Let $\mathsf{ProFinGrp}$ be the category of profinite groups (with continuous homomorphisms). This is equivalent to the Pro-category of $\mathsf{FinGrp}$. Notice that $\widehat{\mathbb{Z}} = \lim_{n>0} \mathbb{Z}/n\mathbb{Z}$ is a cogroup object, since it represents the forgetful functor $U : \mathsf{ProFinGrp} \to \mathsf{Set}$. Although $\mathsf{ProFinGrp}$ has no coproducts (?), the coproduct $\widehat{\mathbb{Z}} \sqcup \widehat{\mathbb{Z}}$ exists, it coincides by formal nonsense with $\widehat{F_2}$, where $F_2$ is the free group on two generators $x,y$. Let us denote the generator of $\mathbb{Z}=F_1$ by $x$. Then the comultiplication of $\widehat{F_1}$ is given by $\widehat{F_1} \to \widehat{F_2}$, $x \mapsto x * y$. There is another cogroup structure on $\widehat{F_1}$, corresponding to the opposite group. The corresponding comultiplication is $\widehat{F_1} \to \widehat{F_2}$, $x \mapsto y * x$.

Question. Are these two the only cogroup structures on $\widehat{\mathbb{Z}}$?

Here is a down-to-earth description what a cogroup structure on $\widehat{\mathbb{Z}}$ is: It is an element $m(x,y) \in \widehat{F_2}$ (a sort of "profinite word" in $x$ and $y$) such that

  • $m(x,1)=m(1,x)=x$ in $\widehat{F_1}$
  • $m(x,m(y,z))=m(m(x,y),z)$ in $\widehat{F_3}$
  • There is some $i \in \widehat{F_1}$ such that $m(x,i(x))=m(i(x),x)=1$.

This is very similar to the definition of a one-dimensional formal group law (which is a cogroup structure on $R[[t]]$ in the category of complete topological $R$-algebras).

Background: An affirmative answer would answer math.SE/656279. It is known that $\mathbb{Z} \in \mathsf{Grp}$ has only two cogroup structures.

  • $\begingroup$ How do the elements of $\widehat{F_2}$ look like explicitly? The proof for $F_2$ is quite easy using the usual normal form, once one dares to write down in detail the associativity rule ... $\endgroup$ – Martin Brandenburg Jan 30 '14 at 18:34
  • $\begingroup$ Dear Martin: Very interesting question. I believe $\widehat{F_2}$ is none other than the étale fundamental group of $\mathbf P^1-\{0, 1, \infty\}$. Deligne has a 200 page paper about it, "Le groupe fondamental de la droite projective moins trois points". I think that by the theory of dessins d'enfants, it has an action of the absolute Galois group of Q on it. In any case it is a very complicated group, I don't think it can be described very explicitly (but I don't know much more about it than what I just said, so I may be wrong). $\endgroup$ – Bruno Joyal Feb 4 '14 at 18:48
  • $\begingroup$ Yes you are right. I hope that at least the elements satisfying the cogroup identities can be classified. I think that associativity is a quite strong restriction. It is quite instructive to see $\widehat{F}_2$ as the limit of all finite $2$-generator groups, where the transition maps preserve the two generators. Actually I am looking for "universal group laws on finite groups". $\endgroup$ – Martin Brandenburg Feb 5 '14 at 23:57

Let $\sigma\in\widehat{\mathbb{Z}}^\times$. It makes sense to raise an element of a profinite group to the $\sigma$ power, and on $\widehat{\mathbb{Z}}$ this is an automorphism. We can build a cogroup structure from the standard one $m(x,y)=xy$ by conjugating by $\sigma$. Explicitly this is $m_{\sigma}(x,y):= (x^{\sigma}y^{\sigma})^{\sigma^{-1}}$. Note that taking $\sigma=-1$ gives your other cogroup law $m_{-1}(x,y)=yx$.

Now $1$ and $-1$ are the only two elements of $\mathbb{Z}^\times$, but $\widehat{\mathbb{Z}}^\times$ has continuum cardinality, so there are lots of other cogroup structures on $\widehat{\mathbb{Z}}$

  • $\begingroup$ sorry for naive question but why are the cogroups thereby built pairwise non-isomorphic? $\endgroup$ – user285001 Dec 27 '17 at 17:10

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