Density of the conjugacy classes of the inertia groups in Galois group of $\mathbb{Q}$

Question

Let $S$ be a finite set of primes and $G_{\mathbb{Q},S}$ the Galois group of maximal extension of $\mathbb{Q}$ which is unramified ouside $S$. Since $\mathbb{Q}$ has no unramified extensions, $G_{\mathbb{Q},S}$ is topologically generated by inertia groups at primes in $S$, i.e. the subgroup generated by inertia groups at primes in $S$ is dense in $G_{\mathbb{Q},S}$. It leads to the following question:

Is the union of conjugates of the inertia groups at primes in $S$ dense in $G_{\mathbb{Q},S}$?

Any comments and suggestions will be appreciated.

Answer 1

Not unless $S$ has one element.

If $S$ contains two primes, $p$ and $q$, the Galois group of the cyclotomic field $\mathbb Q(\mu_{p^\infty q^\infty})$ is a quotient of $G_{\mathbb Q,S}$ and is isomorphic to $\mathbb Z_p^\times \times \mathbb Z_q^\times$. The images of the inertia groups at $p$, $q$, and any other prime are $\mathbb Z_p^\times \times \{1\}, \{1\}\times \mathbb Z_q^\times$, and $\{(1,1)\}$ respectively, and the union of these is not dense.

Answer 2

Not if $S$ has one element.

If $p$ is an irregular prime and $H$ is the Hilbert class field of $\mathbf{Q}(\zeta_p)$ then $\mathrm{Gal}(H/\mathbf{Q})$ has order divisible by $p$ but no inertial element has order $p$.

If $p$ is a regular prime, then $p$ does not divide the class number of $\mathbf{Q}(\zeta_{p^n})$ for any $n$. A theorem of Washington implies that the $\ell$-part of the class group of this tower is uniformly bounded for every $\ell \ne p$. From the class number formula the order of the class groups also become arbitrarily large as $n$ increases. Taken together this implies that the class number of $\mathbf{Q}(\zeta_{p^n})$ is divisible by at least one prime $\ell$ with $(\ell,p(p-1)) = 1$ for sufficiently larger $n$. If $H$ is the Hilbert class field of $\mathbf{Q}(\zeta_{p^n})$ for such an $n$, then $\mathrm{Gal}(H/\mathbf{Q})$ will have order divisible by $\ell$ but no inertial element can have order $\ell$.

Since the prime above $p$ in $\mathbf{Q}(\zeta_{p^n})$ is principal, it (the prime above $p$) splits completely in the Hilbert class field. So in these cases the inertia group above $p$ coincides with the decomposition group. Hnce these examples also demonstrate the stronger claim that not even the union of the decomposition groups above $p$ are dense.