Identifying the Galois Group $G(\mathbb{Q}[\sqrt{2}, \sqrt[3]{2}]/\mathbb{Q})$ I am trying to determine the Galois group $G(\mathbb{Q}[\sqrt{2}, \sqrt[3]{2}]/\mathbb{Q})$.  I am fairly confident I have the correct answer, but I need someone to confirm my work since I have just taught myself this material today.
First, note that $K = \mathbb{Q}[\sqrt{2}, \sqrt[3]{2}]$ is a field extension of degree $6$.  However, it is not a splitting field of some polynomial with coefficients in $\mathbb{Q}$. Thus, the order of the Galois group will be strictly less than $6$ since it is not a Galois extension.  
Consider the polynomial $f(x) = (x^2 - 2)(x^3 - 2)$.  This polynomial has $3$ roots in $K$, namely $\sqrt{2}$, $-\sqrt{2}$, and $\sqrt[3]{2}$.  Any $\mathbb{Q}$-automorphism of $K$ will permute these three roots.  Note that for any such automorphism, $0 = \phi(0) = \phi(\sqrt{2} - \sqrt{2}) = \phi(\sqrt{2})+\phi(-\sqrt{2})$.  Hence, the only two possible automorphisms are the identity and that which swaps $\sqrt{2}$ with $-\sqrt{2}$.  
We conclude that $G(K/\mathbb{Q}) \cong \mathbb{Z}_2$.

Well, $G(\mathbb{Q}[\sqrt{2}, \sqrt[3]{2}]/\mathbb{Q})$  is not Galois. What you have computed here is the automorphism group $Aut(\mathbb{Q}(\sqrt[6]{2})/\mathbb{Q})$ rather than the Galois group (i.e., automorphism group of $(\mathbb{Q}(\sqrt[6]{2}),\sqrt{-3})$, the closure $\mathbb{Q}[\sqrt{2},\sqrt[2]{3}]$ over $\mathbb{Q}$. 
