If $G$ is a residually $p$ group and $G_i$ ANY filtration (i.e. $G_i\subset G_{i-1}$ and $\cap G_i=e$) of normal $p$-power index subgroups, is the corresponding filtration the usual pro-$p$ filtration?

Put differently, if $H$ is any normal $p$-power index subgroup, does it contain one of the $G_i$'s?

The answer to the corresponding question for finite filtrations is of course negative, e.g. the filtration $p^i\Bbb{Z}$ of $\Bbb{Z}$ does not give the profinite topology of $\Bbb{Z}$. On the other hand for abelian groups the answer to the above question is positive.

  • $\begingroup$ I should have added that G is finitely generated. As Tyler correctly pointed out, this is a necessary condition if I want a positive answer. $\endgroup$ Feb 17, 2010 at 9:24

5 Answers 5


Here are three properties a topological group $G$ might have.

(1) The group is topologically finitely generated.

(2) All abstract subgroups with finite index are open.

(3) There are finitely many abstract subgroups with each index.

In 1975, Serre showed that (1) implies (2) for pro-$p$ groups (a proof is in Sect. 4.3.4 of J. S. Wilson, Profinite Groups, Oxford Univ. Press, Oxford, 1998) and he conjectured that (1) implies (2) for all profinite groups. This conjecture was proved in 2007 by N. Nikolov and D. Segal in 2007 using the classification of finite simple groups. (N. Nikolov and D. Segal, On finitely generated profinite groups. I. Strong completeness and uniform bounds, Ann. of Math. 165 (2007), 171--238 and On finitely generated profinite groups. II. Products in quasisimple groups, Ann. of Math. 165 (2007), 239--273.

H. L. Peterson (Discontinuous characters and subgroups of finite index, Pacific J. Math. 44 (1973), 683--691) showed (2) implies (3) for compact Hausdorff groups and M. P. Anderson (Subgroups of finite index in profinite groups, Pacific J. Math. 62 (1976), 19--28) showed (3) implies (2) for pro-solvable groups. Therefore (2) and (3) are equivalent for pro-solvable groups, thus in particular for pro-$p$ groups. That (3) implies (2) for all compact Hausdorff groups (thus all profinite groups) is due to J. S. Wilson (Lemma 6 in Groups satisfying the maximal condition for normal subgroups, Math. Z. 118 (1970), 107--114), while M. G. Smith and J. S. Wilson (On subgroups of finite index in compact Hausdorff groups, Arch. Math. 80 (2003), 123--129) reproved the equivalence of (2) and (3) for compact Hausdorff groups.

At this point we see that (1) implies (2) for profinite groups and (2) and (3) are equivalent for compact Hausdorff groups. To complete the circuit, it is natural to guess that (3) implies (1) for profinite groups (and perhaps even all compact Hausdorff groups). We will check (3) implies (1) for pro-$p$ groups, so (1), (2), and (3) are equivalent when $G$ is a pro-$p$ group.

Let $G$ be a pro-$p$ group satisfying (3). A maximal open subgroup of a pro-$p$ group has index $p$, so (3) implies there are finitely many maximal open subgroups of $G$. The intersection of finitely many open subgroups is open, so the intersection of the maximal open subgroups of $G$ is open. This intersection is the Frattini subgroup $\Phi(G)$, so $\Phi(G)$ is open in $G$. It follows from Prop. 2.8.10 in L. Ribes and P. Zalesskii, Profinite Groups, Springer--Verlag, Berlin, 2000 that $G$ is topologically finitely generated.


No, this is not true (at least assuming the axiom of choice).

Let $G = \prod_{i=0}^\infty \mathbb{Z}/2$ and let $\pi:G \to \mathbb{Z}/2$ be any surjective group homomorphism sending the proper sub-vector-space $\oplus_{i=0}^\infty \mathbb{Z}/2$ to zero. The kernel of this homomorphism is a subgroup $H$ of index $2$.

Let $G_k = \prod_{i=k}^\infty \mathbb{Z}/2 \subset G$. Then the $G_k$ are a filtration of $G$ as you ask, but $H$ does not contain any of the $G_k$. If it did, then the group homomorphism $\pi$ would factor through $G/G_k$ and thus be determined by its values on $\oplus_{i=0}^{k-1} \mathbb{Z}/2 \subset ker(\pi)$.

In the case where $G$ is topologically finitely generated I would guess that you can use the mod-$p$ lower central series to prove that your statement has an affirmative answer - but don't have an argument on hand.



