direct sums of direct products of abelian groups 
I would like to prove that we cannot have an isomorphism $\oplus_i(\prod_jG_{i,j})\cong\prod_j(\oplus_iG_{i,j})$ where the sums and products are infinite and $G_{i,j}$ are non zero abelian groups for infinite indices

I am familiar with the category language, so I tried to prove it using the fact that if by contradiction the statement is true, then the product on the RHS must satisfy the universal property of the coproduct. Without success, I looked for a suitable algebraic invariant but I didn't find one.
 A: The statement relevant to the related question is not that these groups can never be abstractly isomorphic (I think they can if e.g. all of the $G_{i, j}$ are copies of $\mathbb{Q}$) but that the natural map from the LHS to the RHS is (usually) not an isomorphism. 
This map is defined as follows. By the universal property, giving a map into a product $\prod_j$ is the same thing as giving a collection of maps into each term of the product. And the map
$$\bigoplus_i \prod_j G_{i, j} \to \bigoplus_i G_{i, j}$$
is given on each component of the direct sum by the projection $\prod_j G_{i, j} \to G_{i, j}$. 
Why might this map fail to be an isomorphism? Consider the sequence of sequences of groups
$$G_{i, j} = \begin{cases} \mathbb{Q} \text{ if } j \le i \\ 0 \text{ otherwise} \end{cases}.$$
Here the index set is $\mathbb{Z}_{\ge 0}$. Hence $G_{0, j}$ is a single copy of $\mathbb{Q}$ and then zeroes, $G_{1, j}$ is two copies of $\mathbb{Q}$ and then zeroes, etc. So $\prod_j G_{i, j}$ is a finite-dimensional $\mathbb{Q}$-vector space for all $i$, and hence $\bigoplus_i \prod_j G_{i, j}$ is a countable-dimensional $\mathbb{Q}$-vector space. That's the LHS. 
As for the RHS, by construction $\bigoplus_i G_{i, j}$ is always a countable-dimensional $\mathbb{Q}$-vector space, so the product $\prod_j \bigoplus_i G_{i, j}$ is an uncountable-dimensional $\mathbb{Q}$-vector space (because of its cardinality). So not only is the natural map not an isomorphism but the two sides are not even abstractly isomorphic. 
