Q:

Galois Group Calculation

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?

A:

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 $(\mathbb{Z}/2\mathbb{Z})^3$.

