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 $L$?

  • 2
    $\begingroup$ I think you mean $$L=\Bbb Q(\sqrt[15]{15},\omega).$$ $\endgroup$ – awllower Jul 13 '16 at 2:43
  • 1
    $\begingroup$ 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

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 $G_{\Bbb Q}^L$.

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.


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.