Prove that the Galois group $G = Gal(x^5-2; \mathbb{Q})$ is a group metacyclic. Prove that the Galois group $G = Gal(x^5-2; \mathbb{Q})$ is a group metacyclic, that is, $G$ contains a normal cyclic subgroup $N$ whose quotient $G/N$ is also cyclic.
MY ATTEMPTY: I was trying to follow this idea I found, which is:
You have a group $G$ that has a cyclic normal subgroup $K$ such that its quotient $G/K$ is also cyclic. Now take a quotient group $G/N$ of $G$ by a normal subgroup $N$. The image $KN/N$ of $K$ in $G/N$ is clearly also cyclic, as a homomorphic image of the cyclic group $K$. And for the quotient we have
\begin{equation}
(G / N) / (K N / N) \cong G / K N \cong (G /K) / (KN / K),
\end{equation}
which is then also cyclic, as a homomorphic image of the cyclic group $G/K$.
However, first I need to prove that the mentioned Galois Group is metacyclic.
Would you help me?
 A: Hint. Show that $N=\operatorname{Aut}(L/\mathbf{Q}(\zeta_5))$ satisfies these properties using the Galois correspondence and exhibiting the explicit automorphisms in $G=\operatorname{Gal}(\mathbf{Q}(\zeta_5,\sqrt[5]{2})/\mathbf{Q})$.
A: Let $f(x) = x^5 - 2 \in \mathbb{Q}[x]$. Let $\alpha = 2^{1/5}$ and $\omega = e^{2\pi i/5}$. Then the splitting field of $f$ over $\mathbb{Q}$ is $L = \mathbb{Q}(\alpha, \omega)$. Since $f$ is irreducible and $f(\alpha) = 0$, we have $[\mathbb{Q}(\alpha) : \mathbb{Q}] = \deg f = 5$. Also, $5$ is prime so the $5^\text{th}$ cyclotomic polynomial is $x^4 + x^3 + x^2 + x + 1$, which is irreducible, so $[\mathbb{Q}(\omega) : \mathbb{Q}] = 4$. By the tower law, $4$ and $5$ both divide $[L : \mathbb{Q}]$, so $20$ divides $[L : \mathbb{Q}]$. But by the Tower Law, we have
$$
[L:\mathbb{Q}]=[\mathbb{Q}(\alpha, \omega) : \mathbb{Q}] = [\mathbb{Q}(\alpha, \omega) : \mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}].
$$
Since $[\mathbb{Q}(\alpha, \omega) : \mathbb{Q}(\alpha)]$ is at most $4$, and $[\mathbb{Q}(\alpha):\mathbb{Q}] = 5$, we have $[L:\mathbb{Q}]\leq 20$. But since $20$ divides $[L:\mathbb{Q}]$, we have $[L:\mathbb{Q}] = 20$. The extension is a splitting field, so it is normal, and $\operatorname{char}\mathbb{Q} = 0$, so the extension is separable. Therefore $L/\mathbb{Q}$ is a Galois extension and we have $\lvert G \rvert = 20$.
Let $M = \mathbb{Q}(\omega)$. Then $M$ is the splitting field of $x^5 - 1$ over $\mathbb{Q}$, so $M/\mathbb{Q}$ is Galois. By the Fundamental Theorem of Galois Theory $H = \operatorname{Gal}(L/M)$ is a normal subgroup of $G$. Furthermore, we have
$$
G / H \cong \operatorname{Gal}(M/\mathbb{Q}).
$$
Note that $[L : M] = 5$, so $H$ has order $5$, which is prime, so it is cyclic. It remains to show that $\operatorname{Gal}(M/\mathbb{Q})$ is cyclic.
Note that $M$ is the splitting field of the irreducible polynomial $g(x) = x^4 + x^3 + x^2 + x + 1$ over $\mathbb{Q}$, so there is an element $\sigma$ of $H$ with $\sigma(\omega) = \omega^2$ (since these are both roots of $g$, and the Galois group of an irreducible polynomial acts transitively on the roots). Clearly $\sigma^2(\omega) = \omega^4 \neq \omega$, so since $\lvert H \rvert = 4$, we have that $H = \langle \sigma \rangle$ is cyclic.
A: An explicit solution. (I started it, and although there are already good answers, i will submit it, sometimes the pedestrian view may be useful for the reader.)

