Left 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 $x \equiv^l y \pmod H$ denote the relation that $x$ is left congruent modulo $H$ to $y$.

Then the relation $\equiv^l$ is an equivalence relation.


Proof 1

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:

$\tuple {x, y} \in \RR^l_H$

for:

$x \equiv^l y \pmod H$


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

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


We show that $\RR^l_H$ is an equivalence:


Reflexive

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

From Identity of Subgroup:

$e \in H$

Hence:

$\forall x \in G: x^{-1} x = e \in H \implies \tuple {x, x} \in \RR^l_H$

and so $\RR^l_H$ is reflexive.

$\Box$


Symmetric

\(\ds \tuple {x, y}\) \(\in\) \(\ds \RR^l_H\)
\(\ds \leadsto \ \ \) \(\ds x^{-1} y\) \(\in\) \(\ds H\) Definition of Left Congruence Modulo $H$
\(\ds \leadsto \ \ \) \(\ds \tuple {x^{-1} y}^{-1}\) \(\in\) \(\ds H\) Group Axiom $\text G 0$: Closure

But then:

$\tuple {x^{-1} y}^{-1} = y^{-1} x \implies \tuple {y, x} \in \RR^l_H$

Thus $\RR^l_H$ is symmetric.

$\Box$


Transitive

\(\ds \tuple {x, y}, \tuple {y, z}\) \(\in\) \(\ds \RR^l_H\)
\(\ds \leadsto \ \ \) \(\ds x^{-1} y\) \(\in\) \(\ds H\) Definition of Left Congruence Modulo $H$
\(\, \ds \land \, \) \(\ds y^{-1} z\) \(\in\) \(\ds H\) Definition of Left Congruence Modulo $H$
\(\ds \leadsto \ \ \) \(\ds \tuple {x^{-1} y} \tuple {y^{-1} z} = x^{-1} z\) \(\in\) \(\ds H\) Group Properties
\(\ds \leadsto \ \ \) \(\ds \tuple {x, z}\) \(\in\) \(\ds R^l_H\) Definition of Left Congruence Modulo $H$

Thus $\RR^l_H$ is transitive.

$\Box$


So $\RR^l_H$ is an equivalence relation.

$\blacksquare$

Proof 2


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Let $G$ be a group, and let $H$ be a subgroup of $G$.

Then $G$ is a fortiori a monoid.

From Condition for Cosets of Subgroup of Monoid to be Partition, $\RR^l_H$ is a relation induced by partition.

The result follows from Relation Induced by Partition is Equivalence.

$\blacksquare$

Also see


Sources

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