Is it true that $G\cong\bigoplus_{\alpha\in A}\langle g_\alpha\rangle$ implies $G=\langle g_{\alpha}\mid \alpha\in A\rangle$?

Question

Is it true that for a free abelian group $G$ we have that if $(g_\alpha)_{\alpha\in A}\subset G$ is a collection of elements of $G$ and $G\cong\bigoplus_{\alpha\in A}\langle g_\alpha\rangle$ and $\langle g_\alpha\rangle\cong\mathbb{Z}$ then $G=\langle g_{\alpha}\mid \alpha\in A\rangle$? I can prove that if $\varphi$ is an isomorphism between $G$ and $\bigoplus_{\alpha\in A}\langle g_\alpha\rangle$ then $G=\langle \varphi^{-1}\left(i_{\alpha}(g_{\alpha})\right)\mid \alpha\in A\rangle$ where $i_{\alpha}(g_{\alpha})\in\bigoplus_{\alpha\in A}\langle g_\alpha\rangle$ is the vector with all entries equal to the identity element except for $\alpha$ where the entry is $g_\alpha$. It would be reasonable to think that $\varphi^{-1}\left(i_{\alpha}(g_{\alpha})\right)=g_\alpha$ but I don't see wether this is always true and if so how to prove it. I'm somewhat clueless.