The kernel,K, of a homomorphism f: K == {g in G:f(g)=e'}
K is a normal subgroup of G.
proof: forall g,g1,g2 in G:
* f(e)f(g)=f(eg) = f(g) => e in K
* f(g^)f(g)=f(g^g) = f(e) = e'
* g1,g2 in K => g1g2 in K
* associativity satisfied
Proof of normality:
* f(g^Kg)=f(g^)f(K)f(g)=f(g)^ e' f(g) = e'