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'

source

jl@crush.caltech.edu index

representation

homomorph

SU