Do we have such a direct product decomposition of Galois groups? Let $L = \Bbb{Q}(\zeta_m)$ where we write $m = p^k n$ with $(p,n) = 1$. Let $p$ be a prime of $\Bbb{Z}$ and $P$ any prime of $\mathcal{O}_L$ lying over $p$.
Notation: We write $I = I(P|p)$ to denote the inertia group and $D = D(P|p)$ the decomposition group.
Now we have a tower of fields
$$\begin{array}{c} L \\ |\\ L^E \\| \\ L^D \\|   \\ \Bbb{Q}\end{array}$$
and it is clear that $L^E = \Bbb{Q}(\zeta_n)$ so that $E \cong \Bbb{Z}/p^k\Bbb{Z}^\times$.


My question is: I want to identify the decomposition group $D$. So this got me thinking: Do we have a decomposition into direct products $$D \cong E \times D/E?$$
    This would be very convenient because $D/E$ is already known to be finite cyclic of order $f$ while $E$ I have already stated above. Note it is not necessarily given that $(e,f) = 1$ so we can't invoke Schur - Zassenhaus or anything like that.



You can write your $L$ as the compositum of $\mathbb{Q}(\zeta_{p^k})$ and $\mathbb{Q}(\zeta_{n})$. Since $(p,n)=1$, the two are disjoint over $\mathbb{Q}$, and so the Galois group of $L$ is isomorphic to the direct product of the two Galois groups, one of which is $E$. Let's call the other subgroup $H$. Now, every element $g$ of $G$ is uniquely a product of an element $\epsilon$ of $E$ and an element $h$ of $H$. $E$ is contained in $D$, so $\epsilon h$ fixes $P$ if and only if $h$ does. In other words, the decomposition group $D$ is generated by $E$ and the decomposition group of $P$ in ${\rm Gal}(\mathbb{Q}(\zeta_{p^kn})/\mathbb{Q}(\zeta_{p^k}))=H$, so is indeed a direct product.
