Suppose that $f \in K[X]$ is a separable irreducible polynomial. Let $L$ be the splitting field of $f$ over $K$. Is it true that $[L:K] = |Gal(f|K)|$?
Why I think this is because, first of all $|Aut(L|K)| = [L:K]$. Further, since if $\alpha_1, \ldots, \alpha_n$ are the roots of $f$, then $L = K(\alpha_1, \ldots, \alpha_n)$, and every automorphism $\sigma \in Aut(L|K)$ is defined by the image of the $\alpha_i$. Thus, two maps that permute the $\alpha_i$ in the same way, induce the same automorphism.