Editing Quotient group (section)

=== Definition ===
Let <math>N</math> be a normal subgroup of a group {{tmath|1= G }}. Define the set <math>G\,/\,N</math> to be the set of all left cosets of <math>N</math> in {{tmath|1= G }}. That is, {{tmath|1= G\,/\,N = \left\{aN: a \in G\right\} }}. 

Since the identity element {{tmath|1= e \in N }}, {{tmath|1= a \in aN }}. Define a binary operation on the set of cosets, {{tmath|1= G\,/\,N }}, as follows. For each <math>aN</math> and <math>bN</math> in {{tmath|1= G\,/\,N }}, the product of <math>aN</math> and {{tmath|1= bN }}, {{tmath|1= (aN)(bN) }}, is {{tmath|1= (ab)N }}. This works only because <math>(ab)N</math> does not depend on the choice of the representatives, <math>a</math> and {{tmath|1= b }}, of each left coset, <math>aN</math> and {{tmath|1= bN }}. To prove this, suppose <math>xN = aN</math> and <math>yN = bN</math> for some {{tmath|1= x, y, a, b \in G }}. Then
: <math display="inline">(ab)N = a(bN) = a(yN) = a(Ny) = (aN)y = (xN)y = x(Ny) = x(yN) = (xy)N .</math>

This depends on the fact that {{tmath|1= N }} is a normal subgroup. It still remains to be shown that this condition is not only sufficient but necessary to define the operation on {{tmath|1= G\,/\,N }}.

To show that it is necessary, consider that for a subgroup <math>N</math> of {{tmath|1= G }}, we have been given that the operation is well defined. That is, for all <math>xN = aN</math> and {{tmath|1= yN = bN }} for {{tmath|1= x, y, a, b \in G, \; (ab)N = (xy)N }}.

Let <math>n \in N</math> and {{tmath|1= g \in G }}. Since {{tmath|1= eN = nN }}, we have {{tmath|1= gN = (eg)N = (eN)(gN) = (nN)(gN) = (ng)N }}.

Now, <math>gN = (ng)N \Leftrightarrow N = (g^{-1}ng)N \Leftrightarrow g^{-1}ng \in N, \; \forall \, n \in N</math> and {{tmath|1= g \in G }}.

Hence <math>N</math> is a normal subgroup of {{tmath|1= G }}.

It can also be checked that this operation on <math>G\,/\,N</math> is always associative, <math>G\,/\,N</math> has identity element {{tmath|1= N }}, and the inverse of element <math>aN</math> can always be represented by {{tmath|1= a^{-1}N }}. Therefore, the set <math>G\,/\,N</math> together with the operation defined by <math>(aN)(bN) = (ab)N</math> forms a group, the quotient group of <math>G</math> by {{tmath|1= N }}.

Due to the normality of {{tmath|1= N }}, the left cosets and right cosets of <math>N</math> in <math>G</math> are the same, and so, <math>G\,/\,N</math> could have been defined to be the set of right cosets of <math>N</math> in {{tmath|1= G }}.