Subgroups containing kernel of group morphism to an abelian group are normal.

Hint: $G / \ker \varphi$ is abelian. Hence $G' \le \ker \varphi$ where $G'$ is the commutator subgroup.

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$G / \ker \varphi$ is isomorphic to a subgroup of $H$ by the first isomorphism theorem, hence abelian. By the familiar property of $G'$, we have $G' \le \ker \varphi$. It follows that any subgroup $K$ that contains $\ker \varphi$ also contains $G'$. Hence $K$ is normal by another familiar property of $G'$.

If you're not familiar with the property in the last statement, here is how to show it: Since $G/G'$ is abelian and $K/G'$ is a subgroup of $G/G'$, we have $K/G' \trianglelefteq G/G'$. We conclude that $K$ is normal by the fourth isomorphism theorem.

Notice that $\varphi$ induces a bijection between the subgroups of $G$ containing $\ker \varphi$ and the subgroups of $\varphi(G) \leq H$. Therefore, if $K$ is a subgroup of $G$ containing $\ker \varphi$, it is sufficient to show that $\varphi(gKg^{-1})=\varphi(K)$ for all $g \in G$; but the previous equality is clear since $H$ is abelian.

Direct: Let $\ker(\phi) \leq N \leq G$. If $g \in G$, then $\phi(g N g^{-1})=\phi(g) \phi(N) \phi(g)^{-1} = \phi(N)$ shows $g N g^{-1} \subseteq \ker(\phi) N \subseteq N$.

(We only need that every subgroup of $H$ is normal. )