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Let $f(x) = x^{15}-15\in \mathbb{Q}[x]$. By Eisenstein's Criterion (using 3 or 5), $f$ is irreducible. Then $L=\mathbb{Q}(\sqrt[15]{15}, \omega)$ is the splitting field of $f$, where $\omega$ is a primitive $15$th root of unity. We then have that the order of the Galois group $G=Gal(L/\mathbb{Q})$ is just $15\cdot \phi(15) =120$ - the product of the extensions.

My question is, knowing this, can we determine which Sylow subgroups are normal along with their structure? Of course the 3 and 5 Sylow subgroups will be isomorphic to $\mathbb{Z}_3$ and $\mathbb{Z}_5$ respectively, but I'm not sure if there's enough info to determine the 2-Sylow subgroup nor the number of each of them to determine normality. Should I focus on looking at subfields of $L$?

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    $\begingroup$ I think you mean $$L=\Bbb Q(\sqrt[15]{15},\omega).$$ $\endgroup$ – awllower Jul 13 '16 at 2:43
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    $\begingroup$ Just a note: it is not always true that the order would be the product of the extension degrees. $\endgroup$ – Steve D Jul 13 '16 at 4:03
  • $\begingroup$ @awllower ah yes of course, I was typing in a hurry. Sorry about that $\endgroup$ – Curious Jul 13 '16 at 5:39
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First let us see that indeed $[L:\Bbb Q]=120$. To ease notation set $a=15$.

Consider the subfields $K_1=\Bbb Q(\sqrt[3]a,\omega_3)$ and $K_2=\Bbb Q(\sqrt[5]a,\omega_5).$ As $3\nmid[\Bbb Q(\omega_3):\Bbb Q]$, no root of $x^3-a$ lies at $\Bbb Q(\omega_3)$, thus $x^3-a$ is irreducible over $\Bbb Q(\omega_3)$; because of this, hence $[K_1:\Bbb Q]=6.$ Similarly $[K_2:\Bbb Q]=20$ and $x^3-a$ is irreducible over $\Bbb Q(\omega_{15})$

As $[\Bbb Q(\sqrt[3]a,\omega_{15}):\Bbb Q]=24,$ $K_1(\omega_5)=\Bbb Q(\sqrt[3]a,\omega_{15})$ and $[K_1:\Bbb Q]=6$, we obtain $[K_1(\omega_5):K_1]=4.$ Similarly $[K_2(\omega_3):K_2]=2$.

Since $5\nmid[\Bbb Q(\sqrt[3]a,\omega_{15}):\Bbb Q],$ $x^5-a$ is irreducible over $\Bbb Q(\sqrt[3]a,\omega_{15}),$ therefore as $L=\Bbb Q(\sqrt[3]a,\omega_{15})(\sqrt[5]a),$ it follows that $[L:K_1]=20$ and $[\Bbb Q(\sqrt[15]a,\omega_{15}):\Bbb Q]=120$. Similarly $[L:K_2]=6$.

Furthermore, as $[K_1:\Bbb Q]=6=[L:K_2],$ we obtain $K_1$ and $K_2$ are linearly disjoint over $\Bbb Q$. However $K_1\cdot K_2=L$, therefore \begin{equation} G_{\Bbb Q}^L\simeq G_{K_1}^L\times G_{K_2}^L, (1) \end{equation}

since $K_1$ and $K_2$ are normal over $\Bbb Q$.

As $L=K_1(\sqrt[5]{a},\omega_5)$ and $L=K_2(\sqrt[3]{a},\omega_3)$ we may consider $G_{K_1}^L$ and $G_{K_2}^L$ as subgroups of $S_5$ and $S_3$ respectively.

We have $|G_{K_2}^L|=[L:K_2]=6$, in consequence $G_{K_2}^L\simeq S_3$.

It is easy to see $G_{K_1}^L$ has an element $\sigma$ of order $4$; recall that $L=K_1(\sqrt[5]{a},\omega_5)$. As $|G_{K_1}^L|=[L:K_1]=20$, by Sylow's third theorem we must have that $\langle \sigma\rangle$ is normal in $G_{K_1}^L$; otherwise all elements in this group would have order $2$ or $4$. The subgroup of $G_{K_1}^L$ of size $5$ is normal in $G_{K_1}^L$ by the same theorem.

Thus we obtain the Sylow subgroups of $G_{\Bbb Q}^L$ using $(1)$:

  • Let $H_3$ be the Sylow $3$-subgroup of $S_3$, then $\{e\}\times H_3$ is the only Sylow $3$-subgroup of $G_{\Bbb Q}^L$.
  • Let $H_5$ denote the Sylow $5$-subgroup of $G_{K_1}^L$, then $H_5\times\{e\}$ is the only Sylow $5$-subgroup of $G_{\Bbb Q}^L$.
  • If $H_2^{1},H_2^{2}$ and $H_2^{3}$ are the Sylow $2$-subgroups of $S_3$ and $H_2$ is the Sylow $2$-subgroup of $G_{K_1}^L$, then $H_2\times H_2^{i}$ for $i=1,2,3$ are the Sylow $2$ -subgroups of $G_{\Bbb Q}^L$.
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If you think of your full extension as $\Bbb Q\subset\Bbb Q(\omega)=K\subset L$, then $K$ is normal over $\Bbb Q$, so $G^L_K$ is a normal subgroup of the whole group $G^L_{\Bbb Q}$. We do know that $G^L_K\cong\Bbb Z/(15)$. I think this gives you enough information to finish the question off yourself.

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