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

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

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

defines an equivalence relation in $G$.

###### Proof.

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

1. 1.

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

2. 2.

Symmetric: We have

 $\displaystyle a\sim b$ $\displaystyle\Rightarrow ab^{-1}\in H$ $\displaystyle\Rightarrow\left(ab^{-1}\right)^{-1}\in H$ $\displaystyle\Rightarrow ba^{-1}\in H$ $\displaystyle\Rightarrow b\sim a$
3. 3.

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

 $ab^{-1}\in H,\quad\text{and}\quad bc^{-1}\in H$

but then

 $\left(ab^{-1}\right)\left(bc^{-1}\right)\in H$

which gives

 $ac^{-1}\in H$

that is, $a\sim c$.

Title subgroup of a group defines an equivalence relation on the group, proof that a SubgroupOfAGroupDefinesAnEquivalenceRelationOnTheGroupProofThatA 2013-03-22 15:32:46 2013-03-22 15:32:46 Dr_Absentius (537) Dr_Absentius (537) 5 Dr_Absentius (537) Proof msc 20-00