Let be a Group and a Normal subgroup of . Then we may define the quotient group to be the set of left cosets of in , with the group operation that . This is well-defined if and only if is normal.
Proof
normal implies well-defined
Recall that is well-defined if "it doesn't matter which way we represent a coset": whichever coset representatives we use, we get the same answer.
Suppose is a normal subgroup of . We need to show that given two representatives of a coset, and given representatives of another coset, that .
So given an element of , we need to show it is in , and vice versa.
Let ; we need to show that , or equivalently that .
But because ; let . Similarly because ; let .
Then we need to show that , or equivalently that .
Since is closed under conjugation and , we must have that ; and since and is closed under multiplication, we must have as required.
well-defined implies normal
Fix , and consider . Since the quotient is well-defined, this is , which is or (since , because is a subgroup of and hence is closed under the group operation). But that means is the identity element of the quotient group, since when we added it to we obtained itself.
That is, . Therefore .
Since this reasoning works for any , it follows that is closed under conjugation by elements of , and hence is normal.