# Left Congruence Modulo Subgroup is Equivalence Relation

## Theorem

Let $G$ be a group, and let $H$ be a subgroup of $G$.

Let $x, y \in G$.

Let $x \equiv^l y \pmod H$ denote the relation that $x$ is left congruent modulo $H$ to $y$.

Then the relation $\equiv^l$ is an equivalence relation.

## Proof

Let $G$ be a group whose identity is $e$.

Let $H$ be a subgroup of $G$.

For clarity of expression, we will use the notation:

$\tuple {x, y} \in \RR^l_H$

for:

$x \equiv^l y \pmod H$

From the definition of left congruence modulo a subgroup, we have:

$\RR^l_H = \set {\tuple {x, y} \in G \times G: x^{-1} y \in H}$

We show that $\RR^l_H$ is an equivalence:

### Reflexive

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

From Identity of Subgroup:

$e \in H$

Hence:

$\forall x \in G: x^{-1} x = e \in H \implies \tuple {x, x} \in \RR^l_H$

and so $\RR^l_H$ is reflexive.

$\Box$

### Symmetric

 $\ds \tuple {x, y}$ $\in$ $\ds \RR^l_H$ $\ds \leadsto \ \$ $\ds x^{-1} y$ $\in$ $\ds H$ Definition of Left Congruence Modulo $H$ $\ds \leadsto \ \$ $\ds \tuple {x^{-1} y}^{-1}$ $\in$ $\ds H$ Group Axiom $\text G 0$: Closure

But then:

$\tuple {x^{-1} y}^{-1} = y^{-1} x \implies \tuple {y, x} \in \RR^l_H$

Thus $\RR^l_H$ is symmetric.

$\Box$

### Transitive

 $\ds \tuple {x, y}, \tuple {y, z}$ $\in$ $\ds \RR^l_H$ $\ds \leadsto \ \$ $\ds x^{-1} y$ $\in$ $\ds H$ Definition of Left Congruence Modulo $H$ $\, \ds \land \,$ $\ds y^{-1} z$ $\in$ $\ds H$ Definition of Left Congruence Modulo $H$ $\ds \leadsto \ \$ $\ds \tuple {x^{-1} y} \tuple {y^{-1} z} = x^{-1} z$ $\in$ $\ds H$ Group Properties $\ds \leadsto \ \$ $\ds \tuple {x, z}$ $\in$ $\ds R^l_H$ Definition of Left Congruence Modulo $H$

Thus $\RR^l_H$ is transitive.

$\Box$

So $\RR^l_H$ is an equivalence relation.

$\blacksquare$