Let $L/K$ be a Galois extension with group $G = \text{Gal}(L/K)$ of order $n = [L : K]$. Let $\mathfrak{p}$ be a prime of $K$ and $\mathfrak{P}_1, \ldots, \mathfrak{P}_g$ the primes of $L$ above $\mathfrak{p}$.

Step 1: Galois action on primes

The group $G$ acts transitively on $\{\mathfrak{P}_1, \ldots, \mathfrak{P}_g\}$. For $\sigma \in G$, if $\mathfrak{P} | \mathfrak{p}$, then $\sigma(\mathfrak{P}) | \mathfrak{p}$ also.

By orbit-stabilizer: $g = |G| / |D_{\mathfrak{P}_1}|$ where $D_{\mathfrak{P}_1}$ is the decomposition group.

Step 2: Decomposition group and ramification

The decomposition group fits into an exact sequence: $1 \to I_{\mathfrak{P}_1} \to D_{\mathfrak{P}_1} \to \text{Gal}((\mathcal{O}_L/\mathfrak{P}_1)/(\mathcal{O}_K/\mathfrak{p})) \to 1$

The quotient is cyclic of order $f = f(\mathfrak{P}_1|\mathfrak{p})$, generated by the Frobenius element.

The inertia group $I_{\mathfrak{P}_1}$ has order $e = e(\mathfrak{P}_1|\mathfrak{p})$, the ramification index.

Step 3: Computing the orders

From the exact sequence: $|D_{\mathfrak{P}_1}| = |I_{\mathfrak{P}_1}| \cdot f = ef$.

From orbit-stabilizer: $g = n / |D_{\mathfrak{P}_1}| = n / (ef)$.

Step 4: Conclusion

Rearranging: $efg = n$, the fundamental equality.

When $L/K$ is abelian, all decomposition groups are equal as subgroups of $G$, so $e, f, g$ are independent of the choice of prime above $\mathfrak{p}$.