# External Direct Product of Congruence Relations

## Theorem

Let $\struct {A, \odot}$ and $\struct {B, \circledast}$ be algebraic structures.

Let $\RR$ and $\SS$ be congruence relations on $\struct {A, \odot}$ and $\struct {B, \circledast}$ respectively.

Let $\struct {A / \RR, \odot_\RR}$ and $\struct {B / \SS, \circledast_\SS}$ denote the quotient structures defined by $\RR$ and $\SS$ respectively.

Let $\struct {A \times B, \otimes}$ be the external direct product of $\struct {A, \odot}$ and $\struct {B, \circledast}$.

Let $\TT$ be the relation on $\struct {A \times B, \otimes}$ defined as:

$\forall \tuple {u, v}, \tuple {x, y} \in A \times B: \tuple {u, v} \mathrel \TT \tuple {x, y} \iff u \mathrel \RR x \land v \mathrel \SS y$

Then:

$\TT$ is a congruence relation on $\struct {A \times B, \otimes}$

and the mapping $h: \struct {A / \RR, \odot_\RR} \times \struct {B / \SS, \circledast_\SS} \to \struct {\paren {A \times B} / \TT, \otimes_\TT}$ defined as:

$\forall \tuple {\eqclass x \RR, \eqclass y \SS} \in \struct {A / \RR, \odot_\RR} \times \struct {B / \SS, \circledast_\SS}: \map h {\eqclass x \RR, \eqclass y \SS} = \eqclass {\tuple {x, y} } \TT$

is an isomorphism.

## Proof

We have that $\RR$ and $\SS$ are congruence relations on $\struct {A, \odot}$ and $\struct {B, \circledast}$ respectively.

Hence a fortiori $\RR$ and $\SS$ are equivalence relations on $A$ and $B$ respectively.

Thus from Cartesian Product of Equivalence Relations we have that $\TT$ is an equivalence relation.

Let $\tuple {a_1, b_1}, \tuple {a_2, b_2}, \tuple {c_1, d_1}, \tuple {c_2, d_2} \in A \times B$ be such that:

 $\ds \tuple {a_1, b_1}$ $\TT$ $\ds \tuple {a_2, b_2}$ $\ds \tuple {c_1, d_1}$ $\TT$ $\ds \tuple {c_2, d_2}$

Then:

 $\ds a_1$ $\RR$ $\ds a_2$ Definition of $\TT$ $\, \ds \land \,$ $\ds c_1$ $\RR$ $\ds c_2$ $\, \ds \land \,$ $\ds b_1$ $\SS$ $\ds b_2$ $\, \ds \land \,$ $\ds d_1$ $\SS$ $\ds d_2$ $\ds \leadsto \ \$ $\ds a_1 \odot c_1$ $\RR$ $\ds a_2 \odot c_2$ Definition of Congruence Relation $\, \ds \land \,$ $\ds b_1 \circledast d_1$ $\SS$ $\ds b_2 \circledast d_2$ $\ds \leadsto \ \$ $\ds \tuple {a_1 \odot c_1, b_1 \circledast d_1}$ $\TT$ $\ds \tuple {a_2 \odot c_2, b_2 \circledast d_2}$ Definition of $\TT$ $\ds \leadsto \ \$ $\ds \tuple {a_1, b_1} \otimes \tuple {c_1, d_1}$ $\TT$ $\ds \tuple {a_2, b_2} \otimes \tuple {c_2, d_2}$ Definition of External Direct Product

Hence by definition $\TT$ is a congruence relation on $\struct {A \times B, \otimes}$.

It remains to be shown that $h$ as defined is an isomorphism.

From Cartesian Product of Equivalence Relations, the equivalence class of $\tuple {a, b} \in A \times B$ is:

$\eqclass {\tuple {a, b} } \TT = \set {\tuple {x, y}: x \in \eqclass a \RR, y \in \eqclass b \SS}$

## 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: Exercise $13.14 \ \text{(b)}$