Equivalence relation and subgroup I am taking abstract algebra now, and there's a lemma: Let $H$ be a subgroup of group $G$, for $a,b \in G$,define $a\sim b$ if $ab^{-1}\in H$, then it is an equivalence. I know how to prove it and how to use it in the prove of Lagrange theorem, but can anyone give me a more mathematical intuition explanation about it ? My book has an example,let $a,b$ belong the congruence class, then $a\sim b$ will be like $a-b \in H$ like $a-b\pmod n$, but it is a particular problem, can someone tell me a more general one? How does the lemma been created? 

Let an equivalence class of any element $x$ be the set of elements with which $x$ is related.
Suppose one wants that any element of an equivalence class $E$ acts as a representative of it. So if $a,b$ are in the same class then one wants that their equivalence classes can both be identified with each other. This means that the group can be broken up in many parts and if from one part we can pick up any representative then the class of that representative will be equal to the class of any other element in the same part. That is equivalent to saying $ab^{-1}\in H$. 
With this in hand, one picks up any representative from each equivalence class and and if the original subgroup was a nice one (product of two classes is another class) one do "math" with the representatives. We have the freedom to choose any representative from a part and elements are much easier to work with then classes. Instead of multiplying two classes $E_1,E_2$ one need only multiply their representatives $x,y$ and the product $xy$ will be the representative of the class $E_1E_2$.
This freedom to choose any representative stems from the fact that if two elements were related (i.e. $ab^{-1}\in H$) than their classes were the same.
By the way such nice subgroups are called normal subgroups and the classes are called right cosets.
