I have seen the the thread here related to the computation of the Galois group of the same polynomial. However, my question is not about the computation itself but about the group presentation of the Galois group. I will explain.

I have determined that the polynomial $x^6 +3 \in \Bbb{Q}[x]$ has Galois group of order 6. The splitting field is $\Bbb{Q}(a)$, where $a$ is a root of $x^6 + 3$. One can take $a = \sqrt[6]{3}\zeta$ where $\zeta = e^{2\pi i/6} = e^{\pi i/6}$.

Now I have determined the rest of the roots to be:

$$\begin{array}{ccccc} \alpha_1 &=& a &&& \alpha_4 = -\alpha_1 \\ \alpha_2 &=& \frac{a^4 + a}{2}= \zeta a &&& \alpha_5 = -\alpha_2 \\ \alpha_3 &=& \frac{a^4 - a}{2} = \zeta^2 a &&& \alpha_6 = - \alpha_3 \end{array}. $$

I have also computed some automorphisms of the Galois group, for example the automorphism $\tau : \alpha_1 \mapsto \alpha_4$ that has order 2, $\sigma : \alpha_1 \mapsto \alpha_2$ that has order 2 and $\rho : \alpha_1 \mapsto \alpha_3$ that has order 3. The presence of two automorphisms of order 2 tells me that the Galois group is isomorphic to $S_3$

However the problem now is if I want to identify my $\tau,\sigma$ and $\rho$ as cycles in $S_6$, I get the cycles $(14)$, $(12)$ and $(132)$. I don't think these cycles lie in the copy of $S_3$ inside of $S_6$; what am I misunderstanding here?


Edit: I made a mistake in the calculations. We actually need $\alpha_1 = a = \sqrt[3]{3}\zeta$ where $\zeta = e^{\pi i/6}$. I did not take a primitive 6-th root of unity earlier. Now if I write $a = \sqrt[6]{3}e^{\pi i/6}$, then $a^3 = \sqrt{3}i$ and so $\frac{1 + a^3}{2} = \frac{1 + \sqrt{3}i}{2} = \zeta^2$. So indeed with the redefined $\zeta$ and $a$, the equations now are

$$\begin{array}{ccccc} \alpha_1 &=& a &&& \alpha_4 = -\alpha_1 \\ \alpha_2 &=& \frac{a^4 + a}{2}= \zeta^2 a &&& \alpha_5 = -\alpha_2 \\ \alpha_3 &=& \frac{a^4 - a}{2} = \zeta^4 a &&& \alpha_6 = - \alpha_3. \end{array} $$

After all this mess, I have got the automorphisms $\sigma = (12)(45)(36)$ and $\gamma = (135)(246)$. We check that $\sigma\gamma = \gamma^2\sigma$. $\sigma\gamma = (12)(36)(45)(135)(246) = (16)(25)(34)$.

$\gamma^2\sigma = (153)(264)(12)(36)(45) = (16)(25)(34)$ so indeed $\sigma\gamma = \gamma^2\sigma$. Hence the Galois group has elements

$$\{1,\gamma,\gamma^2,\sigma,\sigma\gamma, \sigma\gamma^2\} = \{1, (135)(246),(153)(264),(12)(45)(36),(14)(23)(56),(16)(34)(25)\}.$$

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    $\begingroup$ Hmm. With your $a$ I get $a^6=3$. But weren't we supposed to have $a^6=-3$? $\endgroup$ – Jyrki Lahtonen Jun 9 '12 at 11:47
  • $\begingroup$ @JyrkiLahtonen Yes I think you're right. I have a feeling that I want that $\zeta$ to actually be $e^{\pi i/6}$ $\endgroup$ – user38268 Jun 9 '12 at 11:48
  • $\begingroup$ @JyrkiLahtonen I have edited the the post. $\endgroup$ – user38268 Jun 9 '12 at 11:58
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    $\begingroup$ Take any element $x$ of order 2, and any element $y$ of order 3. If $xy=y^2x$ then you've got yourself a copy of $S_3$. $\endgroup$ – Gerry Myerson Jun 9 '12 at 12:15
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    $\begingroup$ You've got a typo in your last equation, right side, 4th element of the set. $\endgroup$ – Gerry Myerson Jun 9 '12 at 13:02

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 $S_3$.


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 $\,a:=\sqrt[6]{3}\,e^{\pi i/6}\,$


Your $\tau$ isn't $(14)$; it's $(14)(23)(56)$. I'd check the other ones, too, if I were you.


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