[proofplan] We first show that $G$ acts on the set of primes above $\mathfrak{p}$ (each $\sigma \in G$ permutes these primes), then prove transitivity by contradiction using the Chinese Remainder Theorem. Transitivity immediately yields the equality of all ramification indices and residue degrees via the orbit-stabiliser theorem. Part (iii) follows from the general fact that stabilisers of elements in the same orbit are conjugate. [/proofplan] [step:Verify that $G$ acts on the set of primes lying above $\mathfrak{p}$] Let $\sigma \in G$ and let $\mathfrak{P}$ be a prime of $\mathcal{O}_K$ lying above $\mathfrak{p}$, i.e., $\mathfrak{P} \cap \mathcal{O}_k = \mathfrak{p}$. We show $\sigma(\mathfrak{P})$ is also a prime of $\mathcal{O}_K$ lying above $\mathfrak{p}$. Since $\sigma \in \operatorname{Gal}(K/k)$, the map $\sigma: K \to K$ is a $k$-automorphism. In particular, $\sigma$ fixes $k$ pointwise, hence fixes $\mathcal{O}_k$ pointwise (since $\sigma$ preserves integrality: if $\alpha \in \mathcal{O}_K$ satisfies a monic polynomial over $\mathcal{O}_k$, then $\sigma(\alpha)$ satisfies the same polynomial, so $\sigma(\alpha) \in \mathcal{O}_K$). Therefore $\sigma(\mathcal{O}_K) = \mathcal{O}_K$. Since $\sigma: \mathcal{O}_K \to \mathcal{O}_K$ is a ring automorphism, the image $\sigma(\mathfrak{P})$ is a prime ideal of $\mathcal{O}_K$. Moreover, $\sigma(\mathfrak{P}) \cap \mathcal{O}_k = \sigma(\mathfrak{P} \cap \mathcal{O}_k) = \sigma(\mathfrak{p}) = \mathfrak{p}$, where the last equality holds because $\sigma$ fixes $\mathcal{O}_k$ pointwise. So $\sigma(\mathfrak{P})$ lies above $\mathfrak{p}$. This defines a group action $G \curvearrowright \{\mathfrak{P}_1, \ldots, \mathfrak{P}_g\}$ by $\sigma \cdot \mathfrak{P}_i = \sigma(\mathfrak{P}_i)$. The action is well-defined since $\sigma(\mathfrak{P}_i)$ is a prime above $\mathfrak{p}$, hence equals some $\mathfrak{P}_j$. The group action axioms ($\operatorname{id} \cdot \mathfrak{P} = \mathfrak{P}$ and $(\sigma\tau) \cdot \mathfrak{P} = \sigma \cdot (\tau \cdot \mathfrak{P})$) follow from the corresponding properties of function composition. [/step] [step:Prove that $G$ acts transitively using the Chinese Remainder Theorem] We prove part (i): the action is transitive, i.e., for any $i, j \in \{1, \ldots, g\}$ there exists $\sigma \in G$ with $\sigma(\mathfrak{P}_i) = \mathfrak{P}_j$. Suppose for contradiction that the orbit of $\mathfrak{P}_1$ under $G$ does not contain all primes above $\mathfrak{p}$. After relabelling, assume $\{\mathfrak{P}_1, \ldots, \mathfrak{P}_s\}$ is the orbit of $\mathfrak{P}_1$ with $s < g$, so that $\mathfrak{P}_{s+1}, \ldots, \mathfrak{P}_g$ are not in this orbit. By the [Chinese Remainder Theorem](/theorems/734) (for the pairwise coprime ideals $\mathfrak{P}_1, \ldots, \mathfrak{P}_g$ of the Dedekind domain $\mathcal{O}_K$), there exists $\alpha \in \mathcal{O}_K$ satisfying: \begin{align*} \alpha &\equiv 0 \pmod{\mathfrak{P}_1}, \\ \alpha &\equiv 1 \pmod{\mathfrak{P}_j} \quad \text{for } j = s+1, \ldots, g. \end{align*} (The primes $\mathfrak{P}_i$ are pairwise coprime since distinct prime ideals in a Dedekind domain are coprime.) Consider the norm element $N_{K/k}(\alpha) = \prod_{\sigma \in G} \sigma(\alpha)$. Since $\alpha \in \mathfrak{P}_1$ and the orbit of $\mathfrak{P}_1$ is $\{\mathfrak{P}_1, \ldots, \mathfrak{P}_s\}$, for each $\sigma \in G$ we have $\sigma(\mathfrak{P}_1) = \mathfrak{P}_j$ for some $j \in \{1, \ldots, s\}$, and $\sigma(\alpha) \in \sigma(\mathfrak{P}_1) = \mathfrak{P}_j$. In particular, every factor $\sigma(\alpha)$ lies in some $\mathfrak{P}_j$ with $j \le s$. Now $N_{K/k}(\alpha) \in \mathcal{O}_k$ (the norm maps $\mathcal{O}_K$ to $\mathcal{O}_k$), and $N_{K/k}(\alpha) = \prod_{\sigma \in G} \sigma(\alpha) \in \mathfrak{P}_1$ (since $\alpha \in \mathfrak{P}_1$ and $\mathfrak{P}_1$ is an ideal, the factor $\sigma = \operatorname{id}$ contributes $\alpha \in \mathfrak{P}_1$, and the entire product lies in $\mathfrak{P}_1$). Since $N_{K/k}(\alpha) \in \mathcal{O}_k \cap \mathfrak{P}_1 = \mathfrak{p}$, we have $N_{K/k}(\alpha) \in \mathfrak{p} \subset \mathfrak{P}_{s+1}$. On the other hand, $\alpha \equiv 1 \pmod{\mathfrak{P}_{s+1}}$. For any $\sigma \in G$, $\sigma(\mathfrak{P}_{s+1})$ is a prime above $\mathfrak{p}$, say $\sigma(\mathfrak{P}_{s+1}) = \mathfrak{P}_\ell$. Then $\sigma(\alpha) \equiv \sigma(1) = 1 \pmod{\sigma(\mathfrak{P}_{s+1})} = \mathfrak{P}_\ell$. But we need to track $\sigma(\alpha)$ modulo $\mathfrak{P}_{s+1}$, not modulo $\sigma(\mathfrak{P}_{s+1})$. We refine the argument. Consider instead the element $\beta = \prod_{\sigma \in G} \sigma(\alpha)$. We have $\beta = N_{K/k}(\alpha) \in \mathfrak{p}$. Now examine $\beta$ modulo $\mathfrak{P}_{s+1}$. Since $\mathfrak{p} \subset \mathfrak{P}_{s+1}$, we have $\beta \in \mathfrak{P}_{s+1}$, i.e., $\prod_{\sigma \in G} \sigma(\alpha) \equiv 0 \pmod{\mathfrak{P}_{s+1}}$. Since $\mathfrak{P}_{s+1}$ is prime, at least one factor $\sigma(\alpha)$ must lie in $\mathfrak{P}_{s+1}$, i.e., $\sigma(\alpha) \in \mathfrak{P}_{s+1}$ for some $\sigma \in G$. This means $\alpha \in \sigma^{-1}(\mathfrak{P}_{s+1})$. Now $\sigma^{-1}(\mathfrak{P}_{s+1})$ is a prime above $\mathfrak{p}$, and by our assumption it lies in the orbit of $\mathfrak{P}_{s+1}$, hence $\sigma^{-1}(\mathfrak{P}_{s+1}) = \mathfrak{P}_\ell$ for some $\ell \in \{s+1, \ldots, g\}$ (since the orbit of $\mathfrak{P}_{s+1}$ is disjoint from $\{\mathfrak{P}_1, \ldots, \mathfrak{P}_s\}$ by hypothesis). But $\alpha \equiv 1 \pmod{\mathfrak{P}_\ell}$ for every $\ell \in \{s+1, \ldots, g\}$, so $\alpha \notin \mathfrak{P}_\ell$. This contradicts $\alpha \in \sigma^{-1}(\mathfrak{P}_{s+1}) = \mathfrak{P}_\ell$. Therefore the orbit of $\mathfrak{P}_1$ contains all primes above $\mathfrak{p}$, and $G$ acts