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? 
 A: If $H$ is what is called a normal subgroup then the intuition is the following: There exists a homomorphism (that is, a map which preserves the group structure) $\phi: G\rightarrow K$ such that $a\sim b$ if and only if $\phi(a)=\phi(b)$. The subgroup $H$ is precisely the elements of $G$ which map to the identity of $K$, $\phi(h)=1_K$ for all $h\in H$. This is a very fundamental and important notion in group theory.
For example, $\mathbb{Z}$ forms a group under addition, and there is a subgroup $H_n=\{ni; i\in\mathbb{Z}\}$. So addition modulo $n$ corresponds to a homomorphism from $\mathbb{Z}$ with kernel $H_n$.
A normal subgroup, written $H\lhd G$, is one where $g^{-1}Hg=H$ for all $g\in G$, and a (group) homomorphism is a map $\phi: G\rightarrow K$ such that $\phi(g)\phi(h)=\phi(gh)$. The theorem which connects these two concepts is called the First Isomorphism Theorem.
A: 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.
A: The basic idea of that equivalence relation is that the equivalence classes of $\sim$ form what are called cosets. One way to talk about cosets is as equivalence classes of the relation you're talking about. The other way is to define, with $H \subset G$ a subgroup and $g \in G$, $gH = \{gh : h \in H\}$, i.e., the set of all products $gh$, with $g$ fixed and $h$ ranging over the subgroup $H$. It's worthwhile, and not too hard, to show that those two definitions of cosets are equivalent. 
As an example, let $G = \mathbb{Z}$, and $H = 10 \mathbb{Z}$. Then the coset $(1 + H) = \{1, 11, 21, 31, \dots$ } (note that we're using additive notation.) Hopefully that clears up some of your confusion. Assuming you're in an algebra class now, you'll be using cosets extensively, so you'll see their use soon.
A: The relation $a\sim b$, defined as $ab^{-1}\in H$, is true if and only if $a$ and $b$ are both in the same left coset of $H$.
$ab^{-1}\in H$ is the same as $ab^{-1}=h$ for some $h\in H$.  And $ab^{-1}=h$ if and only if $a=hb$.  That means $a\in\{gb : g\in H\}$, i.e. $a$ is in the right coset of $H$ to which $b$ belongs.
Then one has $(ab^{-1})^{-1} = h^{-1} \in H$ since $H$, being a subgroup, is closed under inversion, and that's the same as $ba^{-1}\in H$, or $b\sim a$.
If you know that distinct right cosets of $H$ are disjoint and their union is the whole group, then there you have a partition, and hence and equivalence relation.
BTW, you don't need to write $a$~$b$.  TeX and LaTeX and MathJax are not so primitive.  I changed it to $a\sim b$ in your question.
