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