We define a pro-$p$ group to be a projective (i.e. inverse) limit of $p$-groups.

My question is exactly as stated in the title:

If a subgroup $H$ of a pro-$p$ group $G$ has finite quotient, must $|G/H|$ be some power of $p$?

If we restrict ourselves to open subgroups, then I believe this is the case because the open subgroups in the projective limit form an open basis of the identity. However for general finite index subgroups I am not sure.


No such subgroup exists. Pro-$p$ groups with the second definition (i.e. inverse limit of discrete, finite $p$-groups) can be easily shown to be equivalent to the first definition:

$$G/N\cong P$$

where $N$ is open, normal and $P$ is a finite $p$-group.

How so? Since $N$ is normal it is the union of some open subgroups from a basis of open normal subgroups around the identity, however we know that if

$$G=\varprojlim_{i\in I} G_i$$

with each $G_i$ a discrete, finite $p$-group, then such a basis is given by



$$N=\bigcup_{i\in J\subseteq I} U_i$$

select any $i_0\in J$ then we have a surjective homomorphism:

$$G/U_{i_0}\to G/N$$

with kernel $N/ U_{i_0}$, hence $G/N$ is the homomorphic image of a finite $p$-group, and is hence a $p$-group.

(Edit) For I think, following the comments, I should include the nitty gritty of the reduction to the closed case since there's enough confusion to merit it. Throughout we use $|\cdot|$ for the order of an element and of a subgroup, understood in the generalized sense of profinite groups (i.e. supernatural numbers)

The basic idea: just use the Lagrange theorem for profinite groups.

The problem: indices are only defined for closed subgroups.

If $|A|=p^nm$ with $(p,m)=1$ and select $a\in A$ such that $|a|\big| m$, which is possible by Cauchy's theorem. Name the projection map $\pi:G\to G/N\cong A$ (first isomorphism theorem) is surjective, we may select a lift $\stackrel{\sim}{a}\in G$, and by definition $\left|\pi\left(\stackrel{\sim}{a}\right)\right|=|a|$, we have just changed our context from $A$ to a subgroup of order coprime to $A$--namely $\langle a\rangle$--so that we may assume that $p\not\big| |A|$ rather than the weaker condition $|A|\ne p^n$. Denote by $H$ the closure of $\langle\stackrel{\sim}{a}\rangle$ in $G$. Then $|\stackrel{\sim}{a}|=m$ in $H/H\cap N$, which is finite because $N\cap H$ is clearly relatively open in $H$.

Since $H$ is closed, $m\big||H|$, and by Lagrange $|H|\; \bigg|\; |G|$. However $(m,p)=1$ and the only prime dividing $|G|$ is $p$, hence $m=1$, so that $\langle a\rangle\le A$ is the trivial subgroup and $|A|=p^n$.

  • $\begingroup$ Doesn't the universal property go the other way? $\endgroup$ – Alexander Jan 30 '15 at 4:31
  • $\begingroup$ But doesn't this only work for normal subgroups? $\endgroup$ – Alexander Jan 30 '15 at 5:14
  • $\begingroup$ @Alexander You're assuming that $A$ is the image of $G$ under a homomorphism, and kernels of homomorphisms are always normal... And even so, the open normal subgroups still form a fundamental set of neighborhoods for the topology around the identity, that's how the inverse limit topology is defined. $\endgroup$ – Adam Hughes Jan 30 '15 at 5:16
  • $\begingroup$ Yeah, sorry. I guess I meant open subgroups. Like what if the subgroup is not open but still finite index? $\endgroup$ – Alexander Jan 30 '15 at 6:14
  • $\begingroup$ @Alexander that's impossible, profinite groups are compact, and if $U\subseteq G$ is open then we note that $$\bigcup_{g\in G} gU$$ is an open cover of $G$, so it has a finite subcover. i.e. $G=U\cup g_1U\cup\ldots\cup g_nU$ i.e. $U$ has finite index, since these are all the left cosets of $U$, and they are finite in number. If the group is finite index, then note that it is the complement of the union of the other left cosets, which are all closed, hence it is open. In fact $U$ is finite index, closed iff $U$ is open. $\endgroup$ – Adam Hughes Jan 30 '15 at 6:19

Look at Proposition 4.2.3 in the book by Ribes-Zalesskii, the whole section 4.2. will be interesting for you. Edit: the question is answered affirmatively there.

  • 2
    $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. $\endgroup$ – Edward Jiang Jan 31 '15 at 21:12
  • $\begingroup$ @Edward I agree, the post does not provide an answer, but the cited Proposition does. I did not require any clarification from the author nor did I intend to critisize the original post. To the contrary, I consider this an interesting question. $\endgroup$ – user 59363 Feb 1 '15 at 10:07
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    $\begingroup$ @user59363 can you edit to provide more explicit support for the affirmative? Your answer may be accurate but is not approachably informative. $\endgroup$ – Joffan Feb 1 '15 at 10:20

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