Two finitely generated abelian groups $G,H$ have the same finite quotients (up to isomorphism). Prove they are isomorphic.
According to an earlier result, it seems that I need to use the fact that finitely generated abelian groups are residually finite, meaning that the intersection of all normal subgroups of finite index is trivial.
Denote the two respective intersections by $N_G$ and $N_H$. I thought of trying to apply one of the isomorphism theorems (e.g., by considering $(G/N_G)/(N/N_G)$), but I can't see how they would be helpful, since both $N_H, N_G$ are trivial. What am I missing?
We know that, as $G$ and $H$ are both finitely generated, we have $G \cong T(G) \times S(G)$, where all elements in $T(G)$ have finite order and every non-trivial element in $S$ has infinite order. Same with $H$, $H \cong T(H) \times S(H)$. Now by our assumption, $T(G)\cong T(H)$ and $S(G) \cong S(H)$, hence $G\cong H$. (We actually didn't use the fundamental theorem of finitely generated abelian groups.)