Endomorphisms on Entropic Structure whose Pointwise Product is Identity Automorphism
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Theorem
Let $\struct {S, \odot}$ be an entropic structure.
Let $\alpha$ and $\beta$ be endomorphisms on $S$ such that:
- $\alpha \odot \beta$ is the identity automorphism on $S$
where $\alpha \odot \beta$ denotes the pointwise product of $\alpha$ with $\beta$:
- $\forall x \in S: \map {\paren {\alpha \odot \beta} } x = \map \alpha x \odot \map \beta x$
Let $\otimes$ be the operation on $S$ defined as:
- $\forall x, y \in S: x \otimes y := \map \alpha x \odot \map \beta y$
Then $\struct {S, \otimes}$ is an entropic idempotent structure, and hence self-distributive.
Proof
Let $a, b, c, d \in S$ be arbitrary.
\(\ds a \otimes a\) | \(=\) | \(\ds \map \alpha a \odot \map \beta a\) | Definition of $\otimes$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {\alpha \odot \beta} } a\) | Definition of Pointwise Operation | |||||||||||
\(\ds \) | \(=\) | \(\ds a\) | Definition of Identity Mapping |
Hence $\struct {S, \otimes}$ is an idempotent structure.
\(\ds \paren {a \otimes b} \otimes \paren {c \otimes d}\) | \(=\) | \(\ds \map \alpha {a \otimes b} \odot \map \beta {c \otimes d}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \alpha {\map \alpha a \odot \map \beta b} \odot \map \beta {\map \alpha c \odot \map \beta d}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\map \alpha {\map \alpha a} \odot \map \alpha {\map \beta b} } \odot \paren {\map \beta {\map \alpha c} \odot \map \beta {\map \beta d} }\) | Definition of Endomorphism | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\map \alpha {\map \alpha a} \odot \map \beta {\map \alpha c} } \odot \paren {\map \alpha {\map \beta b} \odot \map \beta {\map \beta d} }\) | Definition of Entropic Structure | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\map \alpha a \otimes \map \alpha c} \odot \paren {\map \beta b \otimes \map \beta d}\) | Definition of $\otimes$ |
![]() | This needs considerable tedious hard slog to complete it. In particular: Can't see where to go from here. If we can prove $\map \alpha a \otimes \map \alpha c = \map \alpha {a \otimes c}$ we're home and dry, but it's not clear that $\alpha$ and $\beta$ are homomorphisms for $\struct {S, \otimes}$ like they are for $\struct {S, \odot}$. 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. |
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.27 \ \text {(c)}$