# Congruence Relation induces Normal Subgroup

## Contents

## Theorem

Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Let $\mathcal R$ be a congruence relation for $\circ$.

Let $H = \left[\!\left[{e}\right]\!\right]_\mathcal R$, where $\left[\!\left[{e}\right]\!\right]_\mathcal R$ is the equivalence class of $e$ under $\mathcal R$.

Then:

- $(1): \quad \left({H, \circ \restriction_H}\right)$ is a normal subgroup of $G$

- $(2): \quad \mathcal R$ is the equivalence relation $\mathcal R_H$ defined by $H$

- $(3): \quad \left({G / \mathcal R, \circ_\mathcal R}\right)$ is the subgroup $\left({G / H, \circ_H}\right)$ of the semigroup $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$.

## Proof

### Proof of Normal Subgroup

We are given that $\mathcal R$ is a congruence relation for $\circ$.

From Congruence Relation iff Compatible with Operation, we have:

- $\forall u \in G: x \mathop {\mathcal R} y \implies \left({x \circ u}\right) \mathop {\mathcal R} \left({y \circ u}\right), \left({u \circ x}\right)\mathop {\mathcal R} \left({u \circ y}\right)$

#### Proof of being a Subgroup

We show that $H$ is a subgroup of $G$.

First we note that $H$ is not empty:

- $e \in H \implies H \ne \varnothing$

Then we show $H$ is closed:

\(\displaystyle x, y\) | \(\in\) | \(\displaystyle H\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle e\) | \(\mathcal R\) | \(\displaystyle x\) | ||||||||||

\(\, \displaystyle \land \, \) | \(\displaystyle e\) | \(\mathcal R\) | \(\displaystyle y\) | by definition of $H$ | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \left({e \circ e}\right)\) | \(\mathcal R\) | \(\displaystyle \left({x \circ y}\right)\) | $\mathcal R$ is compatible with $\circ$ | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle x \circ y\) | \(\in\) | \(\displaystyle H\) | by definition of $H$ |

Next we show that $x \in H \implies x^{-1} \in H$:

\(\displaystyle x\) | \(\in\) | \(\displaystyle H\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle e\) | \(\mathcal R\) | \(\displaystyle x\) | by definition of $H$ | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \left({x^{-1} \circ e}\right)\) | \(\mathcal R\) | \(\displaystyle \left({x^{-1} \circ x}\right)\) | $\mathcal R$ is compatible with $\circ$ | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle x^{-1}\) | \(\mathcal R\) | \(\displaystyle e\) | Group properties | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle x^{-1}\) | \(\in\) | \(\displaystyle H\) | by definition of $H$ |

Thus by the Two-Step Subgroup Test, $H$ is a subgroup of $G$.

$\Box$

#### Proof of Normality

Next we show that $H$ is normal in $G$.

Thus:

\(\displaystyle x\) | \(\in\) | \(\displaystyle H\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle e\) | \(\mathcal R\) | \(\displaystyle h\) | for some $h \in H$, by definition of $H$ | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \left({x \circ e}\right)\) | \(\mathcal R\) | \(\displaystyle \left({x \circ h}\right)\) | $\mathcal R$ is compatible with $\circ$ | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \left({x \circ e \circ x^{-1} }\right)\) | \(\mathcal R\) | \(\displaystyle \left({x \circ h \circ x^{-1} }\right)\) | $\mathcal R$ is compatible with $\circ$ | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle e\) | \(\mathcal R\) | \(\displaystyle \left({x \circ h \circ x^{-1} }\right)\) | Group properties | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle x \circ h \circ x^{-1}\) | \(\in\) | \(\displaystyle H\) | by definition of $H$ |

Thus from Subgroup is Normal iff Contains Conjugate Elements, we have that $H$ is normal.

$\blacksquare$

### Proof of Equality of Relations

Let $\mathcal R_N$ be the equivalence defined by $N$.

Then:

\(\displaystyle x\) | \(\mathcal R\) | \(\displaystyle y\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle e\) | \(\mathcal R\) | \(\displaystyle \left({x^{-1} \circ y}\right)\) | $\mathcal R$ is compatible with $\circ$ | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \left({e \circ e}\right)\) | \(\mathcal R\) | \(\displaystyle \left({x^{-1} \circ y}\right)\) | Group properties | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle x^{-1} \circ y\) | \(\in\) | \(\displaystyle N\) | Definition of $N$ |

But from Congruence Class Modulo Subgroup is Coset:

- $x \mathop {\mathcal R_N} y \iff x^{-1} \circ y \in N$

Thus:

- $\mathcal R = \mathcal R_N$

$\blacksquare$

### Proof that Quotient Structure is Quotient Group

Let $\left[\!\left[{x}\right]\!\right]_\mathcal R \in G / \mathcal R$.

By Congruence Relation on Group induces Normal Subgroup:

- $\left[\!\left[{x}\right]\!\right]_\mathcal R = x N$

where $x N$ is the (left) coset of $N$ in $G$.

Similarly, let $y N \in G / N$.

Then from Normal Subgroup induced by Congruence Relation defines that Congruence:

- $y N = \left[\!\left[{x}\right]\!\right]_\mathcal R$

where:

- $\left[\!\left[{x}\right]\!\right]_\mathcal R$ is the equivalence class of $y$ under $\mathcal R$
- $\mathcal R$ is the equivalence relation defined by $N$.

Hence the result.

$\blacksquare$

## Also see

- Congruence Modulo Normal Subgroup is Congruence Relation, the converse of this result

## Sources

- 1965: Seth Warner:
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