Right 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 $x \equiv^r y \pmod H$ denote the relation that $x$ is right congruent modulo $H$ to $y$

Then the relation $\equiv^r$ 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:
 * $\left({x, y}\right) \in \mathcal R^l_H$

for:
 * $x \equiv^l y \pmod H$

From the definition of left 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 an equivalence:

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

From Identity of Subgroup:
 * $e \in H$

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

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

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

Thus $\mathcal R^r_H$ is symmetric.

Transitive
Thus $\mathcal R^r_H$ is transitive.

So $\mathcal R^r_H$ is an equivalence relation.

Also see

 * Definition:Right Coset
 * Definition:Right Coset Space