# Abstract Algebra/Group Theory/Homomorphism/Kernel of a Homomorphism is a Normal Subgroup

## Theorem

Let f be a homomorphism from group G to group K. Let eK be identity of K.

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

## Proof

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