Let $K\subset \Bbb C$ be the field generated by the roots of the poylnomial
$$
X^5-2\in\Bbb Q[X]\ .
$$
Then one root is $a:=\sqrt[5]2\in\Bbb R$. (The usage of analysis is done rather to have clear notations.) The other roots are of the shape
$$
a\zeta^k\ ,\qquad k\in\{0,1,2,3,4\}\ .
$$
Here, $\zeta=\zeta_5=\cos\frac{2\pi}5+i\sin\frac{2\pi}5$ is a primitive root of unity (in $\Bbb C$, in case analysis is not wanted, take $\zeta$ in some theoretical algebraic closure).
Then a subfield of $K$ is the cyclotomic field $\Bbb Q(\zeta)$. And
$$
K=\Bbb Q(\zeta,a)\ .
$$
A Galois substitution $s\in G$ in the Galois group $G:=\operatorname{Gal}(K:\Bbb Q)$ is determined by the images of the generators $\zeta$ and $a$, which are respectively conjugated to $\zeta$ and $a$. So:
$$
\begin{aligned}
s(\zeta) &=\zeta^k&&\text{ for some unique $k$ among $1,2,3,4$, and}\\
s(a) &= a\zeta^j&&\text{ for some unique $j$ among $0,1,2,3,4$ .}
\end{aligned}
$$
Let us denote by $s(k, j)\in G$ the substitution determined by the above relations.
Then $G$ is the set of all $4\cdot 5=20$ substitutions $s(k,j)$, $k,j$ as above. (A vector space basis of $K$ over $\Bbb Q$ has also $20$ elements, $a^j\zeta^k$, $j$ among $0,1,2,3,4$, and $k$ among... $0,1,2,3$ usually, since $1+\zeta+\zeta^2+\zeta^3+\zeta^4=0$, but $1,2,3,4$ instead is also possible.)
Let
$\sigma=s(2,0)$ and $\tau=s(0,1)$. So

*

*$\sigma$ maps $\zeta\to\zeta^2$ and $a\to a$. (It invariates $a$, has order four since $\sigma^4\zeta=\zeta^{2^4}=\zeta$, and this is the minimal $k>0$ with $\zeta^{2^k}=\zeta$, the inverse $\sigma^{-1}=\sigma^3$ also invariates $a$ and maps $\zeta$ to $\zeta^{2^3}=\zeta^3$. It generates a cyclic (abelian) group of order four  $S<G$.


*$\tau$ maps $\zeta\to\zeta$ and $a\to a\zeta$. It invariates $\zeta$, and has order five, $\tau^j a=a\zeta^j$. Its inverse is $\tau^4$, also invariates $\zeta$ and maps
$a\to a\zeta^4$. So $\tau$ generates a cyclic (abelian) group of order five $T<G$.
So $\sigma, \tau$ generate $G$.
The two mophisms do not commute, but we have.
$$
\begin{aligned}
\sigma^{-1}\tau\sigma\;a
&=
\sigma^{-1}\tau\;a=
\sigma^{-1}\;a\zeta
=a\cdot\sigma^{-1}\zeta
=a\zeta^3
=
\tau^3\; a\ ,
\\
\sigma^{-1}\tau\sigma\;\zeta 
&=
\sigma^{-1}\tau\;\zeta^2=
\sigma^{-1}\;\zeta^2=\zeta
=
\tau^3\; \zeta\ .
\qquad\text{This gives:}\\
\sigma^{-1}\tau\sigma &= \tau^3\ ,\\
\sigma^{-1}T\sigma &= T\ .
\end{aligned}
$$
So $T\triangleleft G$ is a normal subgroup, cyclic, with five elements.
The quotient $G/T$ has four elements,
$T=[1]$,
$\sigma T=[\sigma]$,
$\sigma^2 T=[\sigma^2]=[\sigma]^2$,
$\sigma^3 T=[\sigma^3]=[\sigma]^3$, it is a commutative group with four elements isomorphic to $S\cong \Bbb Z/4$, thus cyclic.
