Right Congruence Modulo Subgroup is Equivalence Relation

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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:

$\tuple {x, y} \in \mathcal R^l_H$

for:

$x \equiv^l y \pmod H$


From the definition of left congruence modulo a subgroup, we have:

$\mathcal R^l_H = \set {\tuple {x, y} \in G \times G: x^{-1} y \in H}$


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:

$\forall x \in G: x x^{-1} = e \in H \implies \tuple {x, x} \in \mathcal R^r_H$

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

$\Box$


Symmetric

\(\displaystyle \tuple {x, y}\) \(\in\) \(\displaystyle \mathcal R^r_H\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x y^{-1}\) \(\in\) \(\displaystyle H\) Definition of Right Congruence Modulo $H$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \tuple {x y^{-1} }^{-1}\) \(\in\) \(\displaystyle H\) Group Axiom $G \, 0$: Closure

But then:

$\tuple {x y^{-1} }^{-1} = y x^{-1} \implies \tuple {y, x} \in \mathcal R^r_H$

Thus $\mathcal R^r_H$ is symmetric.

$\Box$


Transitive

\(\displaystyle \tuple {x, y}, \tuple {y, z}\) \(\in\) \(\displaystyle \mathcal R^r_H\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x y^{-1}\) \(\in\) \(\displaystyle H\) Definition of Right Congruence Modulo $H$
\(\, \displaystyle \land \, \) \(\displaystyle y z^{-1}\) \(\in\) \(\displaystyle H\) Definition of Right Congruence Modulo $H$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \tuple {x y^{-1} } \tuple {y z^{-1} } = x z^{-1}\) \(\in\) \(\displaystyle H\) Group Properties
\(\displaystyle \leadsto \ \ \) \(\displaystyle \tuple {x, z}\) \(\in\) \(\displaystyle R^r_H\) Definition of Right Congruence Modulo $H$

Thus $\mathcal R^r_H$ is transitive.

$\Box$


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

$\blacksquare$


Also see


Sources