Proof
The important steps in the proof are the following two.
The map f' is well defined.
f' is injective.
Suppose
g' 
gN.
Then there is
n N with g' = gn.
Consequently,
f(g') = f(gn) = f(g)f(n) =
f(g)e = f(g).
gN.
Then there is
n N with g' = gn.
Consequently,
f(g') = f(gn) = f(g)f(n) =
f(g)e = f(g).
Thus f'(gN) does not depend on the choice of g' gN.
Suppose g G satisfies
f'(gN) = e.
Then f(g) = e, so g N,
whence gN = N, which is the unit element of G/N.
We have shown that Ker(f') is trivial. By a previous theorem, f'
is injective.
G satisfies
f'(gN) = e.
Then f(g) = e, so g N,
whence gN = N, which is the unit element of G/N.
We have shown that Ker(f') is trivial. By a previous theorem, f'
is injective.