Proof
Clearly, p is surjective. We need to verify:
p is a morphism of groups;
Ker(p) = N.
For all g, h 
G we
have
p(gh) = ghN = gN
* hN = p(g)* p(h).
G we
have
The kernel of p consists of the elements g
G satisfying p(g) = N, that is,
gN = N. Since gN = N is equivalent to
g N, we find ker(p) =
N.
G satisfying p(g) = N, that is,
gN = N. Since gN = N is equivalent to
g N, we find ker(p) =
N.