Let \(K\) be the splitting field over \(\mathbb{Q}\) of the polynomial \(f(x) = (x^3-2)(x^2-3)\).
Show that \([K:\mathbb{Q}] = 12\).
Take as given that every non-Abelian group of order twelve is isomorphic to either \(A_4\), the dihedral group \(D_{12}\), or else the semidirect product \(\mathbb{Z}/3\mathbb{Z} \rtimes_\phi \mathbb{Z}/4\mathbb{Z}\) where \(\phi\) sends the generator of \(\mathbb{Z}/4\mathbb{Z}\) to the non-identity automorphism of \(\mathbb{Z}/3\mathbb{Z}\). For each of these three groups, calculate what a Sylow-\(2\) subgroup is up to isomorphism, and whether or not it is normal in the group.
Which group of order twelve must the Galois group \(\mathrm{Gal}(K/\mathbb{Q})\) be?
Solution
Let \(\alpha=\sqrt[3]{2}\) and let \(\omega\) be a primitive cube root of unity. The splitting field of \(x^3-2\) is \[ E=\mathbb Q(\alpha,\omega). \]
1. We have \([K:\mathbb Q]=12\).
The polynomial \(x^3-2\) is Eisenstein at \(2\), hence irreducible over \(\mathbb Q\). Its discriminant is \[ -27\cdot 2^2=-108, \] which is not a square in \(\mathbb Q\). Therefore its splitting field \(E\) has degree \(6\) and \[ \operatorname{Gal}(E/\mathbb Q)\cong S_3. \] The unique subgroup of index \(2\) in \(S_3\) is \(A_3\), so \(E\) has a unique quadratic subfield. It is \[ \mathbb Q(\omega)=\mathbb Q(\sqrt{-3}). \] In particular \(\sqrt3\notin E\). Hence \[ [E(\sqrt3):E]=2. \] Since \(K\) is the compositum of \(E\) with the splitting field \(\mathbb Q(\sqrt3)\) of \(x^2-3\), we obtain \[ [K:\mathbb Q]=[K:E][E:\mathbb Q]=2\cdot6=12. \]
2. The Sylow-\(2\) subgroups of the three listed nonabelian groups of order \(12\) are as follows.
For \(A_4\), the Klein four subgroup \[ V_4=\{e,(12)(34),(13)(24),(14)(23)\} \] has order \(4\), so it is a Sylow-\(2\) subgroup; it is normal.
For the dihedral group \(D_{12}=\langle r,s\mid r^6=s^2=1,\ srs=r^{-1}\rangle\), a Sylow-\(2\) subgroup is \[ \langle r^3,s\rangle\cong V_4. \] There are three such subgroups, so it is not normal.
For \(C_3\rtimes C_4\) with the generator of \(C_4\) acting on \(C_3\) by inversion, the displayed \(C_4\) is a Sylow-\(2\) subgroup. It is not normal: if \(a\) generates \(C_3\) and \(b\) generates \(C_4\), with \(bab^{-1}=a^{-1}\), then conjugating \(b\) by \(a\) does not remain in \(\langle b\rangle\).
3. We have\[ \operatorname{Gal}(K/\mathbb Q)\cong D_{12}. \]
Both \(E/\mathbb Q\) and \(\mathbb Q(\sqrt3)/\mathbb Q\) are Galois, and their intersection is \(\mathbb Q\) because \(\sqrt3\notin E\). Hence restriction gives \[ \operatorname{Gal}(K/\mathbb Q) \cong \operatorname{Gal}(E/\mathbb Q)\times \operatorname{Gal}(\mathbb Q(\sqrt3)/\mathbb Q) \cong S_3\times C_2. \] This group has a Sylow-\(2\) subgroup isomorphic to \(C_2\times C_2\), and it is not normal. By 2 it is therefore neither \(A_4\) nor \(C_3\rtimes C_4\), so among the three listed nonabelian groups it is \(D_{12}\). Equivalently, \(S_3\times C_2\cong D_{12}\).