transitively. [guided] We prove transitivity by contradiction, using the Chinese Remainder Theorem to construct an element that leads to a contradiction if the action is not transitive. Suppose the orbit of $\mathfrak{P}_1$ under $G$ is $\{\mathfrak{P}_1, \ldots, \mathfrak{P}_s\}$ with $s < g$. The remaining primes $\mathfrak{P}_{s+1}, \ldots, \mathfrak{P}_g$ form a separate orbit (or union of orbits). By the [Chinese Remainder Theorem](/theorems/734) for the Dedekind domain $\mathcal{O}_K$ (where distinct primes are pairwise coprime), choose $\alpha \in \mathcal{O}_K$ with: \begin{align*} \alpha &\equiv 0 \pmod{\mathfrak{P}_1}, \\ \alpha &\equiv 1 \pmod{\mathfrak{P}_j} \quad \text{for } j = s+1, \ldots, g. \end{align*} Why this choice? The element $\alpha$ vanishes at $\mathfrak{P}_1$ (hence at all primes in its orbit) but is a unit modulo every prime outside the orbit. The norm $N_{K/k}(\alpha) = \prod_{\sigma \in G} \sigma(\alpha)$ will create a contradiction. **Norm computation.** $N_{K/k}(\alpha) \in \mathcal{O}_k$ (norms of algebraic integers are algebraic integers). Since $\alpha \in \mathfrak{P}_1$ and $\operatorname{id} \in G$, the factor $\sigma = \operatorname{id}$ contributes $\alpha \in \mathfrak{P}_1$ to the product. Since $\mathfrak{P}_1$ is a prime ideal, $N_{K/k}(\alpha) = \alpha \cdot \prod_{\sigma \neq \operatorname{id}} \sigma(\alpha) \in \mathfrak{P}_1$. Therefore $N_{K/k}(\alpha) \in \mathfrak{P}_1 \cap \mathcal{O}_k = \mathfrak{p}$. **The contradiction.** Since $\mathfrak{p} \subset \mathfrak{P}_{s+1}$ (as $\mathfrak{P}_{s+1}$ lies above $\mathfrak{p}$), we get $N_{K/k}(\alpha) \in \mathfrak{P}_{s+1}$, i.e., $\prod_{\sigma \in G} \sigma(\alpha) \in \mathfrak{P}_{s+1}$. Since $\mathfrak{P}_{s+1}$ is a prime ideal, some factor $\sigma(\alpha)$ lies in $\mathfrak{P}_{s+1}$. This means $\alpha \in \sigma^{-1}(\mathfrak{P}_{s+1})$. Now $\sigma^{-1}(\mathfrak{P}_{s+1})$ is a prime above $\mathfrak{p}$. Can it equal $\mathfrak{P}_j$ for some $j \le s$? If so, then $\sigma(\mathfrak{P}_j) = \mathfrak{P}_{s+1}$, but $\mathfrak{P}_j$ is in the orbit of $\mathfrak{P}_1$ while $\mathfrak{P}_{s+1}$ is not -- contradiction. So $\sigma^{-1}(\mathfrak{P}_{s+1}) = \mathfrak{P}_\ell$ for some $\ell \ge s + 1$. But $\alpha \equiv 1 \pmod{\mathfrak{P}_\ell}$ by construction, so $\alpha \notin \mathfrak{P}_\ell$, contradicting $\alpha \in \sigma^{-1}(\mathfrak{P}_{s+1}) = \mathfrak{P}_\ell$. This contradiction shows $s = g$, i.e., the action is transitive. [/guided] [/step] [step:Deduce the equality of ramification indices and residue degrees] We prove part (ii). By part (i), the action of $G$ on $\{\mathfrak{P}_1, \ldots, \mathfrak{P}_g\}$ is transitive: for any $i, j$ there exists $\sigma \in G$ with $\sigma(\mathfrak{P}_i) = \mathfrak{P}_j$. We show that $\sigma$ induces an isomorphism between the completions at $\mathfrak{P}_i$ and $\mathfrak{P}_j$ that preserves the ramification data. Since $\sigma: \mathcal{O}_K \to \mathcal{O}_K$ is a ring isomorphism fixing $\mathcal{O}_k$ and $\sigma(\mathfrak{P}_i) = \mathfrak{P}_j$, the map $\sigma$ sends $\mathfrak{P}_i^m$ to $\mathfrak{P}_j^m$ for every $m \ge 1$. Therefore: **Equal ramification indices.