Galois group of the $x^8+8\in\Bbb Q[x]$ I have this problem because I think the extension has degree $16$ but I can't decide the group: I think it could be $\Bbb Z_4\times\Bbb Z_2\times\Bbb Z_2$ but not surely:
First I do complex roots
$$x^8=-8=8e^{\pi i+2k\pi},\;k\in\Bbb Z\;,\;\;and\;then\;\;x_k=8^{1/8}e^{\frac\pi8\left(1+2k\right)},\;k=0,1,2,...,7$$
and I write $w=e^{\pi i/8}\;$ and this is primitive root of order $16$ , and then the roots are $8^{1/8}w^k,\;w=1,3,5,7,9,11,13,15$ . 
Next I think maybe  $8^{1/8}\in\Bbb Q(w)$ but this is not possible because $\;\Bbb Q(w)/\Bbb Q\;$ cyclotomic extension and its order $\phi(16)=8$, and $\;\Bbb Q(8^{1/8})/\Bbb Q\;$ also has the order $8$, and then we could get $\;\Bbb Q(8^{1/8})=\Bbb Q(w)\;$ , but the first is real field and the second has non real complex elements.
I had to do trigonometry but it only took me to $\,8^{1/4}\in\Bbb Q(w)\;$ because
$$\frac1{\sqrt2}=\cos\frac\pi4=2\cos^2\frac\pi8-1\implies\cos^2\frac\pi8=\frac{\frac1{\sqrt2}+1}2=\frac{1+\sqrt2}{2\sqrt2=\sqrt8}\implies\cos\frac\pi8=\frac{\sqrt{1+\sqrt2}}{\sqrt[4]8}$$
and I know also
$$w+w^{-1}=w+\overline w=2\,\text{Re}\,w=2\cos\frac\pi8\in\Bbb Q(w)$$
and this is why I get splitting field of degree $16$ :
$$\left[\Bbb Q(w,8^{1/8})\,:\,\Bbb Q\right]=\left[\Bbb Q(w)(8^{1/8}):\Bbb Q(w)\right]\left[\Bbb Q(w)\,:\,\Bbb Q\right]=2\cdot8=16$$
because $\;x^2-\sqrt[4]8\;$ is the minimal polynomial of $\;\sqrt[8]8\;$ over $\;\Bbb Q(w)\;$ .
Now, I know $\;\text{Gal}\,(\Bbb Q(w)\,/\,\Bbb Q)\cong\left(\Bbb Z/16\Bbb Z\right)^*\cong\Bbb Z_4\times\Bbb Z_2\;$ and is cyclotomic extension. Also, it is clearly $\;\Bbb Q(w,8^{1/8})/\Bbb Q\;$ cyclic extension of order two, and now questions
Question 1 Is there a simpler method to get $\;\sqrt[4]8\in\Bbb Q(w)\;$? It is not hard (only the trigonometry) but it is long and has many calculations and in time of exam perhaps one can use some lemma or theorem...?
Question 2 What is the $\;\text{Gal}\,(\Bbb Q(w,8^{1/8})\,/\,\Bbb Q)\,?$ Is it really $\;\Bbb Z_4\times \Bbb Z_2\times \Bbb Z_2\;$ or some harder group? I read there are 14 groups of order $16$, and only 5 of them are abelian, so is there some lemma or trick to know the group this time?
Question 3 Is my work above right? Any correction, suggestion or comments is very much appreciated, thank you.
 A: Not an elegant answer really, but a few suggestions; there are probably better ways of doing this.
The splitting field is
$\mathbb Q(w^2, w8^{1/8})$. Let's start out by adjoining $w^2$. Since $w^2=(1+i)/\sqrt{2}$, we have that $\mathbb Q(w^2)=\mathbb Q(i,\sqrt{2})$. Thus $f$ now factors $f=(x^4+i8^{1/2})(x^4-i8^{1/2})$.
Since $f$ was irreducible over $\mathbb Q$, the degree of the splitting field is a multiple of $8$, and, in particular, we must adjoin a zero to $\mathbb Q(w^2)$. This gives an extension of degree either $2$ or $4$, and if it were $2$, this would mean that $x^4+i8^{1/2}$ factors into a product of two degree $2$ polynomials over $\mathbb Q(w^2)$. However, this isn't happening; we can factor over $\mathbb C$ and then check that such a factorization would mean that $2^{1/4}\in\mathbb Q(w^2)$, which is clearly not the case because just adjoining $2^{1/4}$ to $\mathbb Q$ is already a degree $4$ extension, which, however, does not contain $i$.
So indeed $[\mathbb Q(w^2, w8^{1/8}):\mathbb Q]=4\cdot 4 = 16$. As for the Galois group: the automorphisms are obtained by mapping $i\mapsto\pm i$, $\sqrt{2}\mapsto\pm\sqrt{2}$ (corresponding to $\mathbb Z_2\times\mathbb Z_2$ so far), and then we map $w8^{1/8}$ to one of its four conjugates, or if $x^4+i8^{1/2}\mapsto x^{1/4}-i8^{1/2}$, then we must map to one of the four zeros of the other polynomial.
It's not immediately clear (to me) what group that is; also, thanks to Jyrki for pointing out a blunder in an earlier version. Here are a few things we can say right away:
(1) (Jyrki) $G$ is not abelian because our extension has the non-normal (over $\mathbb Q$) intermediate field $\mathbb Q(w8^{1/8})$, so $G$ has a non-normal subgroup;
(2) $G$ has a normal subgroup $H\cong\mathbb Z_4$ with $G/H=\mathbb Z_2\times\mathbb Z_2$. This follows by considering the intermediate field $\mathbb Q(w^2)$.

Update: I've got it now, but again it's brute force rather than elegant. Consider the following two automorphisms (I write $R=w8^{1/8}$):
$$
a: i\mapsto i,\quad \sqrt{2}\mapsto -\sqrt{2}, \quad R \mapsto iw^2 R;\\
b: i\mapsto -i, \quad \sqrt{2}\mapsto -\sqrt{2}, \quad R \mapsto R
$$
Then we can check that $a^8=b^2=1$ (and these are the orders), and $ba=a^3b$. Since $b\notin\langle a\rangle$, the elements $a,b$ generate $G$, and the relations we have found are the ones from the presentation of the semidihedral group of order 16.
Alternatively, we can describe this as the semidirect product $\mathbb Z_8 \rtimes\mathbb Z_2$, with the non-identity element of $\mathbb Z_2$ acting on $\mathbb Z_8$ by sending $1\mapsto 3$; this automorphism has order $2$, so this does define a group action. We can then check that $a=(1,0)$, $b=(0,1)$ satisfy the relations.
