Galois group of $L=\mathbb{Q}(i,\sqrt[3]2,\sqrt3)$ over $\mathbb{Q}$ is $D_{12}$

Consider $$L = \mathbb{Q}(i,\sqrt[3]2,\sqrt3)$$. Prove that $$\operatorname{Gal}(L/\mathbb{Q})\cong D_{12}$$.

My attempt:

It is easy to verify that $$[L:\mathbb{Q}]=12$$. In particular, $$L$$ is the splitting field of the separable polynomial $$(x^2+1)(x^3-2)(x^2-3)$$ over $$\mathbb{Q}$$, so $$L/\mathbb{Q}$$ is Galois of degree 12. I'm trying to show that $$\operatorname{Gal}(L/\mathbb{Q})\cong C_6\rtimes C_2$$.

The extension $$\mathbb{Q}(i)/\mathbb{Q}$$ is Galois, and therefore, $$H:=\operatorname{Gal}(L/\mathbb{Q}(i))\trianglelefteq \operatorname{Gal}(L/\mathbb{Q})=: G$$. Then $$[\mathbb{Q}(i):\mathbb{Q}]=2=[G:H] \Rightarrow |H|=6.$$

With the theorem of natural irrationalitiets, we find that $$L/\mathbb{Q}(\sqrt[3]2,\sqrt3)$$ is Galois as well, with $$C_2\cong \operatorname{Gal}(\mathbb{Q}(i)/\mathbb{Q})\cong\operatorname{Gal}(L/\mathbb{Q}(\sqrt[3]2,\sqrt3))=: K\le G.$$

If $$\sigma\in H\cap K$$, then $$\sigma$$ fixes $$L$$, implying $$\sigma=1$$ and $$H\cap K=1$$. Then, by cardinality comparison, we get that $$HK=G$$.

So, it remains to show that $$H\cong C_6$$. This is where I'm stuck. I think $$\sigma\equiv (\sqrt[3]2 \quad \sqrt[3]2\zeta_3\quad\sqrt[3]2\zeta_3^2)(\sqrt3 \quad -\sqrt3)\in H$$ is the element of order 6 that we're looking for, but I'm struggling to write this last reasoning down in a rigorous way. I would just write down the whole original group $$G$$, which is $$G=\{1,(i\quad -i),(\sqrt3\quad -\sqrt3),(i\quad -i)(\sqrt3\quad -\sqrt3), (\sqrt[3]2 \quad\sqrt[3]2\zeta_3),\\(\sqrt[3]2 \quad\sqrt[3]2\zeta_3^2),(i\quad -i)(\sqrt[3]2 \quad\sqrt[3]2\zeta_3), \quad (i\quad -i)(\sqrt[3]2 \quad\sqrt[3]2\zeta_3^2),\\(\sqrt3 \quad -\sqrt3)(\sqrt[3]2 \quad\sqrt[3]2\zeta_3),(\sqrt3 \quad -\sqrt3)(\sqrt[3]2 \quad\sqrt[3]2\zeta_3^2), \\ (i\quad -i)(\sqrt3\quad -\sqrt3)(\sqrt[3]2 \quad\sqrt[3]2\zeta_3),(i\quad -i)(\sqrt3\quad -\sqrt3)(\sqrt[3]2 \quad\sqrt[3]2\zeta_3^2)\}$$ I believe. Then the 6 elements which fix $$i$$ are contained in $$H$$. So, I conclude that $$\sigma\in H$$ of order 6, and therefore $$H\cong C_6$$.

EDIT: I see that my reasoning does not make sense, as $$\sigma$$ is not even contained in $$G$$. So, I definitely need help with that.

Is this alright?

Thanks.

Let $$M=\mathbf{Q}(i)$$. You want to show that $$\operatorname{Gal}(L/M)\cong C_6$$. However, $$M(\sqrt[3]{2})/M$$ is not normal. Indeed, if it were, we would have $$\zeta_3 \sqrt[3]{2}\in M(\sqrt[3]{2})$$, which happens if and only if $$\sqrt{-3}\in M(\sqrt[3]{2})$$. As $$i\in M$$, this occurs if and only if $$\sqrt{3}\in M(\sqrt[3]{2})$$, which is impossible. This shows $$L/M$$ has more than 2 subfields of order 3; as there are only two subgroups of order 6, we have $$\operatorname{Gal}(L/M)\cong S_3$$ instead.
Luckily, this argument shows that we should rather take $$M'=\mathbf{Q}(\sqrt{-3})$$ and consider $$L/M'$$. It then holds that $$L$$ is the compositum of the subfields $$M'(\sqrt{3})$$ and $$M'(\sqrt[3]{2})$$ (both Galois over $$M')$$ and $$M'(\sqrt{3})\cap M'(\sqrt[3]{2})=M'$$. It now holds that $$\operatorname{Gal}(L/M')\cong \operatorname{Gal}(M'(\sqrt{3})/M')\times \operatorname{Gal}(M'(\sqrt[3]{2})/M')\cong C_2\times C_3\cong C_6.$$ I think you can continue from here.
• Amazing answer. Thank you very much. Was it possible to immediately start with $M'$ (i.e., are there some obvious structural clues in the exercise, that would directly make you start with $M'$), or is this really a process of trial-and-error? Jun 13, 2021 at 10:02
• Thank you. Well, you can immediately spot three quadratic subfields, namely $\mathbf{Q}(i)$, $\mathbf{Q}(\zeta_3)=\mathbf{Q}(\sqrt{-3})$ and $\mathbf{Q}(\sqrt{3})$. You can try either three as a base field. Jun 13, 2021 at 20:54