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!)

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    $\begingroup$ Concerning the edit: certainly not (if $C$ is f.g.), see the wikipedia entry on free groups (rank!) $\endgroup$ – t.b. Apr 19 '11 at 10:00
  • $\begingroup$ @Theo I guess I didn't really think that is true! But this is certainly true for groups in general (also from Wikipedia) if $\langle R_G | S_H \rangle$ and $\langle R_H | S_H \rangle$ are group presentations for $G$ and $H$ then $G \ast H = \langle R_G \cup R_H | S_G \cup S_H \rangle$, but I guess this fails if the groups are the same? $\endgroup$ – Juan S Apr 19 '11 at 10:09
  • $\begingroup$ No, this is true in general, but you need to observe that the union you denote by $\cup$ is in fact a disjoint union $\amalg$ of sets (and words). Concerning your actual question, I have never thought about this, but I don't think it is difficult or subtle, as soon as you understand a free group in terms of reduced words. $\endgroup$ – t.b. Apr 19 '11 at 10:11
  • $\begingroup$ By the way, have you ever thought about what you actually mean by writing $G = \langle S \vert R \rangle$? You're taking the map $F(S) \to G$ induced by the map $S \to G$ via the universal property of $F(S)$ and then you prove that this induces an isomorphism $F(S)/\langle\langle R \rangle\rangle \to G$, where $\langle \langle R \rangle\rangle$ is the normal subgroup of $F(S)$ generated by the words in $R$. $\endgroup$ – t.b. Apr 19 '11 at 10:15
  • $\begingroup$ @Theo - OK, thanks - I really should have read the next sentence in wikipedia $\endgroup$ – Juan S Apr 19 '11 at 10:15

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.

  • $\begingroup$ thanks that helps a lot. Actually that makes good sense, when I just think of $\mathbb{Z} \ast \mathbb{Z}$ as the fundamental group of a bouquet of two circles - which clearly has two generators $\endgroup$ – Juan S Apr 19 '11 at 10:31

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.)

  • $\begingroup$ thanks I do suspect that this is the solution the question was intending. Pardon my ignorance, but is this context what do you meant by $G^n$? I actually wrote it up wrong I believe we should consider $\mbox{Hom}(G, \quad )$ (I've just found your post here qchu.wordpress.com/2011/01/21/structures-on-hom-sets - which looks very useful) $\endgroup$ – Juan S Apr 20 '11 at 2:06
  • $\begingroup$ @Qwirk: I mean the product of $n$ copies of $G$. A cogroup object $C$ in a category is precisely an object such that $\text{Hom}(C, D)$ inherits a natural group structure for any $D$, not the other way around. $\endgroup$ – Qiaochu Yuan Apr 20 '11 at 4:06

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.


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