Natural Numbers with Divisor Operation is Isomorphic to Subgroups of Integer Multiples under Inclusion
Theorem
Let $\N_{>0}$ denote the set of strictly positive natural numbers.
For $n \in \N_{>0}$, let $n \Z$ denote the set of integer multiples of $n$.
Let $\struct {\Z, +}$ denote the additive group of integers.
Let $\mathscr G$ be the set of all subgroups of $\struct {\Z, +}$.
Consider the algebraic structure $\struct {\N_{>0}, \divides}$, where $\divides$ denotes the divisor operator:
- $a \divides b$ denotes that $a$ is a divisor of $b$
Let $\phi: \struct {\N_{>0}, \divides} \to \struct {\mathscr G, \supseteq}$ be the mapping defined as:
- $\forall n \in \N_{>0}: \map \phi n = n \Z$
Then $\phi$ is an order isomorphism.
Corollary
Consider the ordered semigroup $\struct {\N_{>0}, \times, \divides}$, where:
- $\divides$ denotes the divisor operator:
- $a \divides b$ denotes that $a$ is a divisor of $b$
- $\times$ denotes integer multiplication.
Let $\phi: \struct {\N_{>0}, \times, \divides} \to \struct {\mathscr G, \times_\PP, \supseteq}$ be the mapping defined as:
- $\forall n \in \N_{>0}: \map \phi n = n \Z$
Then $\phi$ is an ordered semigroup isomorphism.
Proof
We note that from Subgroups of Additive Group of Integers, the subgroups of $\struct {\Z, +}$ are precisely the sets of integer multiples $n \Z$, for $n \in \N_{>0}$.
For each $n \in \N_{>0}$, there is a unique $n \Z \in \mathscr G$.
Hence $\phi$ is a bijection.
It remains to be demonstrated that $\phi$ is order-preserving in both directions.
Thus:
\(\ds a\) | \(\divides\) | \(\ds b\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \exists k \in \Z: \, \) | \(\ds k a\) | \(=\) | \(\ds b\) | Definition of Divisor of Integer |
Then:
\(\ds x\) | \(\in\) | \(\ds b \Z\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \exists m \in \Z: \, \) | \(\ds x\) | \(=\) | \(\ds m b\) | Definition of Set of Integer Multiples | |||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \exists k \in \Z: \, \) | \(\ds x\) | \(=\) | \(\ds \paren {m k} a\) | a priori: $k a = b$ | |||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(\in\) | \(\ds a \Z\) | Definition of Set of Integer Multiples | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds a \Z\) | \(\supseteq\) | \(\ds b \Z\) | Definition of Subset |
Hence the result.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.21$