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Let me briefly sketch what I did (hope there is no mistake):

Let $L$ be the galois extension of $x^5+2$ over $\mathbb{Q}$.
Let $a$ be symbol such that $a^5 = -2$ and $\xi$ be $5^{th}$ primitive root of unity.

Now, $\{a, \xi a, \xi^2 a, \xi^3 a, \xi^4 a \}$ are roots of $x^5+2$ and $L = \mathbb{Q}(a,\xi)$.
Since $x^5+2$ is irreducible, $[\mathbb{Q}(a),\mathbb{Q}]=5$ and $[\mathbb{Q}(\xi):\mathbb{Q}]=\varphi(5) = 4$.
It follows by $(4,5)=1$ that $[L:\mathbb{Q}]=20=2^25$.

Let $\sigma: a \mapsto \xi a$ and $\tau: \xi \mapsto \xi^2$. $N=<\sigma>$ and $K=<\tau>$.
By Sylow's Theorem, $N$ is normal. $N\cap K = e$ hence galois group is given by semidirect product $N \rtimes K$.

Question How can I determine all normal subgroups?

$N$(group of order $5$) is normal.

Since $\tau\sigma\tau^{-1} = \sigma^2$, group is not abelian. Therefore, $K$[$\cong C_4$] (group of order $2^2$) cannot be unique.

How can I argue about subgroups of order $2$?

If subgroup of order $10$ exists, it has index $2$ hence necessarily normal. Is it trivial that it exists?

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  • $\begingroup$ The subgroup generated by $\sigma, \tau^2$ has order 10. $\endgroup$ – arkeet Jun 18 '18 at 18:04
  • $\begingroup$ @arkeet thank you! $\endgroup$ – Jo' Jun 18 '18 at 18:08
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If $H$ is a normal subgroup of $G = \operatorname{Gal}(L/\mathbb{Q})$, then since $|N|$ is prime, $H \cap N = N$ or $H \cap N = \{e\}$.

If $H \cap N = \{e\}$, let $g = \sigma^m \tau^n$ be an arbitrary element of $H$. Since $H$ is normal, it contains $(\sigma^{-1} g \sigma) g^{-1} = \sigma^{2^n-1} \in N$ (using the relation $\tau^n \sigma = \sigma^{2^n} \tau^n$). Since $H \cap N = \{e\}$ this implies $\sigma^{2^n - 1} = e$, and therefore $4 \mid n$. So $g = \sigma^m \in H \cap N$ and $H$ is trivial.

So a normal subgroup of $G$ is either trivial or contains $N$. The normal subgroups containing $N$ correspond 1-1 with normal subgroups of $G/N \cong K$, which are easy to describe since $K$ is cyclic.

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  • $\begingroup$ Many thanks @arkeet $\endgroup$ – Jo' Jun 18 '18 at 20:29
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    $\begingroup$ Note that the same argument works for any irreducible polynomial $x^p-c$ over $\mathbb{Q}$ where $p$ is prime - take $a$ to be a $p$th root of $c$, $\xi$ a primitive $p$th root of unity, $\sigma \colon (a,\xi) \mapsto (\xi a,\xi)$, $\tau \colon (a,\xi) \mapsto (a,\xi^t)$ where $t$ is a generator of $(\mathbb{Z}/p\mathbb{Z})^\times$. $\endgroup$ – arkeet Jun 18 '18 at 20:30
  • $\begingroup$ Really appreciate the argument. Is it possible that the relation is $\tau^n\sigma = \sigma^{{2}^n}\tau^n$? @arkeet $\endgroup$ – Jo' Jun 19 '18 at 12:45
  • $\begingroup$ Yep that was a typo, thanks! $\endgroup$ – arkeet Jun 19 '18 at 17:34

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