Lemma 47.5.3.label Let $A$ be a finite alphabet, $S$ be the free semigroup of non-empty words in $A$, $T = (A \sqcup \bracsn{*})^{*}$ be the free semigroup of all non-empty words in $A \sqcup \bracsn{*}$, and $V$ be the set of variable words in $A$, then:
- (1)
The inclusions $S \to T$ and $V \to T$ admit extensions into embeddings $\beta S \to T$ and $\beta V \to T$, which allows identifying $\beta T = \beta S \bigsqcup \beta V$. Moreover, $\beta S$ may be identified as a subgroup of $T$. In particular, every idempotent in $\beta S$ is idempotent in $\beta T$.
- (2)
For any minimal idempotent $p \in \beta S \setminus S$, $p$ is not minimal in $\beta T$.
- (3)
For any minimal idempotent $p \in \beta S \setminus S$, there exists a minimal idempotent $q \in \beta S \setminus \beta T$ such that $pq = qp = q$.
Proof. (1): By (U) of the Stone-Čech compactification.
(2): Suppose for contradiction that $p$ is a minimal idempotent in $\beta T$, then $L = (\beta T)p$ is a minimal left ideal in $\beta T$. Since $(\beta T)*p \subset \beta V \cap (\beta T)p$, $(\beta T)p \cap \beta V = (\beta T)p$ by minimality, and $(\beta T)p \subset \beta V$. As such, $(\beta S)p \subset \beta S \cap (\beta T)p \cap \beta V = \emptyset$, which is impossible.
(3): By Theorem 47.4.11, $S \setminus \beta S$ admits a minimal idempotent $p$, which is idempotent in $\beta T$. By the idempotent refinement lemma, there exists a minimial idempotent $q \in \beta T$ such that $pq = qp = q$.
Suppose again for contradiction that $q \in \beta S$, then $pq = q \in (\beta S)p$, which is minimal in $\beta S$. As such, $(\beta S)q \subset (\beta S)p$, and $(\beta S)q = (\beta S)p \subset \beta S$, so $q$ is a minimal idempotent in $\beta S$. By (2), $q$ is not minimal in $\beta T$, which is impossible.$\square$
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