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 $x \equiv^l y \left({\bmod \, H}\right)$ denote the relation that $x$ is left congruent modulo $H$ to $y$.

Similarly, let $x \equiv^r y \left({\bmod \, H}\right)$ denote the relation that $x$ is right congruent modulo $H$ to $y$

Then the relations $\equiv^l$ and $\equiv^r$ are equivalence relations.

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:
 * $\left({x, y}\right) \in \mathcal R^l_H$

for:
 * $x \equiv^l y \left({\bmod \, H}\right)$

and similarly:
 * $\left({x, y}\right) \in \mathcal R^r_H$

for:
 * $x \equiv^r y \left({\bmod \, H}\right)$

From the definition of congruence modulo a subgroup, we have:
 * $\mathcal R^l_H = \left\{{\left({x, y}\right) \in G \times G: x^{-1} y \in H}\right\}$

We show that $\mathcal R^l_H$ is in fact an equivalence:

Reflexive
Through dint of $H$ being a subgroup of $G$, we know from Identity of Subgroup that $e \in H$.

Hence:
 * $\forall x \in G: x^{-1} x = e \in H \implies \left({x, x}\right) \in \mathcal R^l_H$

and so $\mathcal R^l_H$ is reflexive.

Symmetric
But then $\left({x^{-1} y}\right)^{-1} = y^{-1} x \implies \left({y, x}\right) \in \mathcal R^l_H$.

Transitive
So $\mathcal R^l_H$ is an equivalence relation.

The proof that $\mathcal R^r_H$ is also an equivalence follows exactly the same lines.

Also see

 * Coset
 * Coset Space