Decomposition group regulates prime decomposition for non-Galois extensions

Question

Let $L/K$ be a finite extension of number fields (not necessarily Galois), let $N$ be the normal closure of $L$, and let $\mathfrak{P}$ be a prime ideal in the ring of integers $\mathscr{O}_N$ lying over a prime $\mathfrak{p} \subset \mathscr{O}_K$. Let $G = \mathrm{Gal}(N/K), H = \mathrm{Gal}(N/L)$, with $H$ naturally viewed as a subgroup of $G$. Let $D_{\mathfrak{P}} \subseteq G$ denote the decomposition group of $\mathfrak{P}/\mathfrak{p}$.

In I.9, p. 54-55 of Neukirch's Algebraic Number Theory, he claims that one may use the decomposition group $D_{\mathfrak{P}}$ to also describe the prime decomposition of $\mathfrak{p}$ over the intermediate extension $L$. More precisely, letting $P_{\mathfrak{p}}$ be the set of prime ideals in $\mathscr{O}_L$ lying over $\mathfrak{p}$, there is a bijection $$H \backslash G /D_{\mathfrak{P}} \to P_{\mathfrak{p}}$$ $$[\sigma] \mapsto \sigma \mathfrak{P} \cap L.$$ Here, the source $H \backslash G /D_{\mathfrak{P}}$ is a double coset space of $G$ and $[\sigma]$ denotes the class of $\sigma \in G$ in this double coset space.

Neukirch leaves this as an exercise. It is clear to me that this map is well-defined and surjective, but I am having a lot of trouble showing that it is injective, or otherwise completing the proof that this is indeed a bijection. Comparing the size of source and target does not seem helpful, since both of these things seem hard to compute. It would suffice to show that if we have $\sigma \in G$ such that $\sigma \mathfrak{P} \cap L = \mathfrak{P} \cap L$, then we can find $\tau \in H, \rho \in D_{\mathfrak{P}}$ such that $\tau \sigma \rho = 1$.

How does one show that the map defined is injective?

Answer 1

Injectivity can be proven on "elements": (I'll be using notations as in Neukirch).

Assume that $\sigma \mathfrak{P}\cap L = \rho\mathfrak{P}\cap L = q$. Then, $\sigma \mathfrak{P}$ and $\rho \mathfrak{P}$ lie above $q$ in $N$. As $H$ acts transitively on the primes in $N$ that lie above $q$ (Proposition 9.1 in Neukrich), we get a $\tau \in H$: $\tau\sigma\mathfrak{P} = \rho \mathfrak{P}$.

Now, reformulating gives $\rho^{-1}\tau\sigma\mathfrak{P} = \mathfrak{P} \implies \rho^{-1}\tau\sigma \in G_{\mathfrak{P}}$. Thus, there exists $\varphi \in G_{\mathfrak{P}}$, such that $\rho = \tau\sigma\varphi^{-1}$. This proves that $\rho \in H\sigma G_{\mathfrak{P}}$ and thus $H\rho G_{\mathfrak{P}} = H\sigma G_{\mathfrak{P}}$.