** Write $\mathfrak{p} \mathcal{O}_K = \prod_{\ell=1}^{g} \mathfrak{P}_\ell^{e_\ell}$ for the prime factorisation in $\mathcal{O}_K$. Applying $\sigma$ to both sides (and using $\sigma(\mathfrak{p} \mathcal{O}_K) = \mathfrak{p} \mathcal{O}_K$ since $\sigma$ fixes $\mathfrak{p}$): \begin{align*} \mathfrak{p} \mathcal{O}_K = \sigma(\mathfrak{p} \mathcal{O}_K) = \prod_{\ell=1}^{g} \sigma(\mathfrak{P}_\ell)^{e_\ell}. \end{align*} Since $\sigma$ permutes $\{\mathfrak{P}_1, \ldots, \mathfrak{P}_g\}$, this gives a reordering of the prime factorisation. By the [Unique Factorization of Ideals](/theorems/1589), the exponents are uniquely determined, so $e_{\sigma^{-1}(\ell)} = e_\ell$ for all $\ell$. Since the action is transitive, all $e_i$ are equal to a common value $e$. **Equal residue degrees.** The map $\sigma$ induces a ring isomorphism $\mathcal{O}_K / \mathfrak{P}_i \xrightarrow{\sim} \mathcal{O}_K / \mathfrak{P}_j$ that fixes $\mathcal{O}_k / \mathfrak{p}$ (the common residue field of the base). Therefore $[\mathcal{O}_K / \mathfrak{P}_i : \mathcal{O}_k / \mathfrak{p}] = [\mathcal{O}_K / \mathfrak{P}_j : \mathcal{O}_k / \mathfrak{p}]$, i.e., $f_i = f_j$. Since this holds for all $i, j$, all residue degrees equal a common value $f$. **The fundamental identity.** The general identity for prime decomposition in number field extensions states $\sum_{i=1}^{g} e_i f_i = [K : k]$. With $e_i = e$ and $f_i = f$ for all $i$, this becomes $efg = [K : k]$. [/step] [step:Identify the decomposition group and prove the conjugacy statement] We prove part (iii). The decomposition group of $\mathfrak{P}_i$ over $\mathfrak{p}$ is defined as the stabiliser of $\mathfrak{P}_i$ under the $G$-action: \begin{align*} D_{\mathfrak{P}_i/\mathfrak{p}} = \operatorname{Stab}_G(\mathfrak{P}_i) = \{\sigma \in G : \sigma(\mathfrak{P}_i) = \mathfrak{P}_i\}. \end{align*} **Order of $D_{\mathfrak{P}_i/\mathfrak{p}}$.** By part (i), the action of $G$ on $\{\mathfrak{P}_1, \ldots, \mathfrak{P}_g\}$ is transitive, so the orbit of each $\mathfrak{P}_i$ has size $g$. By the [Orbit-Stabiliser Theorem](/theorems/796): \begin{align*} |G| = |G \cdot \mathfrak{P}_i| \cdot |\operatorname{Stab}_G(\mathfrak{P}_i)| = g \cdot |D_{\mathfrak{P}_i/\mathfrak{p}}|. \end{align*} Since $|G| = [K : k] = efg$ (by the [Order of the Galois Group](/theorems/3325) and part (ii)), we obtain $|D_{\mathfrak{P}_i/\mathfrak{p}}| = efg / g = ef$. **Conjugacy.