How can I directly see that this subgroup has this subfield as its fixed field?

I'm looking at the Galois group for $$F = \mathbb{Q}(\sqrt[3]{2}, \omega)$$ where $$\omega$$ is the first primitive third root of unity. I know this is the splitting field for $$x^3 - 2$$ and, I'm told, that $$G = Gal(F/\mathbb{Q}) \cong S_3$$ (which, for now, I believe). I can also directly see how most of the subgroups hold their corresponding fixed fields fixed. The only one I can't see directly is how the subgroup of $$G$$ isomorphic $$\\{1, (123), (132)\\}$$ holds $$\mathbb{Q}(\omega)$$ constant. Since it must be generated by some $$\sigma$$ I tried

$$\sigma(\omega) = \sigma(\omega \sqrt[3]{2}/\sqrt[3]{2})$$

but I don't have anywhere to go with that. Can someone help me see what $$\sigma$$ looks like and how I'd get that from the subgroup/group element of $$S_3$$? Thanks!

Note that $$\sigma$$ is a field automorphism. It respects addition, subtraction, multiplication, and division.
The automorphism corresponding to $$\sigma$$ fixes every rational, and sends $$\sqrt[3]{2}$$ to $$\omega\sqrt[3]{2}$$, sends $$\omega\sqrt[3]{2}$$ to $$\omega^2\sqrt[3]{2}$$, and sends $$\omega^2\sqrt[3]{2}$$ to $$\sqrt[3]{2}$$. Therefore, it sends $$\omega$$ to \begin{align*} \sigma(\omega) &= \sigma\left(\frac{\omega\sqrt[3]{2}}{\sqrt[3]{2}}\right)\\ &= \frac{\sigma(\omega\sqrt[3]{2})}{\sigma(\sqrt[3]{2})}\\ &= \frac{\omega^2\sqrt[3]{2}}{\omega\sqrt[3]{2}}\\ &=\omega. \end{align*} So $$\sigma$$ fixes $$\mathbb{Q}(\omega)$$.