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Let me start with what I have so far about the Galois group of $x^{15} - 2$ over $\mathbb{Q}$. After that, I'll list the properties of this group that I'm trying to show.

Let $\alpha = \sqrt[15]{2}$ and $\omega = e^{i\frac{2\pi}{15}}.$ Then $L = \mathbb{Q}(\alpha, \omega)$ is the splitting field of $x^{15} - 2$ over $\mathbb{Q}.$

$x^{15} - 2$ is irreducible over $\mathbb{Q}$ by Eisenstein's Criterion with prime $2$, so $\Gamma(L:\mathbb{Q(\omega)}) \cong \mathbb{Z}_{15}.$

Also, $\Gamma(\mathbb{Q}(\omega):\mathbb{Q}) \cong U_{15} \cong \mathbb{Z}_{2} \times\ \mathbb{Z}_{4}$ since $U_{15}$ is abelian, non-cyclic, order $8$, and has an element with $o(\bar{2}) = 4.$

Thus $|\Gamma(L:\mathbb{Q})| = 15 \cdot 8 = 120$.

Next note that $\Gamma(\mathbb{Q}(\omega):\mathbb{Q}) \cong \Gamma(L:\mathbb{Q})/\Gamma(L:\mathbb{Q(\omega)}).$ Specifically, $\phi: \Gamma(L:\mathbb{Q}) \to \Gamma(\mathbb{Q(\omega):\mathbb{Q}})$ defined by $\sigma \mapsto \sigma|_{\mathbb{Q}(\omega)}$ is onto with $\Gamma(\mathbb{Q(\omega):\mathbb{Q}}) \cong \mathbb{Z}_{2} \times\ \mathbb{Z}_{4}$ and $\ker\phi = \Gamma(L:\mathbb{Q(\omega)}) \cong \mathbb{Z}_{15}.$

Now here are the properties of $\Gamma(L:\mathbb{Q})$ that I'm trying to show.

Show there are elements $\rho, \sigma, \tau \in \Gamma(L:\mathbb{Q})$ such that

(1) $\rho, \sigma, \tau$ generate $\Gamma(L:\mathbb{Q}),$

(2) $\rho^{15} = \sigma^{4} = \tau^{2} = 1,$

(3) $\sigma^{-1}\rho\sigma = \rho^{7},$

(4) $\tau^{-1}\rho\tau = \rho^{14},$

(5) $\tau^{-1}\sigma\tau = \sigma$

Looks like $\rho$ would have to be a generator of $\ker\phi \cong \mathbb{Z}_{15}.$ Then from $\bar{\sigma} \in \Gamma(\mathbb{Q(\omega):\mathbb{Q}})$ with $o(\bar{\sigma}) = 4$ and $\bar{\tau} \in \Gamma(\mathbb{Q(\omega):\mathbb{Q}})$ with $o(\bar{\tau}) = 2$, I think $\sigma$ and $\tau,$ respectively, can be obtained.

Do the properties follow with these $\rho, \sigma, \tau$? I am especially wondering about (3) and (4).

Oscar
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    For the second line of your exposition, you need to know that $x^{15}-2$ is irreducible over ${\bf Q}(\omega)$. – Gerry Myerson Aug 07 '20 at 01:40
  • @GerryMyerson Yes, thanks, and let's see, $x^{15} - 2$ is irreducible over $\mathbb{Q}(\omega)$ by showing $[L:\mathbb{Q}(\omega)] = 15$ which requires a more complicated argument like in the answer to this post and actually that post may provide hints to my question. – Oscar Aug 07 '20 at 02:27
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    Or actually $x^{15} - 2$ is irreducible over $\mathbb{Q}(\omega)$ by showing $[L:\mathbb{Q}(\omega)] = 15$ because $[\mathbb{Q}(\alpha):\mathbb{Q}] = 15$ and $\gcd(15, 8)=1.$ – Oscar Aug 07 '20 at 02:35
  • Alternatively to your comment, if you wanted to stick with "Eisenstein-like" ideas you could use the Newton polygon for any prime lying above $2$ in $\mathbb{Q}(\omega)$ to show that $x^{15} - 2$ is irreducible over $\mathbb{Q}(\omega)$. – Mummy the turkey Aug 07 '20 at 03:15
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    Hint: To write down the $\rho, \sigma, \tau$ you just have to say what they do to $\alpha$ and $\omega$. You've probably seen why the galois group of $x^4 - 2$ is $D_4$, can you imitate that (in particular you have pointed out the correct $\rho, \sigma, \tau$ can you say what they are explicitly)? – Mummy the turkey Aug 07 '20 at 03:20
  • @Mummytheturkey Thanks, and let's see, would it be something like $\omega \mapsto \omega, \alpha \mapsto \alpha\omega$ for $\rho$, $\alpha \mapsto \alpha, \omega \mapsto \omega^2$ for $\sigma$, $\alpha \mapsto \alpha, \omega \mapsto \omega^{14}$ for $\tau$? – Oscar Aug 07 '20 at 04:19
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    Yes, exactly - then you can verify the relations simply by checking what's going on with the $\alpha$ and $\omega$ again :) – Mummy the turkey Aug 07 '20 at 04:45
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    Also in case anyone was wondering, this group is $S_3 \times GA(1, 5)$ – Mummy the turkey Aug 07 '20 at 04:52
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    A more standard description of this group that better lines up with the identification of the Galois groups associated to $n$th roots of unity with [a subgroup of] $(\mathbb Z/n\mathbb Z)^*$ is to associate the elements with 2x2 lower triangular matrices whose entries satisfying: 1 in the top left, corner entry determined by $\sigma(\alpha) = \zeta^i\alpha$, and other diagonal by $\sigma(\zeta) = \zeta^j$. In many cases this identification is an isomorphism, such as in the above. –  Aug 07 '20 at 16:55

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