It is well-known that any finite group is the Galois group of a Galois extension. This follows from Cayley's theorem (as can be seen in this answer). This (linked) answer led me to the following question:

What about infinite groups?

Infinite groups appear as Galois groups on infinite extensions, as when one defines the absolute Galois group of a field $F$. There is a natural topology on the Galois group, called the Krull topology, which turns this group into a profinite group (i.e. a compact, totally disconnected, Hausdorff topological group). It can be proven that any profinite group is the Galois group of an extension (see this short paper by Waterhouse). Therefore, the above question is equivalent to the following:

Can any group be given the structure of a profinite group?

I would like to get more information about this question. In particular, if the answer is no, are there definite restrictions (e.g. cardinality)?

M Turgeon
  • 9,776
  • 3
  • 39
  • 79
  • I guess the answer is true considering every group is the fundamental group of a topological space. – Honghao Aug 30 '12 at 16:59
  • 1
    @Honghao And every fundamental group is profinite? – M Turgeon Aug 30 '12 at 17:02
  • @Honghao No, the Wikipedia page says that fundamental groups in topology are not pro-finite. – Thomas Andrews Aug 30 '12 at 17:02
  • 1
    @Honghao: the analogy to covering spaces cannot be pushed too far. An algebraic field extension is determined by its finite subextensions in a way that a covering map is not. – Qiaochu Yuan Aug 30 '12 at 17:03
  • 2
    Also see the Wikipedia link you gave about the pro-finite completion of a group. You can see that a group can be made pro-finite if and only if its pro-finite completion is isomorphic via the natural map. Then you can see that not even $\mathbb Z$ is pro-finite. – Thomas Andrews Aug 30 '12 at 17:20
  • @Thomas: this appears to be false (that is, it's not true in general that the natural map from a profinite group to its profinite completion is an isomorphism). In other words, there exist profinite groups with non-closed subgroups of finite index. – Qiaochu Yuan Sep 03 '12 at 04:26
  • 1
    Hmm, will have to rethink, @QiaochuYuan. I thought it followed from the universal property of the completion, but I could easily have been wrong. – Thomas Andrews Sep 03 '12 at 19:44
  • 1
    @Thomas: consider $(\mathbb{Z}/2\mathbb{Z})^{\infty}$. There are obvious projection maps to $\mathbb{Z}/2\mathbb{Z}$ and then there are non-obvious ones: quotient by the direct sum and you obtain a vector space over $\mathbb{F}_2$, and any linear functional on this vector space gives you a map to $\mathbb{Z}/2\mathbb{Z}$ whose kernel is not closed. So the profinite completion of $(\mathbb{Z}/2\mathbb{Z})^{\infty}$ as an abstract group is larger than the group itself. – Qiaochu Yuan Sep 03 '12 at 21:20
  • 3
    An infinite group which is countably infinite cannot be given the structure of a profinite group. This is because a non-empty compact Hausdorff space without isolated points is uncountable. If $G$ is a profinite group with an isolated point, then every point is isolated, and the topology on $G$ is discrete, so $G$ is necessarily finite. – Keenan Kidwell Sep 25 '12 at 22:36
  • The "equivalence" is not quite so (I agree the topological version is perhaps the "correct" version). See Manfred Dugas, Rüdiger Göbel. *All infinite groups are Galois groups over any field*, Trans. Amer. Math. Soc., **304 (1)**, (1987), 355-384. – Andrés E. Caicedo Jul 04 '13 at 02:16

2 Answers2


No. Profinite groups are residually finite (in fact a group is residually finite if and only if it embeds into its profinite completion) and many groups are not residually finite. If you don't have any particular restriction on the number of generators, $\mathbb{Q}$ is a simple example.

There are other restrictions. A compact Hausdorff group has a Haar measure of total measure $1$ and no countable group can be equipped with such a measure since the measure of any singleton cannot be $0$ and cannot be positive. This rules out $\mathbb{Q}$ but it also rules out, for example, $\mathbb{Z}$.

Qiaochu Yuan
  • 359,788
  • 42
  • 777
  • 1,145
  • Thank you. I am not familiar with residually finite groups. Do you know an example of a finitely generated group which is not residually finite? – M Turgeon Aug 30 '12 at 17:06
  • 2
    @M Turgeon: yes, but they aren't trivial to construct; for example, some Baumslag-Solitar groups (http://en.wikipedia.org/wiki/Baumslag%E2%80%93Solitar_group) work, and there is a finitely presented example due to Higman: http://en.wikipedia.org/wiki/Higman_group – Qiaochu Yuan Aug 30 '12 at 17:10
  • In the second paragraph you probably mean Haar *probability* measure. Counting measure on a discrete group is a perfectly legitimate Haar measure. @MTurgeon one reason that explains why it is difficult to come up with examples of non-residually finite groups is that linear groups are residually finite. – t.b. Aug 31 '12 at 08:44
  • @t.b.: yes, sorry. When you say "non-residually finite groups" you mean "non-residually finite finitely generated groups." $\mathbb{Q}$ is linear but it is not residually finite. – Qiaochu Yuan Aug 31 '12 at 08:47
  • 1
    Yes, right. One nitpick deserves another one :) Nice answer! – t.b. Aug 31 '12 at 08:48
  • @t.b. and Qiaochu Thank you for the extra comments! – M Turgeon Aug 31 '12 at 12:37
  • 1
    http://mathoverflow.net/questions/4939/is-there-a-compact-group-of-countably-infinite-cardinality/4950#4950 – Georges Elencwajg Sep 25 '12 at 22:51

This cannot be true since a profinite group cannot be countable (see this) so therefore we cannot make $\mathbb{Z}$ into a profinite group.

  • 165
  • 1
  • 4