An example of when a product in a category may not exist. My question is about the category of finitely generated Abelian groups; in particular, I want to show, by definition, that there exists a set of objects $T$ in this category for which there is no product. 
I think I can choose this set $T$ to consist of infinitely many copies of $\mathbb{Z}$. Having assumed that the product actually exists, say $P$, next I have to pick one particular object and a family of morphisms, (probably $\mathbb Z$ itself?) such that if the corresponding diagrams commute, it would lead to a contradiction regarding finitely generatedness of $P$. 
Any hint would be very much appreciated!
 A: Let $(X_i)$ be a diagram of finitely generated abelian groups and assume that it has a limit $\lim_i X_i$ in the category of finitely generated abelian groups. I claim that this is also a limit in the category of abelian groups. In other words, the projections $\lim_i X_i \to X_i$ induce for every abelian group $A$ an isomorphism of sets
$$\hom(A,\lim_i X_i) \to \lim_i \hom(A,X_i).$$
In fact, we may write $A$ as a directed colimit of finitely generated abelian groups $A_j$. Then the left hand side becomes
$$\lim_j \hom(A_j,\lim_i X_i)$$
and the right hand side becomes
$$\lim_i \lim_j \hom(A_j,X_i) \cong \lim_j \lim_i \hom(A_j,X_i).$$
Here we use the general principle that limits commute with limits.
But $$\hom(A_j,\lim_i X_i) \to \lim_i \hom(A_j,X_i)$$ is an isomorphism by assumption. In the limit, we get our desired isomorphism.
It follows that a diagram of finitely generated abelian groups has a limit iff the limit of the underlying abelian groups is finitely generated.
In general one has to be more careful when one wants to compute limits and colimits in subcategories. Consider the category of torsion abelian groups. At first one might think that it has no products, since for example $\prod_{n > 0} \mathbb{Z}/n\mathbb{Z}$ is not torsion (and many mathematicians would agree with this argument; it has also appeared in books). But this only tells us that the product of the underlying abelian groups doesn't work. In fact there is a product, namely the torsion subgroup of the product of the underlying abelian groups. The same works for limits.
