Naturally Ordered Semigroup is Unique/Isomorphism is Unique
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Theorem
Let $\struct {S, \circ, \preceq}$ and $\struct {S', \circ', \preceq'}$ be naturally ordered semigroups.
Let:
- $0'$ be the smallest element of $S'$
- $1'$ be the smallest element of $S' \setminus \set {0'} = S'^*$.
Then the isomorphism $g: \struct {S, \circ, \preceq} \to \struct {S', \circ', \preceq'}$ defined as:
- $\forall a \in S: \map g a = \circ'^a 1'$
is unique.
Proof
Let $f: S \to S'$ be another isomorphism different from $g$.
Aiming for a contradiction, suppose $\map f 1 \ne 1'$.
We show by induction that $1' \notin \Cdm f$.
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... Thus $1' \notin \Cdm f$ which is a contradiction.
Thus $\map f 1 = 1$ and it follows
This needs considerable tedious hard slog to complete it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
that $f = g$.
Thus the isomorphism $g$ is unique.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.28 \ \text {(b)}$