# Free band on $n$ generators is a semilattice of rectangular bands.

Exercise (part of it) comes from Clifford, Preston, Algebraic theory of Semigroups.

By $$\mathbb{FB}_n$$ we denote the free band on $$n$$ generators, that is, a free semigroup generated by a set $$X=\{x_1,...,x_n\}$$ subject to relations $$w^2 = w$$ where $$w$$ is an arbitrary word in the free semigroup on $$X$$.

Denote by $$Y$$ the semilattice of subsets of $$X$$ under union of sets (which is isomorphic to the free commutative band on $$X$$).

Part b) of the exercise tells me that $$\mathbb{FB}_n$$ is a semilattice $$Y$$ of rectangular bands $$S_\alpha$$, $$\alpha\in Y$$, where $$S_\alpha$$ consists of all words $$w$$ such that the set of generators appearing in $$w$$ is $$\alpha$$.

From my understanding, for example $$\alpha = \{x_1, x_2\}$$, then $$S_\alpha$$ is the free band generated by $$\alpha$$. But this is not a rectangular band, since $$x_1x_2x_1 \neq x_1$$.

My question is, what exactly do they mean by $$S_\alpha$$?

• I figured they must mean that words of $S_\alpha$ must contain all elements of $\alpha$, which really does seem like a rectangular band. Commented Sep 27, 2019 at 12:20

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.

• Thank you, it makes sense. I had a similar theorem in my book, that if a semigroup is a sum of groups, it's also a semilattice of completely simple semigroups, each being a $J$ class of $S$ (semilattice consisting of principal ideals). Now I can clearly see that $J$ is always a minimum semilattice congruence in this case as well (from proposition 1). Commented Sep 29, 2019 at 0:23
• This also made me realize, that the intended way was probably to see that each set $S_\alpha$ is contained in a $J$-class (that's where maximal homomorphic image is used) so that they are also completely simple (obvious from $H$-class structure as a rectangular band of groups, and that they are subsemigroups), so they must be rectangular bands. Commented Sep 29, 2019 at 0:41