mu\})$ is a minimal left ideal for any $\mu \in M$, so in particular we do have at least one minimal left ideal $q^{-1}(\{\beta\})$ and that $$\Gamma_{\mathrm{s}}=\mathrm{Eq}(q)$$
It is straightforward that $\mathrm{Eq}(p) \circ \mathrm{Eq}(q)=S \times S$, since for any $x, y \in S$ by setting $p(x)=\lambda, q(y)=\mu$ we obtain $p(x)=p(e_{\lambda \mu}), q(e_{\lambda \mu})=q(y)$. Hence it follows that $$\Gamma_{\top}=\Gamma_{\mathrm{b}}=S \times S$$
and thus that $S$ is bilaterally simple (since $S$ itself is a bilateral-Green class). We can now finally draw the conclusion that $S$ is completely simple. $\Box$
Now we are at last ready to establish the main theorem:
Proof: By the Rees structure theorem characterizing completely simple semigroups we can assume that $S=(\Lambda \times G \times M)_{a}$ for a certain group $G$, sets $\Lambda, M$ and matrix $a \in G^{M \times \Lambda}$. By introducing $F_{\lambda \mu}=\{\lambda\} \times G \times \{\mu\}$ we have that $F_{\lambda \mu} \leqslant_{\mathrm{Gr}}S$. Let us define $H=\left(T \cap F_{\lambda \mu}\right)_{\substack{\lambda \in \Lambda\\ \mu \in M}}$ and notice that $$H_{\lambda \mu}H_{\lambda' \mu'} \subseteq H_{\lambda \mu'}$$ for any $\lambda, \lambda' \in \Lambda, \mu, \mu' \in M$. Therefore, by considering $$\Pi=\{\tau \in \Lambda \times M|\ H_{\tau} \neq \varnothing\}$$ the above relation of multiplication tells us that $\Pi \leqslant_{\mathrm{Sg}}\ _{\mathrm{s}} \Lambda \times M_{\mathrm{d}}$. Through an application of lemma 2 we infer that $\Pi=\Lambda' \times M'$ for some certain $\Lambda' \subseteq \Lambda, M' \subseteq M$.
It is immediate that $H_{|\Pi}=(H_{\lambda \mu})_{\lambda \in \Lambda' \\ \mu \in M'}$ is a partition of $T$ and that for $\lambda \in \Lambda', \mu \in M'$ we have $H_{\lambda \mu}$ as a nonempty torsion subsemigroup of the group $F_{\lambda \mu}$, hence itself a group by virtue of lemma 1. Thus, $T$ satisfies all the conditions of the proposition stated above and is therefore itself a completely simple semigroup. $\Box$