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

Reflexive

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

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.