To be more specific: Let $G$ be a group and let $G'$ be an abelian group. Let $\phi: G \mapsto G'$ be a group homomorphism, and let $K = \text{Ker}\phi$. Further, let $H$ be a subgroup of $G$ containing $K$. Prove that $H \trianglelefteq G$.
My professor provided an explanation that makes sense:
"It is sufficient to show that, if $ghg^{-1} \in H$, $\forall g \in G, h \in H$, then $H \trianglelefteq G$. Then, suppose $g \in G$ and $h \in H$. \begin{align*} ghgh^{-1}K &= (gH)(hK)(g^{-1}K)(h^{-1}K) \\ &= (gK)(hK)(gK)^{-1}(hK)^{-1} \\ &= (gK)(gK)^{-1}(hK)(hK)^{-1} \text{ since G/K is abelian }\\ &= (eK)(eK) \\ &= K \end{align*} Therefore, $ghg^{-1}h^{-1} \in K$. Since $K \leq G$, $ghg^{-1}h^{-1} \in H$. Further, say $ghg^{-1}h^{-1} = h'$ for some $h' \in H$. Then, $ghg^{-1} = h'h \in H$."
MY QUESTION: This makes sense to me, but I am wondering if there is a way to use the fact that "A subgroup $H$ of $G$ is normal in $G$ if there is a group homomorphism $\psi: G \mapsto M$ such that $\ker\psi= H$." It seems to me that if you could find a way to show that $\ker\psi= H$ then this proof would be a lot easier. Perhaps this is not possible but I am asking the people on this forum as they are a lot smarter than I am. Thanks in advance.