Do elements of a Galois group act multiplicatively?

Question

Let F be a number field, let $\tau $ be an element of the Galois group of $F$ over the rationals, and let $\alpha , \beta \in F$. Do we have $\tau (\alpha \beta ) = \tau (\alpha ) \tau (\beta )$ in general, where $\tau (\alpha )$ is the action of $\tau $ on $\alpha $? If so, how can we prove this? If not, what are the conditions on $F$, $\alpha , \beta $, $\tau $ such that the claim does hold?

Answer 1

There is nothing to demonstrate... This is part of the definition of the Galois group.

An element of the Galois group is a field automorphism of $F/\mathbb Q$. Which implies in particular that $\tau(\alpha \beta) = \tau(\alpha)\tau(\beta)$ for any $\alpha, \beta \in F$.