= Solution
Let $L/K$ be a finite <Galois extension> of <number fields>, with <Galois group> $G$, and let $\mathfrak p$ be a nonzero <prime ideal> of $\mathcal O_K$. Write its <prime ideal factorization> as
$$
\mathfrak p\mathcal O_L=\prod_{j=1}^g\mathfrak P_j^{e_j},\qquad
f_j=[\mathcal O_L/\mathfrak P_j:\mathcal O_K/\mathfrak p].
$$
The <Galois group> acts transitively on the primes above $\mathfrak p$. One proof uses the <Chinese remainder theorem>. If $\mathfrak Q$ were outside the orbit of $\mathfrak P$, choose an <integral element> $a$ which is zero modulo $\mathfrak Q$ and one modulo every conjugate of $\mathfrak P$. Its <field norm> $\prod_{\sigma\in G}\sigma(a)$ belongs to $\mathfrak Q\cap\mathcal O_K=\mathfrak p$, hence to $\mathfrak P$. But each factor is one modulo $\mathfrak P$, a contradiction. Transitivity also makes all $e_j$ equal to a common <ramification index> $e$ and all $f_j$ equal to a common <residue degree> $f$.
Fix $\mathfrak P$ above $\mathfrak p$. Its <decomposition group> is the stabilizer
$$
D_{\mathfrak P}=\{\sigma\in G:\sigma\mathfrak P=\mathfrak P\}.
$$
The <orbit-stabilizer theorem> gives $|D_{\mathfrak P}|=|G|/g$. Elements of $D_{\mathfrak P}$ act on the <residue field> $\kappa(\mathfrak P)$ over $\kappa(\mathfrak p)$; the <group kernel> is the <inertia group>
$$
I_{\mathfrak P}=\{\sigma\in D_{\mathfrak P}:\sigma(a)\equiv a\pmod{\mathfrak P}\text{ for all }a\in\mathcal O_L\}.
$$
Thus inertia measures the <automorphisms> invisible after reduction, whereas decomposition measures all <automorphisms> preserving the chosen prime.
The completion $L_{\mathfrak P}/K_{\mathfrak p}$ is a <Galois extension> with <group> $D_{\mathfrak P}$: preservation of $\mathfrak P$ lets an <automorphism> extend continuously to the completion, and the local embeddings of this normal extension arise in this way. The <degree of a field extension> formula is $[L_{\mathfrak P}:K_{\mathfrak p}]=ef$. Reduction gives the <exact sequence>
$$
1\longrightarrow I_{\mathfrak P}\longrightarrow D_{\mathfrak P}
\longrightarrow\operatorname{Gal}(\kappa(\mathfrak P)/\kappa(\mathfrak p))
\longrightarrow1.
$$
For <surjectivity>, lift a primitive residue-field generator by the <Hensel lemma> to obtain the <maximal unramified subextension of a local field extension> of $L_{\mathfrak P}/K_{\mathfrak p}$. Its <automorphisms> are the residue-field <automorphisms>. They extend to $L_{\mathfrak P}$ because the latter is normal. Since a finite <residue field> extension has a cyclic <Galois group> of order $f$, the <exact sequence> yields
$$
\boxed{|D_{\mathfrak P}|=ef,\qquad |I_{\mathfrak P}|=e,\qquad |D_{\mathfrak P}/I_{\mathfrak P}|=f,\qquad [L:K]=gef.}
$$
The <fixed field> of local inertia is the <maximal unramified subextension of a local field extension>, and the extension above it is <totally ramified>. Primes in the same global orbit have conjugate <decomposition groups> and <inertia groups>, so the orders do not depend on the chosen prime.
If $e=1$, inertia is trivial and $D_{\mathfrak P}$ is generated by the unique <Frobenius automorphism> acting on residues by $z\mapsto z^{N\mathfrak p}$. Its order is $f$. In a ramified extension this residue action determines only a coset modulo inertia. A prime splits completely exactly when $e=f=1$, equivalently its <decomposition group> is trivial. A <totally ramified> prime has $g=f=1$ and $I_{\mathfrak P}=D_{\mathfrak P}=G$; an unramified prime has $I_{\mathfrak P}=1$. In an <abelian extension>, conjugation does nothing, so the associated <groups> and unramified <Artin symbol> are independent of the prime above $\mathfrak p$.
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