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Show that the Galois group of the splitting field $F$ of $X^3-7$ over $\mathbb{Q}$ is isomorphic to $S_3$.

I have found that the the Galois group is the following:

$$G=\{\tau_{ij}, i=1,2,3, j=1,2\}$$ where $$\tau_{ij}(\sqrt[3]{7})=\omega^{i-1}\sqrt[3]{7} \\ \tau_{ij}(\omega)=\omega^j \\ i=1,2,3 , j=1,2$$

Is this correct??

How could I continue to show that $G$ is isomorphic to $S_3$ ??

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  • $\begingroup$ I assume that $\omega$ is a primitive cube root of $1$. What you have done looks correct. Hint for the rest: Either prove that you have a non-Abelian group of order $6$ here, or consider the cycle structure of these elements on the three roots. $\endgroup$ – Geoff Robinson Jan 7 '15 at 3:05
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First observe that $|G|=6$, because you've written down 6 automorphisms and we know $|G|\leq 3!$. Next observe $G$ is not cyclic, because each of the $\tau_{ij}$ have order $<6$. Thus $G=S_3$.

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find an element $a$ of order 2 ,an element $b$ of order 3 and check whether $ab=b^2a$ holds

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