Galois Group Calculation

Question

Calculate the Galois Group $G$ of $K$ over $F$ when $F=\mathbb{Q}$ and $K=\mathbb{Q}\big(i,\sqrt2,\sqrt3 \big)$.

My thoughts are as follows:

By the Tower Lemma, we can see that $|\text{Gal}(K/F)|=8$, since $K$ is a degree $8$ extension over $F$.

Now, $\phi \in\text{Gal}(K/F)$ where $\phi$ represents complex conjugation. This satisfies $\phi ^2 =\text{Id}$. Similarly, the map which switches $\sqrt2$ and $\sqrt3$ around is also an automoprhism fixing $\mathbb{Q}$, say $\tau$, satisfying $\tau ^ 2=\text{Id}$.

How can I find the other elements of the Galois Group? Have I missed an easier method?

Answer 1

It's just $(\mathbb{Z}/2\mathbb{Z})^3$.

To see this, indeed complex conjugation is an automorphism of order 2. However, so are the automorphisms sending $\sqrt{2}\mapsto -\sqrt{2}$, and $\sqrt{3}\mapsto-\sqrt{3}$ (and fixing the rest). Hence, you have 3 elements of order 2, which commute and generate your group (check this!). These three automorphisms then give you a nice isomorphism to $(\mathbb{Z}/2\mathbb{Z})^3$.

Answer 2

Though your question has already been answered, I think it's important to emphasize exactly why these roots can only go to their negatives (perhaps for readers in the future).

Take an arbitrary $\mathbb{Q}$-automorphism $\phi$. What if $\phi(\sqrt{2}) = \sqrt{3}$? Well, we get the following contradiction:

$$\phi(2) = \phi((\sqrt{2})^2) = \phi(\sqrt{2}) \phi(\sqrt{2})=\sqrt{3}\sqrt{3} = 3$$

And this is no good because $\phi(2) = 2$ by virtue of $\phi$ fixing $\mathbb{Q}$.