what is exactly the index of a subgroup? 
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*If we have two groups $H$ and $G$ such that $H$ is a subgroup of $G$ we define the index of $H$ in G by the number of all coset of the form $gH$ when $g$ describes $G$ , but I am confused , because if $g$ describe $G$. So, the number of all coset is the order of $G$ ? Therefore $|G:H| = \operatorname{ord}(G)$ (or we eliminate similar coset) ? 

 A: Cosets are identified as sets. The index may not be $|G|$ since $gH = g'H$ as sets even though $g \neq g'$. 
When the group $G$ is finite, you know by Legrange's theorem that $|H|$ divides $|G|$. However, in this case the index is actually $\frac{|G|}{|H|}$. However, if $G$ is infinite, it can still happen that there exists a $H$ such that the index is finite. For example, consider $\mathbb{Z}$ and the subgroup $2\mathbb{Z}$. There are two cosets, $2\mathbb{Z}$ and $1 + 2\mathbb{Z}$. 
A: A (left) coset of $H$ in $G$ is a subset of $G$. The index of $H$ in $G$ is the number of distinct (left) cosets of $H$ in $G$; remember that two sets are equal if and only if they have the same elements, regardless of the "name" we give them.
So what we are looking for is the cardinality of the set $\{gH \mid g\in G\}$, as a set. That means that putting the same element more than once doesn't change the number of elements. For example, the set $\{1,2,2,3,2,2,2,3\}$ has just three elements, because $\{1,2,2,3,2,2,2,3\} = \{1,2,3\}$ as sets. 
So you actually want to count the number of distinct sets of the form $gH$; each of these sets will, in general, have many "names": you have that the set $gH$ is the same as the set $xH$ if and only if $x^{-1}g\in H$. This will hold if $x=g$, but may also hold other times. Even though the coset has many names, you only count it once, just like listing $2$ five times in $\{1,2,2,3,2,2,2,3\}$ doesn't mean we have to count it five times.
