Using a right-exact functor to show isomorphism

Question

I'm working on a homework problem which asks me to show that $A\otimes \mathbb{Z}_m \cong A/mA$ for any abelian group $A$. We are also given a hint to explore right exactness of the tensor product. However, I can't see at all why this hint is relevant. I have no idea how one could derive an isomorphism from an exact sequence. A push in the right direction would be really helpful.

Answer 1

Thanks to the first hint above, I figured it out. For anyone else working on this, consider the sequence $\mathbb{Z}\overset{f}{\to}\mathbb{Z}\overset{\pi}{\to}\mathbb{Z}_m\to 0$ with the appropriate rule of assignment for $f$ which makes the sequence exact. Utilize right exactness of the tensor product with $A$ and the first isomorphism theorem to derive $A/mA\cong (A\otimes\mathbb{Z})/(mA\otimes\mathbb{Z})\cong A\otimes\mathbb{Z}_m$.