Checking if $\mathbb{Q}(5^{1/10},e^{\pi i/5})=\mathbb{Q}(5^{1/10},i)$ While doing some Galois theory I encountered with the field $\mathbb{Q}(5^{1/10},e^{\pi i/5})$, which is a splitting field of the polynomial $f(t)=t^{10}-5$. However, as it can be proved, the degree of this field over $\mathbb{Q}$ is equal to $20$, so I think there should exist a better choice of the generators, so they represent better the structure of the field.
I have thought that $i$ could be a good replacement for $\zeta=e^{\pi i/5}$, but there is a problem; while it is true that $\mathbb{Q}(5^{1/10},\zeta)$ and $\mathbb{Q}(5^{1/10},i)$ have the same order, it is not clear that one has any inclusion between those sets at all; for instance
$$\zeta+\zeta^4=2i\sin(\pi/5)$$
and $i$ would lie in the aforementioned field if $2\sin(\pi/5)$ lies in $\mathbb{Q}(5^{1/10},\zeta)$. However
$$2\sin(\pi/5)=\sqrt{\frac{1}{2}(5 - \sqrt{5})}$$
And whether there exists a crazy linear conmbination in the last field which equals $2\sin(\pi/5)$ or not is something I can't tell at the moment.
Maybe I'm wrong with my intuition and $i$ is a bad choice? Is there some other element $z$ such that $[\mathbb{Q}(z):\mathbb{Q}]=2$ and $\mathbb{Q}(5^{1/10},\zeta)=\mathbb{Q}(5^{1/10},z)$?
Thanks in advance for your help.
 A: Another approach to prove that the Galois group is not abelian could be the following : 
In order to simplify the group let's remember that $\mathbb{Q}(\zeta_{5}+\zeta_{5}^{-1}) = \mathbb{Q}(\sqrt5)$
After doing so we could also notice that $\mathbb{Q}(\sqrt5) \subset \mathbb{Q}(5^{\frac{1}{10}})$
 since $(5^{\frac{1}{10}})^{5} = \sqrt5$, 
This simplifies the expresion of splitting field, because know the splitting field $K$ can be seen as $K = \mathbb{Q}(5^{\frac{1}{5}},\zeta_{5})$, splitting field of $t^5-5$, irreducible over $\mathbb{Q}$ due to Eisenstein's criterion.
It's an other well know result that if $m = [\mathbb{Q}(\alpha) : \mathbb{Q}], n = [\mathbb{Q}(\beta):\mathbb{Q}] \hspace{0.1cm}$ $m,n \in \mathbb{N}$ such that $gcd(m,n) = 1$ then $[\mathbb{Q}(\alpha,\beta) : \mathbb{Q}] = mn$
Since in our case $[\mathbb{Q}(5^{\frac{1}{5}}) : \mathbb{Q}] = 5$ and $[\mathbb{Q}(\zeta_{5}) : \mathbb{Q}] = 4$ we can conclude that $[K: \mathbb{Q}] = 20$.
Now we know that the Galois group is a group of order $20$, which embeds into $S_{5}$ (Since the polynomial $t^{5}-5$ is irreducible over $\mathbb{Q}$ and the Galois group permutes the roots transitively, which give us the action on the set of the roots needed).
In order to continue and represent our group fairly explicitly all we have to do is to notice from the Fundamental theorem of Galois that since the Galois group of $\mathbb{Q}(\zeta_{5})$ over $\mathbb{Q}$ is the splitting field of a separable polynomial, $t^5-1 \in \mathbb{Q}[t]$, we have that the Galois group  of $\mathbb{Q}(\zeta_{5})$ over $\mathbb{Q}$ is a normal subgroup of Gal($K/ \mathbb{Q})$. 
If we look at the index of this subgroup, it follows that it has index $4$, which means order 5, due to this knowledge it must be a $5$-Sylow $P$ of $G= $ Gal$(K/\mathbb{Q})$. 
(And it will be the only $5$-Sylow due to normality)
Since we embedded $G$ into $S_{5}$ it follows that $P$ is a $5$-Sylow of $S_{5}$.
Since we found a group, i.e. $G$, where a $5$-Sylow is normal, it follows by defintion of $N_{S_{5}}$ := Normalizer that $G \subset N_{S_{5}}(P)$, but looking at the cardinality, we can afirm that they are equal.
We've discover that coincide with of the normalizer of a $5$-sylow in $S_{5}$. 
You can easily show with this that such a group is isomorphic to 
$\mathbb{Z}_{5}\rtimes_{\phi} \mathbb{Z}_{4}$, where $\phi$ is a non trivial homomorphism which will show that the group is certainly not abelian.

Edit : Note that $i$ can't lie in $\mathbb{Q}(\zeta_{5})$ because $i = \zeta_{4}$ and we know from cyclotomic extension theory that $\mathbb{Q}(\zeta_{n}) \cap \mathbb{Q}(\zeta_{m}) = \mathbb{Q}(\zeta_{gcd[m,n]})$ 

Hope you could take it from here.
