Congruence Relation induces Normal Subgroup

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

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

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

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:

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


 * First we show that $$H$$ is normal in $$G$$.

First we show that $$H$$ is not empty:

$$e_G \in H \implies H \ne \varnothing$$.

Then we show $$H$$ is closed:

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

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


 * Now:

... thus $$H$$ is normal, as we wanted to prove.


 * 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 $$\mathcal{R} = \mathcal{R}_H$$.