# Subsemigroup of a finite simple semigroup is simple

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.

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:

Theorem: Let $$S$$ be a completely simple semigroup and $$T \subseteq S$$ a non-empty torsion subsemigroup of $$S$$. Then $$T$$ is also completely simple.

Before we proceed with the proof, let us prepare a few preliminary notions and notations:

• Given a map $$f: A \to B$$ we denote by $$\mathrm{Eq}(f)=(f \times f)^{-1}(\Delta_B)$$ the equivalence relation canonically associated to $$f$$.
• We will use $$\mathrm{Sg}, \mathrm{Mon}, \mathrm{Gr}$$ to denote the categories of semigroups, monoids and groups.
• Given a semigroup $$S$$, the notation $$T \leqslant_{\mathrm{Sg}} S$$ will express the fact that $$T$$ is a subsemigroup of $$S$$, and analogously $$G \leqslant_{\mathrm{Gr}} S$$ will mean that $$G$$ is a subgroup of $$S$$. In a similar vein, for arbitrary subset $$X \subseteq S$$ we write $$[X]_{\mathrm{Sg}}$$ to denote the subsemigroup generated by $$X$$ in $$S$$, resepctively $$(X)_{\mathrm{s}}, (X)_{\mathrm{d}}, (X)_{\mathrm{b}}$$ for the left, right and bilateral ideals generated by $$X$$ in $$S$$.
• On given semigroup $$S$$ we will consider the well-known Green's preorders and associated equivalence relations, denoted by $$\leqslant_{\mathrm{s}}$$ and $$\leqslant_{\mathrm{d}}$$ for the left and right preorders, respectively by $$\Gamma_{\mathrm{s}},\ \Gamma_{\mathrm{d}},\ \Gamma_{\bot}=\Gamma_{\mathrm{s}} \cap \Gamma_{\mathrm{d}},\ \Gamma_{\top}=\Gamma_{\mathrm{s}} \circ \Gamma_{\mathrm{d}}=\Gamma_{\mathrm{d}} \circ \Gamma_{\mathrm{s}},\ \Gamma_{\mathrm{b}}$$ for the left, right, lower, upper and bilateral Green equivalences.
• Given arbitrary set $$A$$, we will write $$_{\mathrm{s}}A$$ respectively $$A_{\mathrm{d}}$$ for the left- respectively right-zero semigroups on $$A$$.
• Instead of the conventional notation, we will denote the Rees matrix semigroup on left index set $$\Lambda$$, group $$G$$, right index set $$M$$ and with matrix $$a \in G^{M \times \Lambda}$$ by $$(\Lambda \times G \times M)_{a}$$. Its support set is simply the cartesian product $$\Lambda \times G \times M$$ and the multiplication is given by $$(\lambda, x, \mu)(\lambda', y, \mu')=(\lambda, xa_{\mu \lambda'}y, \mu')$$.

In order to efficiently produce a proof let us arrange a couple of auxiliary results beforehand:

Lemma 1. Let $$G$$ be an arbitrary group and $$S \leqslant_{\mathrm{Sg}} G$$ be a nonempty torsion subsemigroup. Then one actually has $$S \leqslant_{\mathrm{Gr}} G$$.

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 $$S \leqslant_{\mathrm{Sg}}G$$. $$\Box$$

Lemma 2. Let $$A, B$$ be arbitrary sets and $$P \leqslant_{\mathrm{Sg}}\ _{\mathrm{s}}A \times B_{\mathrm{d}}$$. Then there exist $$M \subseteq A, N \subseteq B$$ such that $$P=\ _{\mathrm{s}}M \times N_{\mathrm{d}}$$ (succinctly, any subsemigroup of a rectangular band is a rectangular band).

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. $$\Box$$

Proposition: Let $$S$$ be a nonempty semigroup, $$\Lambda$$, $$M$$ sets and $$G \in \mathscr{P}(S)^{\Lambda \times M}$$ a partition of $$S$$ such that:

1. for any $$\lambda \in \Lambda, \mu \in M$$ we have $$G_{\lambda \mu} \leqslant_{\mathrm{Sg}}S$$
1. for any $$\lambda, \lambda' \in \Lambda, \mu, \mu' \in M$$ we have $$G_{\lambda \mu}G_{\lambda' \mu'} \subseteq G_{\lambda \mu'}$$.

Then $$S$$ is completely simple (in more plastic terms, as you put it, a ''rectangular band of groups'' is completely simple).

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 $$G_{\lambda \mu} \subseteq \Gamma_{\bot}$$; 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 $$p^{-1}(\{\lambda\})=\bigcup_{\mu \in M} G_{\lambda \mu} \subseteq \Gamma_{\mathrm{d}}$$

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 $$\Gamma_{\mathrm{d}} \subseteq (e_{\lambda \beta})_{\mathrm{d}} \subseteq p^{-1}(\{\lambda\})$$. 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. $$\Box$$

I was stuck with this question for a while, funny, I've figured it just now.

We have that $$(T\cap H_{ij})(T\cap H_{i'j'})(T\cap H_{ij'})(T\cap H_{ij})(T\cap H_{i'j'}) \subseteq (T\cap H_{ij})(T\cap H_{i'j'})$$ because for example $$(T\cap H_{ij})(T\cap H_{i'j'})(T\cap H_{ij'})(T\cap H_{ij})\subseteq (T\cap H_{ij}).$$ This shows that $$(T\cap H_{ij})(T\cap H_{i'j'})$$ is a bi-ideal of $$T\cap H_{ij'}$$, but since it's a group, it contains no proper bi-ideal, so $$(T\cap H_{ij})(T\cap H_{i'j'}) = T\cap H_{ij'}$$. Hence we can use the theorem about rectangular band of groups being completely simple.