Let *f* be a homomorphism from group G to group K.
Let *e*_{K} be identity of K.

- ${\text{ker}}~f$ is a normal subgroup.

$f(g\ast n\ast g^{-1})=f(g)\circledast f(n)\circledast f(g^{-1})=f(g)\circledast e_{K}\circledast f(g^{-1})=f(g)\circledast f(g^{-1})=f(g\ast g^{-1})=f(e_{G})=e_{K}$