Determine galois group $Gal \Big(\frac{\mathbb{Q}(\sqrt[3]{3},\sqrt{-3})}{\mathbb{Q}}\Big)$ I've had some hard time determining Galois group  $Gal \Big(\frac{\mathbb{Q}(\sqrt[3]{3},\sqrt{-3})}{\mathbb{Q}}\Big)$ because I didn't know exactly how to compute the order of the elements. See here for the computation of the orders and the order of the group. 
However, I'm arriving at two contradictory conclusions. I get that the order of the group is 6 and that all the non-identity elements are of order 2. However, I know this cannot be the case since there are only the diedric group and the cyclic group of order 6 and in both you find elements of order three. 
What am I doing wrong? 
\begin{array}{|c|c|c|c|}
\hline
 notation & \sqrt[3]{3} & \sqrt{-3} & order \\ \hline
 Id  &  \sqrt[3]{3} & \sqrt{-3} & 1\\ \hline
 & \sqrt[3]{3} & -\sqrt{-3} & 2\\ \hline
 &  \omega \sqrt[3]{3}& \sqrt{-3}& 2\\ \hline
&  \omega \sqrt[3]{3}& -\sqrt{-3}& 2\\ \hline
&  \omega^2 \sqrt[3]{3}& \sqrt{-3}& 2\\ \hline
&  \omega^2 \sqrt[3]{3}& -\sqrt{-3}& 2\\ \hline
\end{array}
I do the computations considering that $\omega = -\frac{1}{2}+\frac{\sqrt{-3}}{2}$. So for instance if I compute the order of the isomorphism that sends $\sqrt[3]{3} \mapsto \omega^2 \sqrt[3]{3}$ and $\sqrt{-3} \mapsto -\sqrt{-3}$ I observe that two applications of this isomorphism on $\sqrt{-3}$ will give the identity and for the other element I have the following chain $\sqrt[3]{3} \mapsto \omega^2 \sqrt[3]{3} \mapsto \omega \omega^2 \sqrt[3]{3}$ observing that $\omega^2 \mapsto \omega$.
Edit
Let me update the table as I go along:
\begin{array}{|c|c|c|c|}
\hline
 notation & \sqrt[3]{3} & \sqrt{-3} & order \\ \hline
 Id  &  \sqrt[3]{3} & \sqrt{-3} & 1\\ \hline
 & \sqrt[3]{3} & -\sqrt{-3} & 2\\ \hline
 &  \omega \sqrt[3]{3}& \sqrt{-3}& 3\\ \hline
&  \omega \sqrt[3]{3}& -\sqrt{-3}& 2\\ \hline
&  \omega^2 \sqrt[3]{3}& \sqrt{-3}& 3\\ \hline
&  \omega^2 \sqrt[3]{3}& -\sqrt{-3}& 2\\ \hline
\end{array}
So I conclude the group is diedric of 6 elements. Thanks everyone. 
 A: Wouldn't it be simpler to remark that by definition, $\omega$ is a primitive cubic root of $1$, so that your field $K$ is just the splitting field of the polynomial $X^3 - 3$ ? As such, $K/\mathbf Q$ is normal, with Galois group $G$ isomorphic to the permutation group of the roots, so $G\cong S_3 \cong D_6$, generated by the transposition $\tau:\omega \to \omega^2, \sqrt [3] 3 \to \sqrt [3] 3$ and the 3-cycle $\sigma: \sqrt [3] 3 \to \omega \sqrt [3] 3, \omega \to \omega$ .
A: Consider the automorphism $\sigma$ such that $\sigma(\sqrt[3]{3}) = \omega \sqrt[3]{3}$ and $\sigma(\sqrt{-3}) = \sqrt{-3}$ and note that by your expression for $\omega$, then $\sigma(\omega) = \omega$:
$$
\sigma(\omega) = \sigma \left(\frac{-1 + \sqrt{-3}}{2}\right) = \frac{-1 + \sigma(\sqrt{-3})}{2} = \frac{-1 + \sqrt{-3}}{2} = \omega \, .
$$
Then
$$
\sigma^2(\sqrt[3]{3}) = \sigma(\omega \sqrt[3]{3}) = \sigma(\omega) \sigma(\sqrt[3]{3}) = \omega \sigma(\sqrt[3]{3}) = \omega \omega \sqrt[3]{3} = \omega^2 \sqrt[3]{3} \neq \sqrt[3]{3}
$$
which shows that $\sigma^2 \neq \mathrm{id}$, so $\sigma$ can't have order $2$.
