Congruence Relation induces Normal Subgroup

Theorem
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Let $\mathcal R$ be an equivalence on $G$ compatible with $\circ$.

Let $H = \left[\!\left[{e}\right]\!\right]_\mathcal R$.

Then $H$ is a normal subgroup of $G$ and $\mathcal R$ is the equivalence $\mathcal R_H$ defined by $H$.

Also, $\left({G / \mathcal R, \circ_\mathcal R}\right)$ is the subgroup $\left({G / H, \circ_H}\right)$ of the semigroup $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$.

Proof
From the fact that $\mathcal R$ is compatible with $\circ$, 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 $\mathcal R = \mathcal R_H$.