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.
