A profinite group which is not its own profinite completion? 
Is there a profinite group $G$ which is not its own profinite completion?

Surely not, I thought. But upon looking into it, I found that there is a special name given to a $G$ which is its own profinite completion, namely "strongly complete". And a recent (2003) hard theorem (which according to Wikipedia uses the classification of finite simple groups) due to Nikolov and Segal asserts that, if $G$ is finitely generated (as a topological group), then it is "strongly complete". 
So the $G$ I'm looking for cannot be topologically finitely generated. An equivalent question to the above is:

Is there a profinite group $G$ which admits a non-open subgroup of finite index?

Now here's my problem; the only exposure to profinite groups I've had has been in the context of number theory, absolute Galois groups, local fields, etc. In particular, the only non-topologically-finitely-generated profinite group I'm aware of is the absolute Galois group of a number field, say $\mathbb{Q}$. But I reckon the Krull topology demands that the finite index subgroups of $Gal(\bar{\mathbb{Q}}/\mathbb{Q})$ be open. 
Maybe there is a more 'exotic' example of such a $G$...
 A: It's hard to pick many non-finitely-generated pro-$p$ groups which are isomorphic their own profinite completions. Partially as non-finitely-generated pro-$p$ groups aren't generally that easy to construct. 
My paper here http://arxiv.org/abs/1101.3005 outlines the construction of infinitely many topologically non-isomorphic pro-$p$ groups isomorphic to $\prod_i C_{p^{i}}$. These will thus all have the same profinite completion, but are non-isomorphic.
A: The absolute Galois group $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ has lots of subgroups of finite index that aren't open. See the section "Nonopen subgroups of finite index" in Chapter 7 of Milne's notes on field theory for a construction (it uses the axiom of choice).
A: Example taken from Ribes and Zalesskii's book "Profinite groups". Take an infinite set $I$ and a finite group $T$. You can let $G$ be the profinite group $\prod_I T$. Denote its elements by $(g_i)_{i\in I}$. Let $\mathcal F$ be an ultrafilter which contains the filter of all cofinite subsets of $I$. If you denote $H$ to be the subgroup of elements with $\lbrace i\in I \mid g_i=1\rbrace\in \mathcal F$, it is clear that $H$ is proper normal and that it is not open because it is dense and has finite index $|T|$.
To show that $H$ has index $|T|$ in $G$ consider all elements $a_t=(t,t,\dots)$ for $t\in T$. For any $g\in G$, consider $I_t=\lbrace i\in I \mid g_i=t\rbrace$. Since we have $\bigcup_{t\in T} I_t=I$
then $I_t\in \mathcal F$ for some $t$, and therefore $ga_t^{-1}\in H$.
A: Yes, this is possible. Take as $G$ the product of countable many copies of $\mathbb Z_p$. It has a countable basis of open subgroups, hence only countably many open subgroups. But it has many more subgroups of finite index!
