Galois group of the polynomial over $\mathbb{Q}$

Question

Let $G$ be the Galois group of the splitting field of $x^5-2$ over $\mathbb{Q}$. Then

1) $G$ is cyclic

2) $G$ is non-Abelian

3) $\vert G \vert =20$

4) $G$ has an element of order $4$

Here, The splitting field of $x^5-2$ over $\mathbb{Q}$ is $\mathbb{Q}\big(\rho.2^{\frac{1}{5}}\big )$ where $\rho=e^{\frac{2\pi i}{5}}$.

Hence $\vert G \vert = \vert Gal(\mathbb{Q}\big(\rho.2^{\frac{1}{5}}\big ) : \mathbb{Q} )\vert$= $[\mathbb{Q}\big(\rho.2^{\frac{1}{5}}\big ) : \mathbb{Q} )]$=$20$

So, 3) is true.

1) is clearly false, since $G$ can be realized as a subgroup of $S_5$.

How to prove/disprove 2) and 4) ?

Can you prove that in $G=\text{Gal}(\mathbb{K}:\mathbb{Q})$ there is an element with order $5$? What would happen if $G$ were abelian / if $G$ had an element of order-$4$, too? — Jack D'Aurizio, Jul 16 '17 at 03:51
Also notice that $1)$ is false, but there are plenty of cyclic subgroups of $S_5$. — Jack D'Aurizio, Jul 16 '17 at 03:52
There is no element of order 20 in $S_5$ and hence there is no cyclic subgroup with order 20 — Chinnapparaj R, Jul 16 '17 at 03:54
Yes! there exist an element in $G$ with order $5$ by cauchy theorem — Chinnapparaj R, Jul 16 '17 at 03:56

Dionel Jaime · Accepted Answer · 2017-07-16T04:31:16.393

3

Hint for $1$: The extension $\mathbb{Q}(2^{1/5} )/\mathbb{Q}$ is not Galois.

Hint for $3$: Try and find the generators for $Gal(\mathbb{Q}(2^{1/5},\rho) / \mathbb{Q})$ by writing out the automorphisms. Look specifically at

$\sigma = \begin{cases} 2^{1/5} \to 2^{1/5} \\ \rho \to \rho^2 \end{cases}$

edited Jul 16 '17 at 04:31

answered Jul 16 '17 at 04:07

Dionel Jaime

3,908

Your hint for 1 actually also settles 2. +1 for IMO well thought out hints. – Jyrki Lahtonen Jul 16 '17 at 07:37

Galois group of the polynomial over $\mathbb{Q}$

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