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 $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$.

Also see

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