Structure Induced by Commutative Operation is Commutative
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Theorem
Let $\struct {T, \circ}$ be an algebraic structure, and let $S$ be a set.
Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\circ$.
Let $\circ$ be a commutative operation.
Then the pointwise operation $\oplus$ induced on $T^S$ by $\circ$ is also commutative.
Proof
Let $\struct {T, \circ}$ be a commutative algebraic structure.
Let $f, g \in T^S$.
Then:
\(\ds \map {\paren {f \oplus g} } x\) | \(=\) | \(\ds \map f x \circ \map g x\) | Definition of Pointwise Operation | |||||||||||
\(\ds \) | \(=\) | \(\ds \map g x \circ \map f x\) | $\circ$ is commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {g \oplus f} } x\) | Definition of Pointwise Operation |
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.6: \ 1^\circ$
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.4$