Show that $G$ has a cyclic normal subgroup of order $9$. 
Consider $\mathbb{Q}(\sqrt[9]{5},e^{\frac{2\pi i}{9}})$, which is the splitting field of $f(x) = x^9 - 5$. Let the $G = \text{Gal}_\mathbb{Q} \mathbb{Q}(\sqrt[9]{5},e^{\frac{2\pi}{9}})$. Show that $G$ has a cyclic normal subgroup of order $9$.

I know that the degree of extension if $72$ since
$$\mathbb{Q}\longrightarrow^{\deg = 9} \mathbb{Q}(\sqrt[9]{5})\longrightarrow^{\deg = 6}\mathbb{Q}(\sqrt[9]{5},e^{\frac{2\pi i}{9}})$$
So $|G|=72=3^3 \times 2$, thus $G$ has a subgroup of order $9$. But I don't see why it should be normal and cyclic.
My thought is that to use Fundamental Theorem of Galois Theory. So I need a degree $9$ Galois extension. But $\mathbb{Q}(\sqrt[9]{5})$ is not a normal extension of $\mathbb{Q}$. Should I look for a different intermediate field?
 A: Let us denote $\zeta=e^{2\pi/9}$, $L=\Bbb{Q}(\root9\of5,\zeta)$, $K_1=\Bbb{Q}(\root9\of5)$
and $K_2=\Bbb{Q}(\zeta)$. From the theory of cyclotomic polynomials we know that $[K_2:\Bbb{Q}]=6$. Eisenstein's criterion gives that $[K_1:\Bbb{Q}]=9$.
It is not immediately clear, but we can show that $[L:\Bbb{Q}]=54$. The reason is that the polynomial $f(x)=x^9-5$ remains irreducible over $K_2$. For if we had $f(x)=g(x)h(x)$ non-trivially in $K_2[x]$, then the zeros of $g$ would be a non-trivial subset of the numbers $\zeta^j\root9\of5$ with $j\in\{0,1,2,\ldots,8\}$. Therefore the constant term $A=g(0)$ of $g(x)$ would be of the form
$A=5^{\ell/9}\zeta^m$ with $0<\ell<9$. Therefore $5^{\ell/9}\in K_2$. This is a contradiction because $K_2/\Bbb{Q}$ is abelian, and hence the intermediate field
$M=\Bbb{Q}(5^{\ell/9})$ would be Galois also. But the real field $M$ is not Galois over $\Bbb{Q}$ unless $\ell=0$, because it does not contain all the conjugates of $5^{\ell/9}$.
So we can conclude that $|G|=54$. The rest is easy. Let
$$
H=\operatorname{Gal}(L/K_2)\le G
$$
be the subgroup of $G$ Galois corresponding to the intermediate field $K_2$. Because $K_2/\Bbb{Q}$ is Galois, we deduce that $H\unlhd G$. But
$$
|H|=[L:K_2|=\frac{[L:\Bbb{Q}]}{[K_2:\Bbb{Q}]}=9.
$$
Furthermore, $H$ is generated by the $K_2$-automorphism determined by $\root9\of5\mapsto\zeta\root9\of 5$. Therefore $H$ is cyclic and we are done.
