Why does $x^3-7$ have Galois group isomorphic to $S_3$?

Question

I'm not concerned with showing that the order of the Galois group is $6$; I've already done that. I'm more concerned with the structure of the Galois group. So $x^3-7$ has the roots

\begin{equation} 7^{1/3}, \hspace{0.2cm} 7^{1/3} \zeta, \hspace{0.2cm} 7^{1/3} \zeta^2 \tag{1} \end{equation}

for $\zeta$ a primitive third root of unity. I understand intuitively that the fact that the Galois group isn't cyclic has to do with the notion that since the splitting field is $\mathbb{Q}(7^{1/3}, \zeta)$, then on the one hand an automorphism $\sigma$ of $\mathbb{Q}(7^{1/3}, \zeta)$ can map $7^{1/3}$ to any of the three roots in $(1)$ since it has minimal polynomial $x^3-7$. But $7^{1/3}\zeta$ and $7^{1/3}\zeta^2$ can only be mapped to $7^{1/3}\zeta$ or $7^{1/3}\zeta^2$.

I'm having trouble recalling how to find the order of an automorphism. $($ yes, from the context of groups, the order of $g \in G$ is the smallest $n \in \mathbb{N}$ so that $g^n=1)$. But with automorphisms, I am forgetting what I have to check. Does it suffice to merely check every $\sigma$'s action on the three roots in $(1)$ and if none of them have order $6$, then we're done?

The Galois group can be identified with its action on the roots. Here that implies that the Galois group is a subgroup of $S_3$. On the other hand, the group has six elements. — Jyrki Lahtonen, Jun 02 '24 at 04:22
@JyrkiLahtonen Correct. So $\zeta$ and $\zeta^2$ are bound to $\Phi_3(x)$, the cyclotomic polynomial. But is there a more clever way to show the Galois group is not cyclic besides brutal exhaustion? — Aram Nazaryan, Jun 02 '24 at 04:25
'But $7^{1/3}\zeta$ and $7^{1/3}\zeta^2$ can only be mapped to $7^{1/3}\zeta$ or $7^{1/3}\zeta^2$': no. $G$ acts on the 3 roots by any permutation. — Anne Bauval, Jun 02 '24 at 05:40
You only need to find two automorphisms, which do not commute. So the group of order $6$ cannot be $C_6$, and hence must be $S_3$. For the solution of the whole exercise see this duplicate. — Dietrich Burde, Jun 02 '24 at 11:17

Why does $x^3-7$ have Galois group isomorphic to $S_3$?

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