Let $E/F$ be a finite Galois extension: this means that $F$ is a finite-index subfield of $E$ and that the size of the group $\mathrm{Aut}(E/F)$ of automorphisms of $E$ fixing every element of $F$ is at least (in fact, equal to) the index $[E:F]$. In this case, $G\coloneqq\mathrm{Aut}(E/F)$ is called the Galois group of the Galois extension $E/F$.