# Galois group of $\mathbb{Q}(i, \sqrt2, \sqrt3)$

I'm trying to find the Galois group of $K = \mathbb{Q}(i, \sqrt2, \sqrt3)$ over $\mathbb{Q}$ and the subgroups corresponding to various subfields but I'm having some difficulty.

I think I have found that the degree of $K$ is 8, and it has 3 elements of order 2. The elements that send $i \to -i$, $\sqrt2 \to -\sqrt2$ and $\sqrt3 \to -\sqrt3$. And hence the Galois group must be $C_4 \times C_2$ I think!

Now I am trying to find the subgroups corresponding to the fields $\mathbb{Q} (\sqrt2, i)$ and $\mathbb{Q}(\sqrt2, i\sqrt3)$. My problem is I seem to have far too many degree 4 subfields, but $C_4 \times C_2$ has only 3 degree 2 subgroups? I have found that adjoining any two elements gives a degree 4 extension (e.g $\mathbb{Q}(\sqrt(2), i)$) and there are 3 ways to do that. But then looking at $\mathbb{Q}(\sqrt2, i\sqrt3)$ this is also a degree 4 extension and now I've ran out of subgroups?

Thanks for any help

## 1 Answer

This happens because the Galois group is not $C_4\times C_2$ but is $C_2^3$. In fact you can see that every element of the Galois group has order $2$, and the only group of order $8$ with such a property is $C_2^3$.