MHB Arjun's question at Yahoo Answers (Equivalence relation)

[Fernando Revilla]

Here is the question:

Let G be a group and H be a subgroup of G. We say that a is congruent to b modulo H if and only if a* (b inverse) is an element of H. Show that congruence modulo H is an equivalence relation on G.

Here is a link to the question:

Equivalence Relation question? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello Arjun,

Our relation is $a\sim b\Leftrightarrow ab^{-1}\in H$. This is an equivalence relation on $G$.

$(i)$ Reflexive. For all $a\in G$ is verified $aa^{-1}=e$, and $e\in H$ because $H$ is subgroup of $G$, so $a\sim a$.

$(ii)$ Symmetric. Using that $H$ is subgroup, $(xy)^{-1}=y^{-1}x^{-1}$ and $(x^{-1})^{-1}=x$:

$a\sim b\Rightarrow ab^{-1}\in H\Rightarrow (ab^{-1})^{-1}\in H\Rightarrow (b^{-1})^{-1}a^{-1}\in H\Rightarrow ba^{-1}\in H\Rightarrow b\sim a$

$(iii)$ Transitive. $a\sim b$ and $b\sim c$ implies $ab^{-1}\in H$ and $bc^{-1}\in H$. As $H$ is subgroup the product of these elements is in $H$, that is:
$$(ab^{-1})(bc^{-1})=a(b^{-1}b)c^{-1}=aec^{-1}=ac^{-1}\in H$$ which implies $a\sim c$. $\qquad \square$