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? 

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.
