I am trying to prove this result:
Let {$A_{i}$}$_{i \in I}$ be a family of abelian groups of exponent $m$ (duals will be with respect to the cyclic group of order $m$), then $(\bigoplus_{i \in I} A_{i})^{\land} \cong \prod_{i \in I}(A_{i}^{\land})$.
I wrote my proof without using the property of the direct sum, which is for any $(x_{i})_{i \in I}$ with $x_{i} \in A_{i}$ all $x_{i}=0$ but a finite number of indices. In other word, I view the direct sum as the product of groups. It is possible? Or that property is necessary for proving this result? I am quite confused about that.
And I noticed in Serge Lang's Algebra page 47, there is a theorem: If $A$ is a finite abelian group can be expressed as a product $A = B \times C$, then $A^{\land} \cong B^{\land} \times C^{\land}$. In the proof of this theorem, finiteness of $A$ is not mentioned, it is still ture for infinite abelian group?