Galois group of $x^6 + 3$ isomorphic to a copy of $S_3$ inside $S_6$

Solution 1:

By Galois theory we know that there exists an automorphism $\sigma$ with the property $$\sigma:\alpha_1\mapsto\alpha_2=\frac{\alpha^4+\alpha}2.$$ From this we can deduce that $$ \sigma(a^3)=\left(\frac{a+a^4}2\right)^3=\frac{a^{12}+3a^9+3a^6+a^3}8. $$ Here in the numerator $a^{12}=-3a^6$ and $a^9=-3a^3$, so this simplifies to $-a^3$ and hence $$ \sigma(\zeta_6)=\sigma\left(\frac{1+a^3}2\right)=\frac{1-a^3}2=\zeta_6^{-1}. $$ Therefore $\sigma(\alpha_2)=\alpha_1$, and by computing the images of the other roots we see that $\sigma$ corresponds to the permutation $(12)(36)(45)$.

Complex conjugation $\rho$ will also be an automorphism of your field, and by plotting the roots on the complex plane we see that $\rho=(16)(25)(34)$.

As products of these we get the other non-trivial automorphisms as permutations of roots: $$ \rho\sigma=(153)(264),\quad \sigma\rho=(135)(246), \quad\sigma\rho\sigma=\rho\sigma\rho=(14)(23)(56). $$ The generators $\sigma$ and $\rho$ satisfy the relations $\sigma^2=\rho^2=(\sigma\rho)^3=1$, so they generate a copy of $S_3$.

Solution 2:

The problem is that the order of the subgroup of $\,S_6\,$ generated by $\,(1\,4)\,,\,(1\,2)\,,\,(1\,3\,2)\,$ is way larger than $\,6\,$ ...Either you made some mistake when calculating the order of the Galoi group (which you didn't since the extension you're talking about is simple with minimal polynomial of degree $\,6\,$), or else you didn't calculate correctly the cycles corresponding to the elements of the group.

Added Seeing the comment by Jyrki I realize your mistake appears, apparently, before all your computations: indeed $\,a^6\neq -3\,$ as it should! You'd in fact need $\,a:=\sqrt[6]{3}\,e^{\pi i/6}\,$