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Let $K$ be a number field and $C_K$ be its idele class group. Exercise 5.11 in Milne's notes asks me to show that there is a finite-index subgroup $H$ of $C_K$ which is not open. I haven't found an example, and am having trouble believing it for the following reason:

If $K=\mathbb{Q}$, there is an isomorphism of topological groups

$C_\mathbb{Q} \cong \mathbb{R}^\times_{>0} \times \hat{\mathbb{Z}}^\times$.

Since $\mathbb{R}^\times_{>0}$ has no nontrivial finite-index subgroups, $H$ must contain it, so $H \mathbin/ \mathbb{R}^\times_{>0}$ is a finite-index subgroup of $\hat{\mathbb{Z}}^\times$. However, finite-index implies open for the latter.

What am I missing?

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Contrary to the statement in the OP, the profinite group $\widehat{\mathbb Z}^{\times}$ admits non-open finite index subgroups.

For example, it admits a surjection onto $\prod_{n = 1}^{\infty} \mathbb Z/2\mathbb Z$, and this product has non-open finite index subgroups. (They correspond to non-principal ultrafilters on $\mathbb N$.)

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  • $\begingroup$ Sorry, but how this surjection onto $\prod\limits_{n=1}^{\infty}\mathbb Z/2\mathbb Z$ looks like? $\endgroup$ – Anna Abasheva Jan 5 '18 at 10:28
  • $\begingroup$ For $p > 2$, consider the reduction map $\mathbf{Z}_p^\times \rightarrow \mathbf{F}_p^\times \simeq \mathbf{Z}/(p-1) \mathbf{Z}$ Since $p - 1$ is even, this surjects onto $\mathbf{Z}/2\mathbf{Z}$. Since we can do this for each prime $p > 2$, projecting on each component (and sending $\mathbf{Z}_2$ to $1$) gives the desired surjection. $\endgroup$ – Dorebell Feb 14 '18 at 7:14

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