A question about groups as categories. I am currently reading about categories. Groups as categories are so described:
Let $G$ be a finite group. Then we can form an object $B_G$, which is nothing but $*$, an arbitrary element. $Hom(B_G,B_G)$ contains all elements $g\in G$ such that $g$ maps $*\to *$. Also, the operation on $Hom(B_G,B_G)$ is nothing but the binary operation on $G$. 
I don't understand how $g\in G$ maps $*\to *$. Wouldn't it be more intuitive to have $G$ as the object $B_G$ and all its automorphisms as $Hom(B_G,B_G)$, with composition as the operator defined for automorphisms?
Thanks in advance!
 A: Well, the observation is that a group $G$ may be identified with an abstract category which has only one object and which the set of endomorphisms of this object coincides with $G$.
You have to keep in mind that a category is abstractly defined as a class of objects endowed with sets of morphisms (for each ordered pair of objects) and a composition satisfying some axioms ( see http://en.wikipedia.org/wiki/Category_theory ).
This definition is as abstract as the definition of a group. We don't say anything about the nature of an element of group. Similarly, we don't say anything about the nature of the objects or the nature of the morphisms in a category. We say that a group is a set endowed with a binary operation satisfying some conditions. And, on the other hand, in the definition of a category, for each ordered pair $ (A,B) $ of objects in a given category, we have a set of morphisms $ Hom(A,B) $.
So, you may define abstract categories, as you can define abstract groups (we can define a category defining the composition, the class of objects and the sets of morphisms).
Given a group $G$ , you may construct a category with only one object $ \ast $ such that $ Hom (\ast , \ast ) = G $ and the composition is the product of the group $ G $.
To prove this, you need only to verify whether the axioms of a category are satisfied.... Indeed, you may prove that the category of categories with only one object and which all morphisms are isomorphisms is equivalent to the category of groups. 
Furthermore, it is possible to construct a "more concrete" category from a group $ G $. Using the Cayley theorem, we may prove that each group $G$ is isomorphic to a subgroup of Bijections(G). So, you may consider the category with only one object $G$ and the morphisms being the maps $ f_g: G\to G $, $f_g(h) = g h $ (for each $g\in G $).
A: A group "is the same thing as a category with one object, such that all morphisms are invertible." By this I mean given a group $G$ there is a category with one object X with Hom(X, X) = G, unique up to isomorphism [not just equivalence!] of categories. Conversely, given a category with one object such that all morphisms are invertible, the (only) hom set in the category is certainly a group.
I've never heard of giving X the same name as the multiplication operation on G. That's an odd convention -- are you sure you are reading it right? I am sure that if it is in a textbook, there must be a cool reason to think of it that way, but I can't think of one. The point is that it doesn't matter what you name the only object in the category.
