Congruence Relation induces Normal Subgroup

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:
 * $(1): \quad H$ 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
We are given that $\mathcal R$ is a congruence relation for $\circ$.

From Congruence Relation iff Compatible with Operation, we have:


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

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:

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 Normal Subgroup Equivalent Definitions, we have that $H$ is normal, as we wanted to prove.

Proof of Equality of Relations
Now we need to show that $\mathcal R_H$, the equivalence defined by $H$, is actually $\mathcal R$.

But from Congruence Class Modulo Subgroup is Coset:
 * $x \ \mathcal R_H \ y \iff x^{-1} \circ y \in H$

Thus:
 * $\mathcal R = \mathcal R_H$

Also see

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