 Solution

PutThe four roots are $\alpha ,-\alpha ,\beta ,-\beta$. If $\delta \in \mathbb{Q}(\alpha )$, thenbecause $g$ is irreducible. Hence the Galois group has order four and acts transitively, hence regularly, on the roots.

Let $\tau$ be the automorphism taking $\alpha$ to $\beta$. Since $\tau (\delta )=\pm \delta$,If $\tau (\delta )=\delta$, then $\tau$ is an involution. Together with the distinct involution taking $\alpha$ to $-\alpha$, it givesIn the ordered root set $(\alpha ,-\alpha ,\beta ,-\beta )$ these generators act as double transpositions.

If $\tau (\delta )=-\delta$, thenso $\tau$ has order four andThis is the degree-four case in the Galois group of an irreducible even quartic.