$\DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\Aut}{Aut}
\DeclareMathOperator{\Gal}{Gal}
\newcommand{\Z}{{\Bbb Z}}
\newcommand{\Q}{{\Bbb Q}}
\newcommand{\Fbar}{{\overline F}}
\newcommand{\G}{\Gamma}
$Let $F=\Q(\sqrt{7}\,)$, and consider the $\Gal(\Fbar/F)$-module $\mu_8$.
The Galois group $\Gal(\Fbar/F)$ acts on $\mu_8$ via a surjective homomorphism
$$\alpha\colon \Gal(\Fbar/F)\to\Aut(\mu_8)=(\Z/8\Z)^\times.$$
Write $\G=(\Z/8\Z)^\times$, and let $E/F$ be the finite Galois extension in $\Fbar$
corresponding to $\ker\alpha$; then $\Gal(E/F)=\G$.
Note that $\G$ is a non-cyclic group of order 4.
The group $\G$ naturally acts on $E$ and on the set of places of $E$.
For each place $v$ of $F$, consider the decomposition group $\G_v\subset \G$,
the stabilizer of $w$ in $\G$ where $w$ is a place of $E$ over $v$.
Question. Is it true that all decomposition groups $\G_v$ are cyclic?
If yes, I would be grateful for a reference or a proof.
Best Answer
If I understand this, $E = \mathbf Q(\sqrt{7},\zeta_8) = F(\zeta_8)$. Since $\mathbf Q(\zeta_8)/\mathbf Q$ ramifies only at $2$ (I am ignoring infinite places), $E/F$ can ramify only at a place over $2$, and there is just one such place in $F$ since $2$ is totally ramified in $F$: $(2) = \mathfrak p^2$ where $\mathfrak p = (2,1+\sqrt{7})$. For all other finite places $v$, the decomposition group is cyclic since inertia is trivial (so the decomposition group is isomorphic to the Galois group of the residue field extension).
To figure out the ramification and decomposition groups at $\mathfrak p$, we can pass to completions and compute there. We have $F_{\mathfrak p} = \mathbf Q_2(\sqrt{7})$, which equals $\mathbf Q_2(i)$ since $-7$ is a square in $\mathbf Q_2$. Thus $$ F_{\mathfrak p}(\zeta_8) = \mathbf Q_2(\sqrt{7},\zeta_8) = \mathbf Q_2(i,\zeta_8)= \mathbf Q_2(\zeta_8), $$ which is a quadratic totally ramified extension of $F_{\mathfrak p} = \mathbf Q_2(i)$. So $\Gamma_\mathfrak p$ has order $ef = 2$: it is cyclic.