subgroup of a group defines an equivalence relation on the group, proof that a

Let $H$ be a subgroup of $G$. Then

$$a\sim b⟺a{b}^{-1}\in H$$

defines an equivalence relation in $G$.

Proof.

We need to show that the relation is reflexive, symmetric and transitive.

1. Reflexive: $a{a}^{-1}=e\in H$ therefore $a\sim a$.

2. Symmetric: We have

   $a\sim b$ $\Rightarrow a{b}^{-1}\in H$

   $\Rightarrow {\left(a{b}^{-1}\right)}^{-1}\in H$

   $\Rightarrow b{a}^{-1}\in H$

   $\Rightarrow b\sim a$

3. Transitive: If $a\sim b$ and $b\sim c$ then we have that

   $$a{b}^{-1}\in H,\text{and}\mathit{\hspace{1em}}b{c}^{-1}\in H$$

   but then

   $$\left(a{b}^{-1}\right)\left(b{c}^{-1}\right)\in H$$

   which gives

   $$a{c}^{-1}\in H$$

   that is, $a\sim c$.

∎

Title	subgroup of a group defines an equivalence relation on the group, proof that a
Canonical name	SubgroupOfAGroupDefinesAnEquivalenceRelationOnTheGroupProofThatA
Date of creation	2013-03-22 15:32:46
Last modified on	2013-03-22 15:32:46
Owner	Dr_Absentius (537)
Last modified by	Dr_Absentius (537)
Numerical id	5
Author	Dr_Absentius (537)
Entry type	Proof
Classification	msc 20-00