# Congruence Relation on Group 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:

$\left({H, \circ \restriction_H}\right)$ is a normal subgroup of $G$

where $\circ \restriction_H$ denotes the restriction of $\circ$ to $H$.

## Proof

We are given that $\mathcal R$ is a congruence relation for $\circ$.

From Congruence Relation iff Compatible with Operation, we have:

$\forall u \in G: x \mathop {\mathcal R} y \implies \left({x \circ u}\right) \mathop {\mathcal R} \left({y \circ u}\right), \left({u \circ x}\right)\mathop {\mathcal R} \left({u \circ y}\right)$

#### Proof of being a Subgroup

We show that $H$ is a subgroup of $G$.

First we note that $H$ is not empty:

$e \in H \implies H \ne \varnothing$

Then we show $H$ is closed:

 $\displaystyle x, y$ $\in$ $\displaystyle H$ $\displaystyle \implies \ \$ $\displaystyle e$ $\mathcal R$ $\displaystyle x$ $\, \displaystyle \land \,$ $\displaystyle e$ $\mathcal R$ $\displaystyle y$ by definition of $H$ $\displaystyle \implies \ \$ $\displaystyle \left({e \circ e}\right)$ $\mathcal R$ $\displaystyle \left({x \circ y}\right)$ $\mathcal R$ is compatible with $\circ$ $\displaystyle \implies \ \$ $\displaystyle x \circ y$ $\in$ $\displaystyle H$ by definition of $H$

Next we show that $x \in H \implies x^{-1} \in H$:

 $\displaystyle x$ $\in$ $\displaystyle H$ $\displaystyle \implies \ \$ $\displaystyle e$ $\mathcal R$ $\displaystyle x$ by definition of $H$ $\displaystyle \implies \ \$ $\displaystyle \left({x^{-1} \circ e}\right)$ $\mathcal R$ $\displaystyle \left({x^{-1} \circ x}\right)$ $\mathcal R$ is compatible with $\circ$ $\displaystyle \implies \ \$ $\displaystyle x^{-1}$ $\mathcal R$ $\displaystyle e$ Group properties $\displaystyle \implies \ \$ $\displaystyle x^{-1}$ $\in$ $\displaystyle H$ by definition of $H$

Thus by the Two-Step Subgroup Test, $H$ is a subgroup of $G$.

$\Box$

#### Proof of Normality

Next we show that $H$ is normal in $G$.

Thus:

 $\displaystyle x$ $\in$ $\displaystyle H$ $\displaystyle \implies \ \$ $\displaystyle e$ $\mathcal R$ $\displaystyle h$ for some $h \in H$, by definition of $H$ $\displaystyle \implies \ \$ $\displaystyle \left({x \circ e}\right)$ $\mathcal R$ $\displaystyle \left({x \circ h}\right)$ $\mathcal R$ is compatible with $\circ$ $\displaystyle \implies \ \$ $\displaystyle \left({x \circ e \circ x^{-1} }\right)$ $\mathcal R$ $\displaystyle \left({x \circ h \circ x^{-1} }\right)$ $\mathcal R$ is compatible with $\circ$ $\displaystyle \implies \ \$ $\displaystyle e$ $\mathcal R$ $\displaystyle \left({x \circ h \circ x^{-1} }\right)$ Group properties $\displaystyle \implies \ \$ $\displaystyle x \circ h \circ x^{-1}$ $\in$ $\displaystyle H$ by definition of $H$

Thus from Subgroup is Normal iff Contains Conjugate Elements, we have that $H$ is normal.

$\blacksquare$