Inductive Semigroup whose Inductive Elements Commute is Commutative Semigroup
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Theorem
Let $\struct {S, \circ}$ be a semigroup.
Let there exist $\alpha, \beta \in S$ which fulfil the condition for $\struct {S, \circ}$ to be an inductive semigroup:
- the only subset of $S$ containing both $\alpha$ and $x \circ \beta$ whenever it contains $x$ is $S$ itself.
Let $\alpha$ and $\beta$ commute.
Then $\struct {S, \circ}$ is a commutative semigroup.
Proof
Suppose $\struct {S, \circ}$ is a semigroup.
Suppose there exist $\alpha, \beta \in S$ such that $\struct {S, \circ}$ is an inductive semigroup.
That is, suppose there exist $\alpha, \beta \in S$ such that the only subset of $S$ containing both $\alpha$ and $x \circ \beta$ whenever it contains $x$ is $S$ itself.
That is:
- $\exists \alpha, \beta \in S: \forall A \subseteq S: \paren {\alpha \in A \land \paren {\forall x \in A: x \circ \beta \in A} } \implies A = S$
By Form of Elements of Inductive Semigroup, the elements of $S$ are of the form:
- $\alpha \circ \beta \circ \beta \circ \cdots \circ \beta$
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Then:
- $\forall x, y \in S: x \circ y = y \circ x$
Hence, $\struct {S, \circ}$ is a commutative semigroup, as required.
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.9 \ \text {(a)}$