Definition of groups in a more abstract way I'm trying to understand the definition of group objects in categories, this is an extract of Paolo Aluffi's book:


QUESTIONS


*

*Can I say that $e(1)$ is the identity in our group $G$ we have just built?

*Why $1$ should be a final object? in another words, why the morphism $G\to 1$ have to be unique?

*Why this category has to have finite products? it's not suffice just products with two components? (for the multiplication)


Thanks in advance
 A: *

*What do you mean by "$e(1)$"? The identity in your group is the morphism $e$ itself.

*In any category $C$ with a terminal object $1$, a morphism $1 \to c$ for an object $c \in C$ is called a global point of $c$. In some, but not all, categories, this often gives a reasonable "underlying set" functor $\text{Hom}(1, -) : C \to \text{Set}$, although it may not be the obvious one: for example, if $C$ is the category $G\text{-Set}$ of $G$-sets for a group $G$, then a global point is a fixed point.
So this is one reason you might not think it's totally crazy to ask that picking out a particular element should abstractly correspond to picking out a morphism from $1$. But there are at least two better reasons. 
First, $e$ should describe the outcome of taking the product of zero elements of $G$, and the domain of this operation should be the product of zero copies of $G$, which is the terminal object. 
Second, using the Yoneda lemma, you can show that a group object in a category $C$ is precisely the same thing as a "representable presheaf of groups on $C$," or more precisely a functor $C^{op} \to \text{Grp}$ which is representable in the sense that it is representable after composing with the forgetful functor $\text{Grp} \to \text{Set}$. Meditate in particular on the example of the group object $\text{GL}_n(-)$ in affine schemes, which is the functor which, given a commutative ring $R$, returns the group $\text{GL}_n(R)$. 

*As Najib pointed out in the comments to Marco's answer, having a terminal object and binary products is equivalent to having all finite products. But finite products is morally the correct condition anyway, because the morally correct definition involves all finite products: see this blog post and this blog post for details. 
A: It may be useful to see an example on a more interesting category.
Let $X$ be a topological space, and consider the category of all bundles over $X$. Recall that a bundle over $X$ is the same thing as a continuous map $E \to X$; $E$ is called the "total space" of the bundle. A morphism of between two bundles $E \to X$ and $E' \to X$ is a continuous map $E \to E'$ that fills in a commutative triangle.
In this category, $\mathbf{1}$ is the bundle $1_X : X \to X$. The product of two bundles $E \to X$ and $E' \to X$ is the one whose total space is the pullback $E \times_X E'$ (and whose structure map is the obvious map to $X$).
A simple example of this is the bundle $A$ defined by the map $\pi_1 : \mathbf{R}^3 \times \mathbf{R}^3 \to \mathbf{R}^3$ given by the projection map onto the first coordinate.
The maps $\mathbf{1} \to A$ are global sections of this bundle: continuous functions $s : \mathbf{R}^3 \to \mathbf{R}^3 \times \mathbf{R}^3$ that are the identity on the first coordinate. These, of course, can be identified with the composite $\pi_2 \circ s$, which you might know these under the more familiar name of a "vector field", and you may know the total space under the more familiar name of the "tangent bundle to $\mathbf{R}^3$".
If you grind out that the usual algebraic structure on the tangent bundle (where we only consider $+$ and $-$ and not scalar multiplication, since we're talking about groups), you'll see that it is an example of this abstract group in the category of bundles over $\mathbf{R}^3$.
And the group identity is not an "element" of the total space: it is a section of the bundle. (although, you'll eventually learn that sections provide a very good notion of the concept of "element" in categories like this)
A: 1.- You can not talk about the "identity" of $G$, because $G$ may not be a set. But thats the intuition.
2.- $1$ is a final object because thats the definition, not a consequence. The morphism $G \to 1$ is unique because $1$ is a final object.
3.- The two assertions are equivalent. A category has finite products if and only if has the product of every two objects and a terminal object.
