# Congruence Relation induces Normal Subgroup

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

Thus:

$\RR = \RR_N$

$\blacksquare$

### Proof that Quotient Structure is Quotient Group

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

$\eqclass x \RR = x N$

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

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

$y N = \eqclass x \RR$

where:

$\eqclass x \RR$ is the equivalence class of $y$ under $\RR$
$\RR$ is the equivalence relation defined by $N$.

Hence the result.

$\blacksquare$