What is the Galois group of $27x^8 - 72x^4 - 16$ over the rationals? Consider the polynomial $f = 27x^8 - 72x^4 - 16$. My question is: what is the Galois group of the splitting field of $f$ over the rational numbers?
When I tried to calculate this, I found that there are $32$ elements of the Galois group, but Sage and Magma both say that there are only $16$ elements. I struggle, however, to see why this is the case because this example is rather subtle. Any suggestions / corrections would be most welcome.
What I've done so far:
I've found that the $8$ roots of $f$ are as follows:
$$\pm \sqrt{\dfrac{2}{3}}\sqrt[4]{3 \pm 2 \sqrt{3}}, \, \, \, \,\,\,\,\, \pm i \sqrt{\dfrac{2}{3}}\sqrt[4]{3 \pm 2 \sqrt{3}}.  $$
An automorphism of $K$ is determined by where it sends the two elements $\sqrt{\dfrac{2}{3}}\sqrt[4]{3 + 2 \sqrt{3}}$ and $\sqrt{\dfrac{2}{3}}\sqrt[4]{3 - 2 \sqrt{3}}$. Further, these two elements, which I will call $A$ and $B$ respectively, satisfy the following algebraic relation with $\mathbf{Q}$-coefficients:
$$ A^4 B^4 = -\dfrac{4}{3}.$$
This means that if $A$ is sent to an element with a $+$ sign underneath the $4^{\rm{th}}$ root, then $B$ is sent to an element with a $-$ sign underneath the $4^{\rm{th}}$ root, and vice versa. Because if both their images have the same sign under the $4^{\rm{th}}$ root, when we take the fourth power and multiply them, we won't get a rational number, so the above algebraic relation won't hold anymore.
The upshot:
There are $8$ possible choices to send $A$ to. For each of those, there are $4$ choices for $B$, since given the algebraic relation written above, $A$ and $B$ are sent to numbers with opposite signs under the $4^{\rm{th}}$ root, and there are $4$ of these. This means that there are $8 \cdot 4 = 32$ automorphisms. But this is wrong! There are only $16$ automorphisms.
I think there is another more elusive algebraic relation these numbers have to satisfy, which constrains the automorphism group to be half as small as expected. But I struggle to find what this other algebraic relation is. If anyone could share an algebraic relation that I'm missing here / share some suggestions for how to move forward, that would be much appreciated! Thanks.
 A: The roots are easily found to be $$\pm\sqrt[4]{\frac{4(3+2\sqrt{3})}{9}}=\pm a\text{ (say)} ,\\ \pm i \sqrt[4]{\frac{4(3+2\sqrt{3})}{9}},\\ \pm(1\pm i) \sqrt[4]{\frac{2\sqrt{3}-3}{9}}=\pm(1\pm i) b\text{ (say)} $$ with $a, b$ being positive real numbers.
The given polynomial is irreducible over $\mathbb {Q} $ and hence $[\mathbb {Q} (a):\mathbb {Q}] =8$. The splitting field $L$ of the polynomial contains $a, ai$ and hence $i\in L$. But $i\notin\mathbb {Q} (a) $ and hence $[\mathbb {Q} (a, i) :\mathbb {Q}] =16$.
It can be proved with some effort that $L=\mathbb {Q} (a, i) $ is the splitting field of the polynomial and hence the Galois group is of order $16$.
We can note that $$ab=\frac{\sqrt[4]{12}}{3}=\frac{c}{3}\text { (say)} $$ and next we show that $c\in L$. Since $c$ is real we have in fact $c\in\mathbb {Q} (a) $. We can observe that $$c^2=2\sqrt{3}=\frac{9a^4-12} {4}\in\mathbb {Q} (a) $$ and
\begin{align}
9a^4&=4c^2+12\notag\\
&=4c^2+c^4\notag\\
&=c^2(c^2+4)\notag\\
&=c^2(4+2\sqrt{3})\notag\\
&=c^2(1+\sqrt{3})^2\notag
\end{align}
This implies $$3a^2=c(1+\sqrt{3})$$ ie $$c=\frac{6a^2}{2+c^2}\in \mathbb {Q} (a) $$ It now follows that $b=c/(3a)\in\mathbb{Q } (a) \subset L$ and $L$ is the desired splitting field.
The Galois group is $D_{16}$ (dihedral group of order $16$) as explained in another answer.

Based on some calculations in another answer here by user dan_fulea one can check that $a/b=1+\sqrt{3}$ which obviously lies in $\mathbb {Q} (a) $. Perhaps checking ratios instead of product of conjugates also helps.
A: First for intuition let us graph the roots of the polynomial:

This suggests the following automorphisms:

*

*$\sigma : z \mapsto \bar{z}$ (flip vertically)

*$\tau : z \mapsto -z$ (flip horizontally)

*$\lambda : z \mapsto i z$ (rotation by 90 degrees)

*$\omega : \sqrt{3} \mapsto - \sqrt{3}$ (mysterious automorphism that swaps the inner wheel with the outer points)

We can label the outer roots $r1$ to $r4$ and the inner roots $r5$ to $r7$ we can then come up with the following permutation representations:

*

*$\sigma = (2\,4)(5\,8)(6\,7)$

*$\tau = (1\,3)(5\,6)(7\,8)$

*$\lambda = (1\,2\,3\,4)(5\,6\,7\,8)$

*$\omega = (2\,8)(3\,7)(4\,6)(1\,5)$ (we do not know this yet, this is calculated later)

and GAP software can recognize this group as D16:
G := Group(
 (2,4)(5,8)(6,7),
 (1,3)(5,6)(7,8),
 (1,2,3,4)(5,6,7,8),
 (1,5)(2,8)(3,7)(4,6)
 );
Order(G);
StructureDescription(G);


To know that the order of the Galois group is 16 we calculate the degree of the splitting field. The splitting field is $$L = \mathbb Q(r_1,r_2,r_3,r_4,r_5,r_7,r_6,r_7,r_8)$$ but we have seen that $r_1$ can be mapped to any of the roots $r_2,r_3,r_4$ by our simple transformations, same with $r_5$ so $L = \mathbb Q(i,r_1,r_5)$ we also have (by raising $r_1$ to the 4th power) $\sqrt{3}$ in our field $L = \mathbb Q(i,\sqrt{3},r_1,r_5)$.
Now you can calculate $r_1 r_5 = 2 \sqrt[4]{\frac{-1}{27}}$ so  $L = \mathbb Q(i,\sqrt{3},r_1,\sqrt[4]{\frac{-1}{27}}) = \mathbb Q(i,\sqrt[4]{-3},r_1)$
Over $\mathbb Q$, $i$ has degree 2. Over $\mathbb Q(i)$, $\sqrt[4]{-3}$ has degree 4 and over $\mathbb Q(i,\sqrt[4]{-3})$ adjoining $r_1$ is a simple quadratic extension. so $[\mathbb Q:L] = 2 \cdot 4 \cdot 2 = 16$.
By watching the effect of $r \mapsto 2 \sqrt[4]{-\frac{1}{27}}/r$ on roots $r_1$ to $r_4$ you can find the exact permutation representation of $\omega$.
