# 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:

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

This needs considerable tedious hard slog to complete it.In particular: intuitively obvious, tedious proving it rigorouslyTo discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

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