Congruence Relation on Group 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:
 * $\struct {H, \circ \restriction_H}$ is a normal subgroup of $G$

where $\circ \restriction_H$ denotes the restriction of $\circ$ to $H$.

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

From Equivalence Relation is Congruence 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:

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

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

Proof of Normality
Next we show that $H$ is normal in $G$.

Thus:

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

Also see

 * Congruence Relation induces Normal Subgroup


 * Normal Subgroup induced by Congruence Relation defines that Congruence
 * Quotient Structure on Group defined by Congruence equals Quotient Group