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?

  • $\begingroup$ 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

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.

  • $\begingroup$ Many thanks @arkeet $\endgroup$ – Jo' Jun 18 '18 at 20:29
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.