** For any $i, j$, transitivity gives $\tau \in G$ with $\tau(\mathfrak{P}_i) = \mathfrak{P}_j$. We claim $D_{\mathfrak{P}_j/\mathfrak{p}} = \tau D_{\mathfrak{P}_i/\mathfrak{p}} \tau^{-1}$. Let $\sigma \in D_{\mathfrak{P}_i/\mathfrak{p}}$, so $\sigma(\mathfrak{P}_i) = \mathfrak{P}_i$. Then: \begin{align*} (\tau \sigma \tau^{-1})(\mathfrak{P}_j) = \tau(\sigma(\tau^{-1}(\mathfrak{P}_j))) = \tau(\sigma(\mathfrak{P}_i)) = \tau(\mathfrak{P}_i) = \mathfrak{P}_j. \end{align*} So $\tau \sigma \tau^{-1} \in D_{\mathfrak{P}_j/\mathfrak{p}}$, giving $\tau D_{\mathfrak{P}_i/\mathfrak{p}} \tau^{-1} \subset D_{\mathfrak{P}_j/\mathfrak{p}}$. Since both groups have the same order $ef$ (by the calculation above), the inclusion is an equality: $D_{\mathfrak{P}_j/\mathfrak{p}} = \tau D_{\mathfrak{P}_i/\mathfrak{p}} \tau^{-1}$. [guided] Part (iii) connects the decomposition group to the orbit-stabiliser structure. The decomposition group $D_{\mathfrak{P}_i/\mathfrak{p}} = \{\sigma \in G : \sigma(\mathfrak{P}_i) = \mathfrak{P}_i\}$ is exactly the stabiliser of $\mathfrak{P}_i$ under the transitive $G$-action on $\{\mathfrak{P}_1, \ldots, \mathfrak{P}_g\}$. **Computing the order.** The [Orbit-Stabiliser Theorem](/theorems/796) states that for a finite group $G$ acting on a set $X$ with $x \in X$: \begin{align*} |G| = |G \cdot x| \cdot |\operatorname{Stab}_G(x)|. \end{align*} The orbit $G \cdot \mathfrak{P}_i = \{\mathfrak{P}_1, \ldots, \mathfrak{P}_g\}$ has size $g$ (by transitivity). The [Order of the Galois Group](/theorems/3325) gives $|G| = [K : k]$. By part (ii), $[K : k] = efg$. Therefore: \begin{align*} |D_{\mathfrak{P}_i/\mathfrak{p}}| = \frac{|G|}{g} = \frac{efg}{g} = ef. \end{align*} **Conjugacy of stabilisers.** This is a general fact about group actions: stabilisers of elements in the same orbit are conjugate. Here is the explicit computation. Given $\tau \in G$ with $\tau(\mathfrak{P}_i) = \mathfrak{P}_j$, take any $\sigma \in D_{\mathfrak{P}_i/\mathfrak{p}}$. Then $\sigma(\mathfrak{P}_i) = \mathfrak{P}_i$, and we compute: \begin{align*} (\tau \sigma \tau^{-1})(\mathfrak{P}_j) = \tau(\sigma(\underbrace{\tau^{-1}(\mathfrak{P}_j)}_{= \mathfrak{P}_i})) = \tau(\underbrace{\sigma(\mathfrak{P}_i)}_{= \mathfrak{P}_i}) = \tau(\mathfrak{P}_i) = \mathfrak{P}_j. \end{align*} So $\tau \sigma \tau^{-1} \in D_{\mathfrak{P}_j/\mathfrak{p}}$, and thus $\tau D_{\mathfrak{P}_i/\mathfrak{p}} \tau^{-1} \subset D_{\mathfrak{P}_j/\mathfrak{p}}$. By the same argument with $\tau^{-1}$ (noting $\tau^{-1}(\mathfrak{P}_j) = \mathfrak{P}_i$), $\tau^{-1} D_{\mathfrak{P}_j/\mathfrak{p}} \tau \subset D_{\mathfrak{P}_i/\mathfrak{p}}$, giving the reverse inclusion. Hence $D_{\mathfrak{P}_j/\mathfrak{p}} = \tau D_{\mathfrak{P}_i/\mathfrak{p}} \tau^{-1}$. This conjugacy means the decomposition groups of different primes above $\mathfrak{p}$ are "the same up to conjugation." In particular, all decomposition groups are isomorphic, and any property invariant under conjugation (such as being abelian, being cyclic, or having a particular composition series) is shared by all of them. [/guided] [/step]