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:
- for any $\lambda \in \Lambda, \mu \in M$ we have $G_{\lambda \mu} \leqslant_{\mathrm{Sg}}S$
- 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 $G_{\lambda \mu} \subseteq \Gamma_{\bot}<e_{\lambda \mu}>$; 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. $\Box$
Let me first recall a few definitions. A band is an idempotent semigroup
and a semilattice is an idempotent and commutative semigroup. A semilattice congruence on a semigroup $S$ is a congruence $\equiv$ on $S$ such that the quotient semigroup $S/{\equiv}$ is a semilattice. The minimum semilattice congruence on $S$ is the intersection of all semilattice congruences on $S$. Alternatively, it is the congruence generated by the relations $s \sim s^2$ and $st \sim ts$ for all $s, t \in S$. Finally,
two elements of a semigroup are
$\mathcal{J}$-related if they generate the same ideal.
The following results hold:
Proposition 1. Let $S$ be a band, let $T$ be a semilattice and let $h:S \to T$ be a semigroup morphism. If $a \mathrel{\mathcal{J}} b$, then $h(a) = h(b)$.
Proof. Since a semigroup morphism preserves $\mathcal{J}$-classes, $h(a) \mathrel{\mathcal{J}} h(b)$. But in a semilattice, the $\mathcal{J}$-relation is the equality. Thus $h(a) = h(b)$.
Proposition 2. Let $S$ be a band. Then the $\mathcal{J}$-relation
is a congruence on $S$ and the quotient $S/{\mathcal{J}}$ is a semilattice.
Proof. Let $a, b \in S$. Then $ab = (ab)^2 = a(ba)b$ and $ba = (ba)^2 = b(ab)b$. Thus $ab \mathrel{\mathcal{J}} ba$. Suppose now that $a \mathrel{\mathcal{J}} b$ and let $c \in S$. Then $b = xay$ and $a = ubv$ for some $x,y,u,v \in S^1$. Therefore
$$
ca = c(ubv) = (cu)bv \leqslant_\mathcal{J} (cu)b \mathrel{\mathcal{J}} b(cu) \leqslant_\mathcal{J} bc \mathrel{\mathcal{J}} cb
$$
and similarly, $cb \leqslant_\mathcal{J} ca$, whence $cb \mathrel{\mathcal{J}} ca$. It follows that $\mathcal{J}$ is a left congruence and a dual argument would show that it is a right congruence. Finally, $S/{\mathcal{J}}$ is idempotent and since $ab \mathrel{\mathcal{J}} ba$, it is also commutative.
Corollary 1. In a band, the minimum semilattice congruence is equal to $\mathcal{J}$.
Proof. Let $S$ be a band, let $\sim$ be its minimum semilattice congruence and let $p:S \to S/{\sim}$ be the quotient morphism. Proposition 1 shows that if $a \mathrel{\mathcal{J}} b$, then $p(a) = p(b)$, whence $a \sim b$. Moreover, Proposition 2 shows that $\mathcal{J}$ is a semilattice congruence and hence contains $\sim$. Thus $a \sim b$ implies $a \mathrel{\mathcal{J}} b$, which concludes the proof.
Corollary 2. In $\mathbb{FB}_n$, the $\mathcal{J}$-relation is the minimum semilattice congruence and the quotient $\mathbb{FB}_n/{\mathcal{J}}$ is the free semilattice $\mathbb{FSl}_n$.
Back to your question. The free semilattice $\mathbb{FSl}_n$ on $X$ can be identified to the semigroup $(\mathcal{P}(X), \cup)$, where $\mathcal{P}(X)$ is the set of nonempty subsets of $X$. Consider the following quotient morphisms
$$
X^+ \xrightarrow{q} \mathbb{FB}_n \xrightarrow{p} \mathbb{FSl}_n = (\mathcal{P}(X), \cup) \quad \text{and let} \quad c = p \circ q: X^+ \to (\mathcal{P}(X), \cup)
$$
Then for each word $w \in X^+$, $c(w)$ is the content of $w$, that is, the set of letters occurring in $w$. Since $c(uv) = c(u) \cup c(v)$, the content map is indeed a semigroup morphism from $X^+$ to $(\mathcal{P}(X), \cup)$.
Furthermore, $q(u) \mathrel{\mathcal{J}} q(v)$ if and only if $c(u) = c(v)$.
Finally, for each subset $Y$ of $X$, $p^{-1}(Y)$ is the $\mathcal{J}$-class
consisting of the elements of the form $q(u)$, where $c(u) = Y$.
It is a rectangular band, and one of the $S_\alpha$ of your question.
Example. If $X= \{a,b\}$, there are three $\mathcal{J}$-classes in the free band on $X$: $\{a\}$, $\{b\}$ and $\{ab, aba, bab, ba\}$ (the elements of content $\{a,b\}$. You can verify by hand that each of them is a rectangular band.
Best Answer
Consider $S=(\{1,2,3,\cdots\},\mathrm{max})$. Every positive integer $e$ is idempotent and is an identity with respect to the upper set $\{e,e+1,e+2,\cdots\}$. We can say $eSe=S$ for $e=1$.
Then $S^1\cong(\{0,1\cdots,\},\mathrm{max})$ with $0$ the external identity, and $S^1\cong S$ via $n\mapsto n+1$.