# 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:

 $\ds \paren {x \circ y} \circ \paren {x' \circ y'}^{-1}$ $=$ $\ds \paren {x \circ y} \circ \paren {y'^{-1} \circ x'^{-1} }$ Inverse of Group Product $\ds$ $=$ $\ds \paren {\paren {x \circ y} \circ y'^{-1} } \circ x'^{-1}$ Group Axiom $\text G 1$: Associativity $\ds$ $=$ $\ds \paren {x \circ \paren {y \circ y'^{-1} } } \circ x'^{-1}$ Group Axiom $\text G 1$: Associativity
$x \circ x'^{-1} \in N$ 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:

 $\ds x \circ H \circ x'^{-1}$ $=$ $\ds H \circ x \circ x'^{-1}$ Definition of Normal Subgroup $\ds$ $\subseteq$ $\ds H \circ H$ Definition of Subset Product: $x \circ x'^{-1} \in H$ $\ds$ $=$ $\ds H$ Product of Subgroup with Itself

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.

$\blacksquare$