Restriction of Commutative Operation is Commutative
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Theorem
Let $\struct {S, \circ}$ be an algebraic structure.
Let $T \subseteq S$.
Let the operation $\circ$ be commutative on $\struct {S, \circ}$.
Then the restriction $\circ {\restriction_T}$ of $\circ$ to $T$ is also commutative.
Proof
\(\ds T\) | \(\subseteq\) | \(\ds S\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall a, b \in T: \, \) | \(\ds a, b\) | \(\in\) | \(\ds S\) | Definition of Subset | |||||||||
\(\ds \leadsto \ \ \) | \(\ds a \mathop {\circ {\restriction_T} } b\) | \(=\) | \(\ds a \circ b\) | |||||||||||
\(\ds \) | \(=\) | \(\ds b \circ a\) | $\circ$ is commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds b \mathop {\circ {\restriction_T} } a\) |
$\blacksquare$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 5.1$. Subsets closed to an operation
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets