# Congruence Relation induces Normal Subgroup

## Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $\RR$ be a congruence relation for $\circ$.

Let $H = \eqclass e \RR$, where $\eqclass e \RR$ is the equivalence class of $e$ under $\RR$.

Then:

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

- $(2): \quad \RR$ is the equivalence relation $\RR_H$ defined by $H$

- $(3): \quad \struct {G / \RR, \circ_\RR}$ is the subgroup $\struct {G / H, \circ_H}$ of the semigroup $\struct {\powerset G, \circ_\PP}$.

## Proof

### Proof of Normal Subgroup

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

From Congruence Relation iff Compatible with Operation, we have:

- $\forall u \in G: x \mathrel \RR y \implies \paren {x \circ u} \mathrel \RR \paren {y \circ u}, \paren {u \circ x} \mathrel \RR \paren {u \circ y}$

#### Proof of being a Subgroup

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

First we note that $H$ is not empty:

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

Then we show $H$ is closed:

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

\(\ds \leadsto \ \ \) | \(\ds e\) | \(\RR\) | \(\ds x\) | |||||||||||

\(\, \ds \land \, \) | \(\ds e\) | \(\RR\) | \(\ds y\) | by definition of $H$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \paren {e \circ e}\) | \(\RR\) | \(\ds \paren {x \circ y}\) | $\RR$ is compatible with $\circ$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds x \circ y\) | \(\in\) | \(\ds H\) | by definition of $H$ |

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

\(\ds x\) | \(\in\) | \(\ds H\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds e\) | \(\RR\) | \(\ds x\) | Definition of $H$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \paren {x^{-1} \circ e}\) | \(\RR\) | \(\ds \paren {x^{-1} \circ x}\) | $\RR$ is compatible with $\circ$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds x^{-1}\) | \(\RR\) | \(\ds e\) | Group properties | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds x^{-1}\) | \(\in\) | \(\ds H\) | 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:

\(\ds x\) | \(\in\) | \(\ds H\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds e\) | \(\RR\) | \(\ds h\) | for some $h \in H$, by definition of $H$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \paren {x \circ e}\) | \(\RR\) | \(\ds \paren {x \circ h}\) | $\RR$ is compatible with $\circ$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \paren {x \circ e \circ x^{-1} }\) | \(\RR\) | \(\ds \paren {x \circ h \circ x^{-1} }\) | $\RR$ is compatible with $\circ$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds e\) | \(\RR\) | \(\ds \paren {x \circ h \circ x^{-1} }\) | Group properties | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds x \circ h \circ x^{-1}\) | \(\in\) | \(\ds H\) | 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 $\RR_N$ be the equivalence defined by $N$.

Then:

\(\ds x\) | \(\RR\) | \(\ds y\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds e\) | \(\RR\) | \(\ds \paren {x^{-1} \circ y}\) | $\RR$ is compatible with $\circ$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \paren {e \circ e}\) | \(\RR\) | \(\ds \paren {x^{-1} \circ y}\) | Group properties | ||||||||||

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

But from Congruence Class Modulo Subgroup is Coset:

- $x \mathrel {\RR_N} y \iff x^{-1} \circ y \in N$

Thus:

- $\RR = \RR_N$

$\blacksquare$

### Proof that Quotient Structure is Quotient Group

Let $\eqclass x \RR \in G / \RR$.

By Congruence Relation on Group induces Normal Subgroup:

- $\eqclass x \RR = x N$

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:

- $y N = \eqclass x \RR$

where:

- $\eqclass x \RR$ is the equivalence class of $y$ under $\RR$
- $\RR$ 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:
*Modern Algebra*... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Theorem $11.5$