Definition:Congruence Modulo Subgroup

Theorem
Let $$G$$ be a group, and let $$H \le G$$.

Then we can use $$H$$ to define an equivalence relation on $$G$$:


 * $$\mathcal{R}^l_H = \left\{{\left({x, y}\right) \in G \times G: x^{-1} y \in H}\right\}$$

When $$\left({x, y}\right) \in \mathcal{R}^l_H$$, we write $$x \equiv^l y \left({\bmod \, H}\right)$$.

This is called left congruence modulo $$H$$.

Similarly, we can use $$H$$ to define another equivalence relation on $$G$$:


 * $$\mathcal{R}^r_H = \left\{{\left({x, y}\right) \in G \times G: x y^{-1} \in H}\right\}$$

When $$\left({x, y}\right) \in \mathcal{R}^r_H$$, we write $$x \equiv^r y \left({\bmod \, H}\right)$$.

This is called right congruence modulo $$H$$.

Proof
We need to 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.