Quotient Structure on Group defined by Congruence equals Quotient Group

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

Let $\mathcal R$ be a congruence relation for $\circ$.

Let $\left({G / \mathcal R, \circ_\mathcal R}\right)$ be the quotient structure defined by $\mathcal R$.

Let $N = \left[\!\left[{e}\right]\!\right]_\mathcal R$ be the normal subgroup induced by $\mathcal R$.

Let $\left({G / N, \circ_N}\right)$ be the quotient group of $G$ by $N$.

Then $\left({G / \mathcal R, \circ_\mathcal R}\right)$ is the subgroup $\left({G / N, \circ_N}\right)$ of the semigroup $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$.

Proof
Let $\left[\!\left[{x}\right]\!\right]_\mathcal R \in G / \mathcal R$.

By Congruence Relation on Group induces Normal Subgroup:
 * $\left[\!\left[{x}\right]\!\right]_\mathcal R = x N$

where $x N$ is the (left) coset of $N$ in $G$.

Similarly, let $y N \in G / N$.

Then from Normal Subgroup induced by Congruence Relation defines that Congruence:
 * $y N = \left[\!\left[{x}\right]\!\right]_\mathcal R$

where:
 * $\left[\!\left[{x}\right]\!\right]_\mathcal R$ is the equivalence class of $y$ under $\mathcal R$
 * $\mathcal R$ is the equivalence relation defined by $N$.

Hence the result.

Also see

 * Congruence Relation induces Normal Subgroup


 * Congruence Relation on Group induces Normal Subgroup
 * Normal Subgroup induced by Congruence Relation defines that Congruence