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Write $e=e(\mathfrak{q}/\mathfrak{p})$, $f=f(\mathfrak{q}/\mathfrak{p})$ and $g$ for the number of primes above $\mathfrak{p}$. The Galois group acts transitively on those primes, with stabilizer $D_{\mathfrak{q}}$. The fundamental identity for prime decomposition and the orbit-stabilizer theorem therefore giveIt remains to establish that the residue action has image of order $f$, rather than assume this in the desired equality.

Use the decomposition group of a prime and local Galois group identification $D_{\mathfrak{q}}=\mathrm{Gal}(F_{\mathfrak{q}}/K_{\mathfrak{p}})$. A generator of the finite field extension $k_{\mathfrak{q}}/k_{\mathfrak{p}}$ has a separable monic minimal polynomial of an algebraic element. Lift its coefficients to ${\mathcal{O}}_{K_{\mathfrak{p}}}$. The Hensel lemma lifts each of its distinct residue roots to $F_{\mathfrak{q}}$. One such lift has field degree at most $f$, while its residue generates the degree-$f$ finite field extension, so its field degree is exactly $f$ and its ramification index of a prime ideal is one. It therefore generates an unramified extension of degree $f$, and sending it to each of the other lifts gives its $f$ embeddings. Because $F_{\mathfrak{q}}/K_{\mathfrak{p}}$ is a Galois extension, these embeddings extend to its automorphisms. Their reductions realize every automorphism of $k_{\mathfrak{q}}/k_{\mathfrak{p}}$. Thus the residue action is surjective, and

Now let $G$ be cyclic of order ${\ell }^2$, and let $H=\mathrm{Gal}(F/L)$ be its unique subgroup of order $\ell$. In a tower of Galois extensions, inertia maps onto inertia in the intermediate extension. This also follows from the just-proved order formula: its image has order $e(F/K)/e(F/L)=e(L/K)$, using multiplicativity of the ramification index of a prime ideal in a tower. If $\mathfrak{p}$ ramifies in $L/K$, this image in $G/H$ is nontrivial. Every proper subgroup of $G$ is contained in $H$, so inertia upstairs must be $G$. HenceThe fundamental identity for prime decomposition then also forces a single prime above $\mathfrak{p}$ and residue-field degree one.

Let $G=\mathrm{Gal}(L/K)$ and let $\mathcal{P},\mathcal{Q}$ lie over $\mathfrak{p}$. These primes define two extensions to $L$ of the $\mathfrak{p}$-adic absolute value on $K$. The conjugacy of extensions of a valuation to a normal extension says that some $\sigma \in G$ carries the first extension to the second. Equivalently, $\sigma \mathcal{P}=\mathcal{Q}$. This proves the transitivity of the Galois action on primes.

The decomposition group is the stabilizerEach such automorphism extends continuously to the completions, and restriction gives the decomposition group of a prime and local Galois group isomorphism

For the splitting field $L=\mathbb{Q}(\sqrt[3]{7},{\zeta }_3)$ of $X^3-7$, the global Galois group is $S_3$. Since $7\equiv 1(\mathrm{mod}3)$, the field ${\mathbb{Q}}_7$ contains ${\zeta }_3$. The local splitting field is therefore ${\mathbb{Q}}_7(\sqrt[3]{7})$, an Eisenstein, totally ramified cyclic extension of degree three. Hence there areprimes of $L$ above $7$, and the decomposition group of each is the normal subgroup