Showing $|\mathrm{Gal}(f(x))| = \deg(f)$ How can I prove that if $f(x)$ is a separable irreducible poly. and $\mathrm{Gal}(f(x))$ is abelian then $|\mathrm{Gal}(f(x))| = \deg(f)$?

Let $K$ be the ground field and let $a$ be a zero of $f$. Then $K(a)$ is a subfield of the splitting field $L$ of $f$. Since $\mathrm{Gal}(L/K)$ is abelian the subgroup fixing $K(a)$ is normal, so $K(a)/K$ is a normal extension. Since $f$ is irreducible and it has a zero in $K(a)$, it splits completely in $K(a)$. So $L=K(a)$. Therefore $[L:K]=\deg(f)$, which means $|\mathrm{Gal}(L/K)|=\deg(f)$.
