Equivalence Class of Equal Elements of Cross-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:
- $\forall c, d \in S_1 \cap S_2: \tuple {c, c} \boxtimes \tuple {d, d}$
Proof
Note that in order for $\tuple {c, c}$ and $\tuple {d, d}$ to be defined, $c$ and $d$ must be in both $S_1$ and $S_2$.
Hence the restriction given:
- $\forall c, d \in S_1 \cap S_2$
Then:
\(\ds \forall c, d \in S_1 \cap S_2: \, \) | \(\ds c \circ d\) | \(=\) | \(\ds d \circ c\) | Commutativity of $\circ$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {c, c}\) | \(\boxtimes\) | \(\ds \tuple {d, d}\) | Definition of $\boxtimes$ |
Hence the result.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers