Finitely generated free group is a cogroup object in the category of groups I am trying to show that every finitely generated free group is a cogroup object in the category of groups. (Note I believe that this is also true for non-finite free groups, but that is probably much harder to prove - I think it is theorem of Kan's)
I know that the coproduct in the category of groups is just the free product. So then is it just basically show that the free product satisfies the usual diagrams for co-associativity, co-identity and co-inverse?
I guess I am confused by the map $C \to C \coprod C$ (can't get my head around co-multiplication yet). 
This was OK in the category of abelian groups - as the coproduct is the direct product and the there is an obvious map $c \mapsto (c,c)$. But (and maybe I am struggling with the notion of a free product here), if I tried something similar - take an element from a finitely generated free group (say on $n$ generators) and then take the coproduct with itself - what does that give? Wouldn't it just leave a single element? (As all adjacent elements belong to the same set the reduced word cancels everything out?)
Not looking for an answer, I guess just some clarity - I am sure I am not understand something properly.
(I guess one could also do this by showing that $\mbox{Hom}(\quad , G)$ take a group structure?)
Edit: Actually is $C \ast C = C$? ($\ast$ is the free product) (No - as Theo points out!)
 A: When you take a free product $C*C$, words from the second copy of $C$ don't cancel with words from the first copy. So for example to think of $\mathbb Z*\mathbb Z$, suppose that $\mathbb Z$ is multiplicatively generated by $t$. Then it is helpful to rename the generator on the second copy of $\mathbb Z$ to emphasize the fact that they don't cancel. Say the second $\mathbb Z$ is generated by $s$. Then an element of the free product would look like $s^{k_1}t^{\ell_1}\cdots s^{k_m}t^{\ell_m}$, where adjacent powers of $s$ or $t$ can be absorbed together.
So in your definition of the coproduct, it might be helpful to think of the second $C$ as being an isomorphic copy of $C$, say $C'$. Then $c\mapsto c*c'$ doesn't involve any cancellation.
A: I think the easiest way to do this is indeed to show that $\text{Hom}(F_n, G) \cong G^n$ naturally inherits a group structure for a free group $F_n$ on $n$ letters and a group $G$. In fact it naturally inherits the product group structure. The generalization to arbitrary free groups is clear. (This is a simple example where keeping the universal property in mind makes things clearer than trying to work from explicit constructions.)
A: Sketch of an argument:
The forgetful functor from the category of groups to the category of sets is always group-object-valued, as the group objects in the category of sets are precisely the groups.
Hence, the left adjoint of the forgetful functor (if there is one!) must be always co-group-object-valued in the category of groups.
As the free functor is the left adjoint of the forgetful functor, every free group is a co-group object in the category of groups.
