Let $A$ be a left $R$-module and $G$ an abelian group. Make $\rm{Hom}_{\mathbb{Z}}(R,G)$ into a left $R$-module with $$(r.\varphi )(\lambda)=\varphi(\lambda r).$$ Define a map from $\mathrm{Hom}_{\mathbb{Z}}(A,G)$ to $\mathrm{Hom}_{R}(A,\rm{Hom}_{\mathbb{Z}}(R,G))$ as follows:
given $\psi\in \mathrm{Hom}_{\mathbb{Z}}(A,G)$ define $\psi':A\rightarrow \rm{Hom}_{\mathbb{Z}}(R,G)$ by $$\psi'(a):R\rightarrow G, \hskip5mm \psi'(a)(\lambda)=\psi(\lambda a).$$ It is easy to show that $\psi'$ is additive homomorphism. My trouble is in proving that it is $R$-a module homomorphism.
$$\psi'(ra)(\lambda)=\psi(\lambda(ra))=\psi((\lambda r)a)=\psi'(a)(\lambda r).$$ I couldn't proceed here after to prove that $\psi'$ is $R$-module homomorphism. How to proceed?