Cross-Relation is Equivalence Relation
Theorem
Let $\struct {S, \circ}$ be a commutative semigroup with cancellable elements.
Let $\struct {C, \circ {\restriction_C} } \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$, where $\circ {\restriction_C}$ denotes the restriction of $\circ$ to $C$.
Let $\struct {S_1, \circ {\restriction_1} } \subseteq \struct {S, \circ}$ be a subsemigroup of $S$.
Let $\struct {S_2, \circ {\restriction_2} } \subseteq \struct {C, \circ {\restriction_C} }$ be a subsemigroup of $C$.
Let $\left({S_1 \times S_2, \oplus}\right)$ be the (external) direct product of $\struct {S_1, \circ {\restriction_1} }$ and $\struct {S_2, \circ {\restriction_2} }$, where $\oplus$ is the operation on $S_1 \times S_2$ induced by $\circ {\restriction_1}$ on $S_1$ and $\circ {\restriction_2}$ on $S_2$.
Let $\boxtimes$ be the cross-relation on $S_1 \times S_2$, defined as:
- $\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$
Then $\boxtimes$ is an equivalence relation on $\struct {S_1 \times S_2, \oplus}$.
Proof
Reflexivity
- $\forall \tuple {x_1, y_1} \in S_1 \times S_2: x_1 \circ y_1 = x_1 \circ y_1 \implies \tuple {x_1, y_1} \boxtimes \tuple {x_1, y_1}$
So $\boxtimes$ is a reflexive relation.
$\Box$
Symmetry
\(\ds \tuple {x_1, y_1}\) | \(\boxtimes\) | \(\ds \tuple {x_2, y_2}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x_1 \circ y_2\) | \(=\) | \(\ds x_2 \circ y_1\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x_2 \circ y_1\) | \(=\) | \(\ds x_1 \circ y_2\) | $\circ$ is commutative | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {x_2, y_2}\) | \(\boxtimes\) | \(\ds \tuple {x_1, y_1}\) |
So $\boxtimes$ is a symmetric relation.
$\Box$
Transitivity
\(\ds \tuple {x_1, y_1}\) | \(\boxtimes\) | \(\ds \tuple {x_2, y_2}\) | ||||||||||||
\(\, \ds \land \, \) | \(\ds \tuple {x_2, y_2}\) | \(\boxtimes\) | \(\ds \tuple {x_3, y_3}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x_1 \circ y_2\) | \(=\) | \(\ds x_2 \circ y_1\) | |||||||||||
\(\, \ds \land \, \) | \(\ds x_2 \circ y_3\) | \(=\) | \(\ds x_3 \circ y_2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x_1 \circ y_3 \circ y_2\) | \(=\) | \(\ds x_1 \circ y_2 \circ y_3\) | $\circ$ is commutative | ||||||||||
\(\ds \) | \(=\) | \(\ds x_2 \circ y_1 \circ y_3\) | substituting $x_2 \circ y_1$ for $x_1 \circ y_2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x_2 \circ y_3 \circ y_1\) | $\circ$ is commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds x_3 \circ y_2 \circ y_1\) | substituting $x_3 \circ y_2$ for $x_2 \circ y_3$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x_3 \circ y_1 \circ y_2\) | $\circ$ is commutative | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x_1 \circ y_3\) | \(=\) | \(\ds x_3 \circ y_1\) | as $y_2 \in S_2$, therefore $y_2 \in C$, and so is cancellable for $\circ$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {x_1, y_1}\) | \(\boxtimes\) | \(\ds \tuple {x_3, y_3}\) |
So $\boxtimes$ is a transitive relation.
$\Box$
All the criteria are therefore seen to hold for $\boxtimes$ to be an equivalence relation.
$\blacksquare$
Also see
Examples
- Cross-Relation on Natural Numbers is Equivalence Relation
- Cross-Relation on Real Numbers is Equivalence Relation
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers