Congruence Modulo Normal Subgroup is Congruence Relation

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

Let $N$ be a normal subgroup of $G$.

Then congruence modulo $N$ is a congruence relation for the group product $\circ$.

Proof
Let $x \mathop {\mathcal R_N} y$ denote that $x$ and $y$ are in the same coset, that is:
 * $x \mathop {\mathcal R_N} y \iff x \circ N = y \circ N$

as specified in the definition of congruence modulo $N$.

Let $x \mathop {\mathcal R_N} x'$ and $y \mathop {\mathcal R_N} y'$.

To demonstrate that $\mathcal R_N$ is a congruence relation for $\circ$, we need to show that:


 * $\left({x \circ y}\right) \mathop {\mathcal R_N} \left({x' \circ y'}\right)$

So:

By Equal Cosets iff Product with Inverse in Coset, we have that:
 * $x \circ x'^{-1} \in N$ and $y \circ y'^{-1} \in N$

Thus: $\left({x \circ y}\right) \circ \left({x' \circ y'}\right)^{-1} \in x \circ H \circ x'^{-1}$

But we also have that:

That is:
 * $\left({x \circ y}\right) \circ \left({x' \circ y'}\right)^{-1} \in H$

and so:
 * $\left({x \circ y}\right) \mathop {\mathcal R_N} \left({x' \circ y'}\right)$

Hence the result, by definition of congruence relation.

Also see

 * Congruence Relation Gives Rise to Normal Subgroup, the converse of this result