Exercise from Marcus on class groups I stumbled upon this:
Let $p$ be a prime of $K$ (a given number field) and let $m$ be the order of $[p]$ in $Cl(K)$. Suppose $\mathcal{P}|p$ is a prime of $L$ (here, $L/K$ is an extension of $K$). Show that $m/(m, f(\mathcal{P},p))$ divides the order of the class $[\mathcal{P}]$ in $Cl(L)$.
Any help is welcomed, thanks!
 A: Here is an argument avoiding class field theory:
Let $a$ be the order of $\mathcal P$ in the class group of $L$. Then
$\mathcal P^a = (\alpha)$ for some integer $\alpha$ of $L$.  Taking norms
down to $K$, we find that
$p^{f(\mathcal P,p) a}$ is principal.  Thus $m$ divides $a f(\mathcal P,p)$,
and hence $m/(m,f(\mathcal P,p))$ divides $a$, as claimed. 
A: This is closely related to the class-field-theoretic type argument I mentioned in the other answer.  Here's a fast solution (if possibly not the most elementary).  
Note that by class field theory, specifically properties of the Artin symbol, the order $m$ of $[p]$ equals the relative inertia degree of $p$ in the Hilbert class field $K^{(1)}$ of $K$, and similarly for the order of $[\mathcal{P}]$ and the residue degree of $\mathcal{}P$ in $L^{(1)}/L$.  Now consider the following diagram:
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
L & \ra{m':=\text{ order of }\mathcal{P}} & L^{(1)} \\
\da{f:=f(\mathcal{P}/p)} & & \da{}\\
K & \ra{m :=\text{ order of }p} & \,K^{(1)} \\
\end{array}
$$
Here the labels on each arrow are the inertia degrees of a prime above $p$ in each extension.  Now by multiplicativity of inertia degrees in towers, we have $m\mid m'f$, so $\frac{m}{(m,f)}\mid m',$ as desired.

(Apologies for the sloppy diagram, I've never jury-rigged an xypic on MSE or MO before.  Can anyone make those arrows "headless"?  And/or move the$f(\mathcal{P}/p)$ to the left of the arrow?)  
A: This is just a hint, that will hopefully help you realize where this $m/(m,f(\mathcal{P},p))$ comes from.
Let $m>1$ be a fixed integer and suppose that $a\in\mathbb{Z}$ is relatively prime to $m$. Then, for any $d\geq 1$, we have:
$$\operatorname{ord}(a^d) = \frac{\operatorname{ord}(a)}{\gcd(\operatorname{ord}(a),d)}.$$
