<s>[INST] What are the intermediate fields of $\mathbb Q(\sqrt[3]2,e^{\frac{2i\pi}{3}})$ (Galois group) The elements of Galois group are
\begin{align*}
\sigma _1:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{2i\pi}{3}},
\end{align*}
\begin{align*}
\sigma _2:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2}e^{\frac{2i\pi}{3}},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{2i\pi}{3}},
\end{align*}
\begin{align*}
\sigma _3:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2}e^{\frac{4i\pi}{3}},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{2i\pi}{3}},
\end{align*}
\begin{align*}
\sigma _4:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{4i\pi}{3}},
\end{align*}
\begin{align*}
\sigma _5:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2}e^{\frac{2i\pi}{3}},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{4i\pi}{3}},
\end{align*}
\begin{align*}
\sigma _6:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2}e^{\frac{4i\pi}{3}},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{4i\pi}{3}}.
\end{align*}
We remark that
\begin{align*}
\sigma _2^3=1\\
\sigma _3^3=1\\
\sigma _4^2=1\\
\sigma _5^2=1\\
\sigma_6^2=1
\end{align*}
But $$\sigma _2\sigma _4(\sqrt[3]2)=\sigma _2(\sqrt[3]2)=\sqrt[3]2e^{\frac{2i\pi}{3}}$$
and
$$\sigma _4\sigma _2(\sqrt[3]2)=\sigma _4(\sqrt[3]2e^{\frac{2i\pi}{3}})=\sqrt[3]2e^{\frac{4i\pi}{3}},$$
therefore $\{\sigma _i\}_{i=1}^6$ is not a commutatif group and thus
$$\text{Gal}(E/\mathbb Q)=\{\sigma _i\}_{i=1}^6\cong \mathfrak S_3.$$
My problem is that it should have a $\sigma _i$ such that $\sigma _i^6=1$ and $\sigma _i^n\neq 1$ for $i\in\{1,2,3,4,5\}$ because one of my intermediate field must be $\mathbb Q$. And if not, is there a mistake in my applications $\sigma _i$?
 [/INST] The subfields are the fields fixed by a subgroup of the Galois group. Not every subgroup is generated by a single element, whence it is no problem that there is no element of order $6$.  
</s>