With $f(x)$ being an irreducible, separable polynomial of finite degree and $G/L$ the Galois extension of $f(x)$. I know that $\exists x\in Gal(G/L)$ such that for all pairs of roots $(\alpha,\beta)$ there is a positive integer n such that $x^n(\alpha)=\beta$.
In terms of the symmetric group, if $\deg(f(x))=m$ then the cycle $(1~2~\dots~m-1~m)$ has to exist in the Galois group, where the Galois group can be represented by permutations of the roots of $f(x)$. I just cannot remember why this is true.
It is not difficult to show that the Galois group has to have an element of order m, but unless the polynomial is of prime degree that doesn't help too much. Does anyone have any ideas?