Let $Q$ be the mapping defined as: :$\forall H \le \mathbb H_1: Q \left({H}\right) = \left\{{\phi \left({h}\right): h \in H}\right\}$ Let $H$ be a subgroup of $G_1$ such that $K \subseteq H$. From Group Homomorphism Preserves Subgroups, $\phi \left({H}\right)$ is a subgroup of $G_2$. This establishes that $Q$ is actually a mapping. Let $N \lhd G_1$. From Group Epimorphism Preserves Normal Subgroups, $\phi \left({N}\right)$ is a normal subgroup of $G_2$. This establishes that: :$\forall N \lhd G_1: Q \left({N}\right) \lhd G_2$ Next it is shown that $Q$ is a bijection. === Injective Nature of $Q$ === Let $H, J \in \mathbb H_1$. Let $Q \left({H}\right) = Q \left({J}\right)$. Let $h \in H$. {{begin-eqn}} {{eqn | l=\phi \left({h}\right)       | o=\in       | r=Q \left({H}\right)       | c= }} {{eqn | ll=\implies       | l=\phi \left({h}\right)       | o=\in       | r=Q \left({J}\right)       | c= }} {{eqn | ll=\implies       | lo=\exists j \in J:       | l=\phi \left({j}\right)       | r=\phi \left({h}\right)       | c= }} {{eqn | ll=\implies       | l=e_{G_2}       | r=\left({\phi \left({j}\right)}\right)^{-1} \phi \left({h}\right)       | c=definition of inverse element }} {{eqn | r=\phi \left({j^{-1} }\right) \phi \left({h}\right)       | c=Group Homomorphism Preserves Inverses }} {{eqn | r=\phi \left({j^{-1} h}\right)       | c=morphism property of $\phi$ }} {{eqn | ll=\implies       | l=j^{-1} h       | o=\in       | r=K       | c=definition of kernel }} {{eqn | ll=\implies       | lo=\exists k \in K:       | l=j^{-1} h       | r=k       | c= }} {{eqn | ll=\implies       | l=h       | r=j k       | c= }} {{eqn | o=\in       | r=J       | c=as $K \subseteq J$ and so $k \in j$ }} {{eqn | ll=\implies       | l=H       | o=\subseteq       | r=J       | c= }} {{end-eqn}} A similar argument shows that $J \subseteq H$. So by definition of set equality: : $H = J$ Thus: : $Q \left({H}\right) = Q \left({J}\right) \implies H = J$ So by definition, $Q$ is injective. {{qed|lemma}} === Surjective Nature of $Q$ === Now let $N' \in \mathbb H_2$. By definition of $\mathbb H_2$, $N'$ is a subgroup of $G_2$. Let $N = \left\{{x: \phi \left({x}\right) = N'}\right\}$. We have from Identity of Subgroup that $e_{G_2} \in N'$. Thus by definition of kernel, $K \subseteq N$. Now suppose $\phi \left({x}\right), \phi \left({y}\right) \in N'$. Then: {{begin-eqn}} {{eqn | l=\phi \left({x y^{-1} }\right)       | r=\phi \left({x}\right) \phi \left({y^{-1} }\right)       | c= }} {{eqn | r=\phi \left({x}\right) \phi \left({y}\right)^{-1}       | c= }} {{eqn | o=\in       | r=N'       | c=One-Step Subgroup Test }} {{eqn | ll=\implies       | l=x y^{-1}       | o=\in       | r=N }} {{end-eqn}} So by the One-Step Subgroup Test, $N$ is a subgroup of $G_1$. It has been established that $K \subseteq N$, and so $N \in \mathbb H_1$. Thus it follows that for all $N' \in \mathbb H_2$ there exists $N \in H_1$ such that $Q \left({N}\right) = N'$. So $Q$ is a surjection. {{qed|lemma}} So $Q$ has been shown to be both an injection and a surjection, and so by definition is a bijection. Finally, it can then be shown that if $N'$ is normal in $G_2$, it follows that $N = Q^{-1} \left({N'}\right)$ is normal in $G_1$. This establishes that:  :$\forall N \lhd G_2: Q^{-1} \left({N}\right) \lhd G_1$ {{finish|if $N'$ is normal in $G_2$, it follows that $N {{=}} Q^{-1} \left({N'}\right)$ is normal in $G_1$.}} {{qed}}