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 $$\mathcal R = \mathcal R_H$$.