Congruence Modulo Normal Subgroup is Congruence Relation

Theorem
Let $\struct {G, \circ}$ be a group.

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

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

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

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

Let $x \mathrel {\RR_N} x'$ and $y \mathrel {\RR_N} y'$.

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


 * $\paren {x \circ y} \mathrel {\RR_N} \paren {x' \circ y'}$

So:

By Cosets are Equal iff Product with Inverse in Subgroup:
 * $x \circ x'^{-1} \in N$ and $y \circ y'^{-1} \in N$

Thus:
 * $\paren {x \circ y} \circ \paren {x' \circ y'}^{-1} \in x \circ H \circ x'^{-1}$

But we also have that:

That is:
 * $\paren {x \circ y} \circ \paren {x' \circ y'}^{-1} \in H$

and so:
 * $\paren {x \circ y} \mathrel {\RR_N} \paren {x' \circ y'}$

Hence the result, by definition of congruence relation.

Also see

 * Congruence Relation induces Normal Subgroup, the converse of this result