I need help understanding this table for a Galois group

Here we are considering the field $E = \mathbb{Q}(\sqrt{2}, \omega)$, where $\omega$ is a primitive cube-root of unity. The following table represents the Galois group $G(E/\mathbb{Q})$. I know that $\sigma, \tau, \alpha, \beta$ etc... are elements of the group and that they are functions which map elements as specified below each letter. For example, the function $\sigma$ maps $\sqrt{2}$ to $\omega$ and fixes everything else (is that correct?), but why are there two that map $\sqrt{2}$ to $\omega$? Is one the identity? If so, why is there another one that's identical? Also, why doesn't $\sqrt{4}$ appear anywhere?

• It seems that $\sigma$ is the set of generators of $E$ but not a $\mathbb Q$-automorphism. You have after $\sigma$ the identiy, and generators $\alpha$ and $\beta$ of the group. This is certainly the case because if not then your group would have $7$ elements which is absurde. – Piquito May 8 '16 at 17:12

The leftmost column is a label, not data. It says that for an automorphism $\sigma$, the second row indicates what $\sqrt2$ goes to and the third row indicates what $\omega$ goes to. As the splitting field is generated by those two elements, that is enough.
The second column is the identity automorphism $\iota$. It fixes both $\sqrt2$ and $\omega$. The third column is $\alpha$, which fixes $\sqrt{3}$ and $\alpha(\omega) = \omega^2$, and so on.
Observe that $E$ is the splitting field of $x^3-2$ over $\mathbb{Q}$ and that $\mathbb{Q}$ has characteristic zero. It follows that $E$ is a Galois extension of $\mathbb{Q}$. So the size of $G$ is $[E:F]=3(2)=6$ by considering the tower $$\mathbb{Q}\subseteq\mathbb{Q}(\sqrt2)\subseteq\mathbb{Q}(\sqrt2,\omega)$$ and using the tower law.
The elements of $G$ are the automorphisms of $E$ which fix $\mathbb{Q}$. Under such an automorphism it is easily seen that $\sqrt{2}$ is sent to another root of its minimal polynomial $x^3-2$ over $\mathbb{Q}$ and the same is true for $\omega$ whose minimal polynomial is $x^2+x+1$ over $\mathbb{Q}$. There are $3$ choices for $\sqrt{2}$ and $2$ for $\omega$, hence, $6$ potential maps in total all of which exhaust $G$.
We can list the maps in $G$ by considering the generators of $G$ namely the maps $\beta$ which fix $\omega$ and send $\sqrt{2}$ to $\sqrt{2}\,\omega$ and $\alpha$ which fix $\sqrt{2}$ and send $\omega$ to $\omega^2$.