Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Could someone help me with an simple example of a profinite group that is not the p-adics integers or a finite group? It's my first course on groups and the examples that I've found of profinite groups are very complex and to understand them requires advanced theory on groups, rings, field and Galois Theory. Know a simple example?

Last, how to prove that that $\mathbb{Z}$ not is a profinite group?

share|cite|improve this question
Well, there's $\hat{\mathbf Z}$, which is in the same vein. – Dylan Moreland Jun 20 '12 at 15:30
The Galois group of any infinite Galois extension is profinite. – Arturo Magidin Jun 20 '12 at 15:33
Dylan. What is $\mathbb{\widehat{Z}}$? – Andres Jun 20 '12 at 15:33
@Andres It's just $\varprojlim \mathbf Z/n\mathbf Z$ directed by divisibility. It follows from the Chinese remainder theorem that $\hat{\mathbf Z} \approx \prod\mathbf Z_p$ [product over all primes], so this may not be so interesting for you. – Dylan Moreland Jun 20 '12 at 15:43

Yes. Let $G = \prod_{i=1}^{\infty} \mathbb{Z}/2\mathbb{Z}$ be the direct product of (countably) infinitely many copies of the cyclic group of order $2$. This profinite group, sometimes called (well, by me at least) the Bernoulli group, occurs naturally in probability theory. As a topological space it is homeomorphic to the standard Cantor set.

(For that matter, any Cartesian product of finite groups is a profinite group, and this is an important example, because any profinite group is a closed subgroup of such a product.)

As for your second question, a profinite group is in particular a compact (Hausdorff) topological group and thus carries a Haar measure, i.e., a translation-invariant probability measure. Thus it cannot be countably infinite, and in particular $\mathbb{Z}$ is not (or more precisely, cannot be endowed with the structure of) a profinite group.

share|cite|improve this answer

One way to get a profinite group is to start with any torsion abelian group $A$, and take $\hom(A, \mathbb{Q}/\mathbb{Z})$. This acquires a topology as the inverse limit topology: it is the inverse limit of $\hom(A_0, \mathbb{Q}/\mathbb{Z})$ for $A_0 \subset A$ a finitely generated (and necessarily torsion) subgroup.

In fact, this gives an anti-equivalence between profinite abelian groups and torsion abelian groups, which is sometimes useful (it means, for instance, that a filtered inverse limit of profinite abelian groups is always exact, since the corresponding fact for filtered direct limits is true in abelian groups).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.