Do there exist pro-$p$ groups with finite quotients of non $p$ power order?

Question

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.

Answer 1

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

$$U_i=\pi_i^{-1}(G_i)$$

If

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

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 $|A|=p^n$.

Answer 2

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.