Congruence Relation on Group 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 $H$ is a normal subgroup of $G$.

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

From Congruence Relation iff Compatible with Operation, we have:


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

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:

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.

Also see

 * Normal Subgroup Induced by Congruence Relation Defines That Congruence
 * Quotient Structure on Group Defined by Congruence Equals Quotient Group