Let $G(n,p)$ be the kernel of $\GL_n({\Bbb Z}_p) \to \GL_n({\Bbb Z}/p)$. This is a pro-$p$ group.

Enough random matrices in $G(n,p)$ generate a free group which has positive probability of being dense. In other words, the induced topology on the free group has completion $G(n,p)$. If we take the same number of matrices in $G(2,p)$ and $G(3,p)$ this gives two topologies on the same free group with different completions, neither of which is the full pro-$p$ topology. Without the randomness, the kernel of $\GL_2({\Bbb Z}) \to \GL_2({\Bbb Z}/p)$ is free, but the topology coming from $G(2,p)$ is not the full pro-$p$ topology.

In a more positive direction, the congruence subgroup property says that every finite index subgroup of $\GL_3({\Bbb Z})$ contains the kernel of reduction $\bmod m$ for some $m$. I think that this implies that the only $p$-topology on the kernel of reduction $\bmod p$ is the one with completion $G(3,p)$.

  • $\begingroup$ Can you elaborate on why this isn't the full pro-$p$ topology? What fails? $\endgroup$ Feb 17, 2010 at 18:50
  • $\begingroup$ G(2,p) has abelianization (Z/p)^4, while the free pro-p group on 4 generators has abelianization (Z_p)^4. (and the kernel of reduction mod p has is free on more generators than 4) $\endgroup$ Feb 19, 2010 at 1:44

Your question, in the finitely generated case, is sort of answered above.

In the infinitely generated case, however, this is a research question I have been looking at for the last year or so. I'm in the middle of redrafting a paper dealing with the abelian case.

If $G$ is any finitely-generated pro-$p$ group, by Serre, any pro-$p$ topology must have all finite index subgroups open, so the answer here is yes. This is not too hard to see. A result of Nikolov and Segal (2006) showed the same holds for profinite groups.

There are two different notions of unique topology you could mean here. If $f:G\to H$ is an abstract isomorphism between two pro-p groups, we have three possibilities:

(1) $f$ must be continuous. (2) $f$ is not continuous, but $G,H$ are isomorphic as pro-$p$ groups. (3) $G,H$ are not isomorphic as pro-$p$ groups.

Which case this is is entirely dependant on the algebraic structure of $G$.

The first case is equivalent to saying all automorphisms of $G$ are continuous. As above, any finitely-generated pro-$p$, indeed, profinite groups have this property. However not every group here must be finitely generated. By consideration of centralisers, which, as kernels of word maps, must be closed in any profinite topology, one can see that any (unrestricted Cartesian) product of finite centreless groups has this property. Moreover, considering centralisers will also give this result for Branch Groups.

If we assume the Generalised Continuum Hypothesis fails, we get uninteresting examples of non-isomorphic abstractly ismorphic pro-$p$ groups - the (unrestricted Cartesian) product of $\aleph$, $\beth$ copies of $C_p$ with $2^\aleph=2^\beth$, for example. It is more interesting to look for examples which do not depend on this. Hence it makes sense to look at countably-based profinite groups - those which are the inverse limit of a countable collection of finite groups, first.

Tyler Lawson gives a good example above of this, but it is not too hard to see that these groups are still abstractly isomorphic.

More interestingly, if $G$ is a countably-based abelian pro-$p$ group with infinite exponent torsion group, then $G$ is abstractly isomorphic to a product of cyclic $p$-groups, and there are uncountably many different pro-$p$ topologies on $G$, which give rise to uncountably many isomorphism classes of pro-$p$ groups.

My paper showing this is mathematically complete but being redrafted. A version should appear on the ArXiV in the next few days. In the meantime, I can e-mail a copy on request.


Contrary to OP's claim, this is false even for finitely generated abelian groups, including $\mathbf{Z}^2$.

Indeed, fix a dense group embedding $i:\mathbf{Z}^2\to\mathbf{Z}_p$ (namely $i(n,m)=n+mt$ for some fixed irrational $t\in\mathbf{Z}_p$). Define $G_n=i^{-1}(p^n\mathbf{Z}_p)$. Then $(G_n)$ is a decreasing sequence of subgroups with intersection reduced to $\{0\}$. Moreover $G_0/G_n$ is isomorphic to $\mathbf{Z}/p^n\mathbf{Z}$. Hence no $G_n$ is contained in $p\mathbf{Z}^2$.


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