Let us write $\phi \left({G}\right)$ to denote the homomorphic image of $\phi$. From Image of Group Homomorphism is Subgroup, $\phi \left({G}\right)$ is a subgroup of $\left({H, +_H}\right)$. For any $\phi \left({g}\right)$ and $\phi \left({g'}\right)$ in $\phi \left({G}\right)$, we have: {{begin-eqn}} {{eqn|l = \phi \left({g}\right) +_H \phi \left({g'}\right)      |r = \phi \left({g +_G g'}\right)      |c = $\phi$ is a homomorphism }} {{eqn|r = \phi \left({g' +_G g'}\right)      |c = $\left({G, +_G}\right)$ is an abelian group }} {{eqn|r = \phi \left({g'}\right) +_H \phi \left({g}\right)      |c = $\phi$ is a homomorphism }} {{end-eqn}} hence $\phi \left({G}\right)$ is an abelian group. Now we can turn to showing that $\phi \left({G}\right)$ is an $R$-module. To do this, we take the $R$-module axioms in turn. === Proof of $(1)$ === It is to be shown that for all $\lambda \in R$ and $\phi \left({g}\right), \phi \left({g'}\right) \in \phi \left({G}\right)$: :$\lambda \circ_H \left({\phi \left({g}\right) +_H \phi \left({g'}\right)}\right) = \left({\lambda \circ_H \phi \left({g}\right)}\right) +_H \left({\lambda \circ_H \phi \left({g'}\right)}\right)$ Compute, using that $\phi$ is a homomorphism repetitively: {{begin-eqn}} {{eqn|l = \lambda \circ_H \left({\phi \left({g}\right) +_H \phi \left({g'}\right)}\right)      |r = \lambda \circ_H \phi \left({g +_G g'}\right) }} {{eqn|r = \phi \left({\lambda \circ_G \left({g +_G}\right)}\right) }} {{eqn|r = \phi \left({\left({\lambda \circ_G g}\right) +_G \left({\lambda \circ_G g'}\right)}\right)      |c = $G$ is an $R$-module }} {{eqn|r = \phi \left({\lambda \circ_G g}\right) +_H \phi \left({\lambda \circ_G g'}\right) }} {{eqn|r = \left({\lambda \circ_H \phi \left({g}\right)}\right) +_H \left({\lambda \circ_H \phi \left({g'}\right)}\right) }} {{end-eqn}} {{qed|lemma}} === Proof of $(2)$ === It is to be shown that for all $\lambda, \mu \in R$ and $\phi \left({g}\right) \in \phi \left({G}\right)$: :$\left({\lambda +_R \mu}\right) \circ_H \phi \left({g}\right) = \left({\lambda \circ_H \phi \left({g}\right)}\right) +_H \left({\mu \circ_H \phi \left({g}\right)}\right)$ Compute, using that $\phi$ is a homomorphism repetitively: {{begin-eqn}} {{eqn|l = \left({\lambda +_R \mu}\right) \circ_H \phi \left({g}\right)      |r = \phi \left({\left({\lambda +_R \mu}\right) \circ_G g}\right) }} {{eqn|r = \phi \left({\left({\lambda \circ_G g}\right) +_G \left({\mu \circ_G g}\right)}\right)      |c = $G$ is an $R$-module }} {{eqn|r = \phi \left({\lambda \circ_G g}\right) +_H \phi \left({\mu \circ_G g}\right) }} {{eqn|r = \left({\lambda \circ_H \phi \left({g}\right)}\right) +_H \left({\mu \circ_H \phi \left({g}\right)}\right) }} {{end-eqn}} {{qed|lemma}} === Proof of $(3)$ === It is to be shown that for all $\lambda, \mu \in R$ and $\phi \left({g}\right) \in \phi \left({G}\right)$: :$\left({\lambda \times_R \mu}\right) \circ_H \phi \left({g}\right) = \lambda \circ_H \left({\mu \circ_H \phi \left({g}\right)}\right)$ Compute, using that $\phi$ is a homomorphism repetitively: {{begin-eqn}} {{eqn|l = \left({\lambda \times_R \mu}\right) \circ_H \phi \left({g}\right)      |r = \phi \left({\left({\lambda \times_R \mu}\right) \circ_G g}\right) }} {{eqn|r = \phi \left({\lambda \circ_G \left({\mu \circ_G g}\right)}\right)      |c = $G$ is an $R$-module }} {{eqn|r = \lambda \circ_H \phi \left({\mu \circ_G g}\right) }} {{eqn|r = \lambda \circ_H \left({\mu \circ_H \phi \left({g}\right)}\right) }} {{end-eqn}} {{qed|lemma}} Having verified that $\phi \left({G}\right)$ satisfies the three axioms, we conclude it is an $R$-module. {{qed}}