Categories in which Coproduct does not coincide with Product According to “An Introduction to  Homological Algebra” written by Rotman, he said that:
In Category Groups, Coproducts and Products are not same, as the coproduct of G,H is their free product, but product is their direct product
I wonder the details in this claim and could anyone give more examples of this situation, and give some explanations?
Thanks in advance!
 A: Rotman's remark is a litte vague. He defines the coproduct of objects $A, B$ as a triple $(A \sqcup B, \alpha : A \to A \sqcup B, \beta : B \to A \sqcup B)$ having a certain universal property. You may think of $A \sqcup B$ as the "coproduct object", but be aware that this object on its own does not contain any information about the morphisms $\alpha, \beta$ which are essential constituents of the coproduct. See for example the proof of Proposition 5.1 where Rotman explicitly says "The statement of the proposition is not complete, for a coproduct requires morphisms $\alpha$ and $\beta$."
In many cases the coproduct object directy suggests what the morphisms $\alpha, \beta$ look like (i.e. there is only one natural choice for such morphisms), and that may be the reason why one sometimes laxly denotes the coproduct object as the coproduct.
Similarly, the product of objects $A, B$ is a triple $(A \sqcap B, p :  A \sqcap B \to A, q : A \sqcap B \to B)$ having a certain universal property.
The concepts of coproduct and product are dual (in the sense that arrows are reversed in their definitions) and it should be clear that on the level of the complete structures $(A \sqcup B, \alpha, \beta)$ and $(A \sqcap B, p, q)$ it does not make any sense to say that they are the same.
However, it may be true that coproduct and product objects are the same, or perhaps we should better say isomorphic. But even if that is the case, there is no way to automatically find $p, q$ if we know $\alpha, \beta$ (or conversely). For such cases one introduces the categorical concept of the  biproduct of $A, B$ which is tupel $(A \diamondsuit B, \alpha, \beta, p, q)$ such that $(A \diamondsuit B, \alpha, \beta)$ is the coproduct of $A, B$ and $(A \diamondsuit B, p, q)$ is the product of $A, B$. See https://en.wikipedia.org/wiki/Biproduct.
In the categories of Abelian groups, $R$-modules and vector spaces over a field $K$ biproducts exist (this is more generally true in additive categories having (co)products). In these categories coproducts are written as $A \oplus B$ (direct sum) and products as $A \times B$ (Cartesian product). This may be confusing for beginners since we have $A \oplus B = A \times B$ - so why two notations for the same thing? The reason is that the distinct notations refer to distinct universal properties - and these are frequently not explicitly mentioned in beginner's courses. That there is a difference becomes clear indirectly when infinite direct sums and infinite Cartesian products are introduced.
In the category of groups it is not true. The coproduct is the free product $A * B$ which is never isomorphic to the product $A \times B$ unless one of $A, B$ is the trivial group. For example, the free product of finite nontrivial groups is always infinite. To understand details you have to consult a textbook on group theory, Rotman is not an adequate source for free products.
Other (simpler) examples are the categories of sets and topological spaces. Here coproducts are disjoint unions $A \sqcup B$ and products are Cartesian products $A \times B$ which are in general not isomorphic.
The category of sets is a nice example to understand the difference between (co)products and (co)product objects. Consider two sets $A, B$ which are not both finite. Then $A \sqcup B$ and $A \times B$ have the same cardinality, i.e. are isomorphic as sets. This means that we could take $A \times B$ as the coproduct object, but this does in general not allow to explicitly specify the inclusions $\alpha : A \to A \times B$ and $\beta : B \to A \times B$. Except for simple cases like $A = B = \mathbb N$ (where we can tediously construct a bijection $\mathbb N \sqcup \mathbb N \to \mathbb N \times \mathbb N$) we need the axiom of choice and do not see anything explicitly.
A: Let $k$ be a fixed field. It is well known that in the category $\textbf{Vect}(k)$ of all vector spaces over $k$ the categorical product coincides with the direct product $\prod V_i$ while the categorical coproduct coincides with the direct sum $\bigoplus V_i$. In particular in the category $\textbf{Vect}_{fin}(k)$ of all finite dimensional vector spaces over $k$ the product and coproduct coincide (if they exist).
