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.


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

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