Finding subfields of degree $3$ Let $F$ be the splitting ﬁeld of the polynomial $X^3−5$ over $Q$. Let $G = Gal(F/Q)$. 
(i) Determine $G$ up to isomorphism. 
(ii) Find all subﬁelds $M$ of $F$ such that $Q⊆M$ and $[M :Q] = 3$. [Give the answers in the form $Q(u)$ or $Q(u,v)$ for appropriate $u,v∈F$.]
[You may use the fact that $[F :Q] = 6$ without proof.]

Roots are: $5^{1/3}$, $5^{1/3}e^{\frac{2\pi}3i}$, $5^{1/3}e^{\frac{4\pi}3i}$.
Let $\xi=e^{\frac{2\pi}3i}$.
So splitting field $F=Q(5^{1/3}, 5^{1/3}\xi, 5^{1/3}\xi^2)=Q(5^{1/3}, \xi)$
I think (i) is $S_3$.
Stuck on (ii): I think we can have $M_1=Q(5^{1/3})$, $M_2=Q(5^{1/3}e^{\frac{2\pi}3i})$, $M_3=Q(5^{1/3}e^{\frac{4\pi}3i})$ will all have degree $3$. But i feel there are more.
Please help.
 A: As $ [F : \mathbb{Q}] = 6 $, we know that the Galois group has order 6. On the other hand, each automorphism is uniquely determined by its values at $ \zeta_3 $ and $ \sqrt[3]{5} $, which have 2 and 3 $\mathbb{Q} $-conjugates, respectively. This means that the Galois group is generated by the automorphisms $ \sigma : \zeta_3 \to \zeta_3^2 $ and $ \tau : \sqrt[3]{5} \to \zeta_3 \sqrt[3]{5} $, and has presentation
$$ G = \langle \textrm{ord}(\sigma) = 2, \textrm{ord}(\tau) = 3, \sigma \tau \sigma = \tau^{-1} \rangle $$
from which we conclude that $ G \cong D_6 $.
The Galois correspondence links subfields of degree 3 with subgroups of $ D_6 $ with index 3, or order 2. To identify these subgroups, it suffices to identify all elements of order 2 in $ D_6 $. These elements are $ \{ \sigma, \sigma \tau, \sigma \tau^2 \} $, and the corresponding fixed fields are $ \mathbb{Q}(\sqrt[3]{5}), \mathbb{Q}(\zeta_3 \sqrt[3]{5}), \mathbb{Q}(\zeta_3^2 \sqrt[3]{5}) $.
A: If the answer to (i) is $S_3$ (it is...), then subfields of $F$ of degree $3$ over $\mathbb{Q}$ correspond to subgroups of $S_3$ of index $3$, which means order $2$. Each such subfield is the fixed field of one of those subgroups. Can you take it from here?
