Let \(K\) be the splitting field of \((x^3-2)(x^2-3)\) over \(\mathbb{Q}\) and \(G\) be the Galois group of \(K/\mathbb{Q}\).
For each prime divisor \(p\) of \(|G|\), decide whether \(G\) has a normal Sylow \(p\)-subgroup.
Show that there are subgroups \(G_1\) and \(G_2\) of \(G\) such that \(G = G_0 \triangleright G_1 \triangleright G_2 \triangleright G_3 = \{1\}\) with \(G_i/G_{i+1}\) cyclic of prime order. Determine the corresponding subfields.
Solution
Choose \(a=\sqrt[3]{2}>0\), \(b=\sqrt3>0\), and \(\zeta=(-1+i\sqrt3)/2\), a primitive cube root of unity. Then \(K=\mathbb Q(a,\zeta,b)\) and \(G\cong S_3\times C_2\). The Sylow \(3\)-subgroup is normal, but no Sylow \(2\)-subgroup is normal.
1. The field \(E=\mathbb Q(a,\zeta)\) has degree \(6\) and Galoisgroup \(S_3\), and its unique quadratic subfield is \(\mathbb Q(\zeta)\).
The roots of \(x^3-2\) are \(a,\zeta a,\zeta^2a\), so their splitting field is \(E\): the ratio \((\zeta a)/a\) recovers \(\zeta\). Eisenstein’s criterion at \(2\) makes \(x^3-2\) irreducible over \(\mathbb Q\), giving \([\mathbb Q(a):\mathbb Q]=3\) [@DF04]. The field \(\mathbb Q(a)\) is real and \(\zeta\) is nonreal, with \(\zeta^2+\zeta+1=0\). Hence \([E:\mathbb Q(a)]=2\) and \([E:\mathbb Q]=6\).
As a splitting field in characteristic zero, \(E/\mathbb Q\) is Galois. Its group acts faithfully on the three roots, since these generate \(E\). It is therefore a subgroup of \(S_3\) of order \(6\), and must equal \(S_3\). This group has exactly one subgroup of order \(3\): its two \(3\)-cycles are inverse and generate the same subgroup. The Galois correspondence thus gives exactly one quadratic subfield [@DF04]. The nonreal root \(\zeta\) of \(x^2+x+1\) has degree \(2\) over \(\mathbb Q\), so that subfield is \(\mathbb Q(\zeta)\).
2. The field \(K\) has degree \(12\) and \(G\cong S_3\times C_2\).
The number \(3\) is not a rational square: in a square of a nonzero rational number the exponent of the prime \(3\) is even, whereas in \(3\) it is one. Thus \(\mathbb Q(b)\) is a real quadratic field. If \(b\in E\), step 1 would force \(\mathbb Q(b)=\mathbb Q(\zeta)\), which is impossible because the latter contains a nonreal element. Therefore \(b\notin E\).
The polynomial \(x^2-3\) remains irreducible over \(E\), since neither of its roots \(\pm b\) lies in \(E\). Consequently \([K:E]=2\) and \([K:\mathbb Q]=12\). The roots of the product polynomial generate \(E\) and contain \(b\), so \(K=E(b)\) is precisely its splitting field, and is Galois over \(\mathbb Q\).
Every element of \(G\) preserves both \(E\) and \(\mathbb Q(b)\), because they are splitting fields of the two individual factors. Restriction gives an injective homomorphism \[ G\longrightarrow \operatorname{Gal}(E/\mathbb Q)\times \operatorname{Gal}(\mathbb Q(b)/\mathbb Q). \] It is injective because \(E\) and \(b\) generate \(K\). Both sides have order \(12\), so it is an isomorphism. The two factors are \(S_3\) and the order-two group generated by \(b\mapsto-b\).
3. The group admits generators with the following actions:\[ \begin{array}{c|ccc} &a&\zeta&b\\\hline r&\zeta a&\zeta&b\\ s&a&\zeta^{-1}&b\\ t&a&\zeta&-b \end{array} \]They satisfy \(r^3=s^2=t^2=1\), \(srs=r^{-1}\), and \(t\) commuteswith \(r,s\).
Since \([E:\mathbb Q(\zeta)]=3\), the minimal polynomial of \(a\) over \(\mathbb Q(\zeta)\) is \(x^3-2\). Sending \(a\) to \(\zeta a\) therefore defines an automorphism of \(E\) fixing \(\zeta\): its image contains \(\zeta\) and \(\zeta a\), hence also \(a\). Complex conjugation on \(E\) fixes \(a\) and sends \(\zeta\) to \(\zeta^{-1}\). These two automorphisms have orders \(3\) and \(2\), and their six products \(r^is^j\) are distinct on \(a,\zeta\). They generate the \(S_3\) factor. The product isomorphism in step 2 lifts them while fixing \(b\), and lifts the nontrivial automorphism of \(\mathbb Q(b)\) while fixing \(E\). These lifts are exactly \(r,s,t\). The relations follow by evaluating on the three field generators, and all elements of \(G\) have the distinct forms \(r^is^jt^k\) with \(0\leq i<3\) and \(j,k\in\{0,1\}\).
4. For part (a), the Sylow \(3\)-subgroup is normal and theSylow \(2\)-subgroups are not normal.
The group \(\langle r\rangle\) has order \(3\), the full power of \(3\) dividing \(12\). It is normal because conjugation by \(s\) inverts \(r\) and conjugation by \(t\) fixes \(r\).
3. Thus \(P\) is not normal. Every Sylow\(2\)-subgroup is conjugate to \(P\) by Sylow’s theorem [@DF04],so none is normal. The only prime divisors of \(|G|\) are \(2,3\),and both have now been decided.
5. For part (b), take \(G_1=\langle r,t\rangle\) and\(G_2=\langle r\rangle\). The subgroup and fixed-field data are\[ \begin{array}{c|c|c|c} i&G_i&|G_i|&K^{G_i}\\\hline 0&G&12&\mathbb Q\\ 1&\langle r,t\rangle&6&\mathbb Q(\zeta)\\ 2&\langle r\rangle&3&\mathbb Q(\zeta,b)\\ 3&\{1\}&1&K \end{array} \]
The commuting generators \(r,t\) give \(G_1\cong C_3\times C_2\), of order \(6\). It has index \(2\) in \(G\), so is normal. Its subgroup \(G_2\) is normal since \(G_1\) is abelian. The three successive quotients have orders \(2,2,3\), respectively, and each prime-order group is cyclic. This proves every requirement on the series.
Both \(r\) and \(t\) fix \(\zeta\), so \(\mathbb Q(\zeta)\subseteq K^{G_1}\). The two fields have the same degree \(2\) over \(\mathbb Q\), since the Galois correspondence gives \([K^{G_1}:\mathbb Q]=12/6=2\) [@DF04]. They are equal.
The generator \(r\) fixes both \(\zeta\) and \(b\). The field \(\mathbb Q(\zeta,b)\) has degree \(4\) over \(\mathbb Q\): \(\mathbb Q(b)\) is real quadratic, and adjoining the nonreal root \(\zeta\) of \(x^2+x+1\) has degree \(2\). Since \([K^{G_2}:\mathbb Q]=12/3=4\), it equals \(\mathbb Q(\zeta,b)\). The endpoint fields are \(K^G=\mathbb Q\) and \(K^{\{1\}}=K\).
Thus the corresponding increasing field tower is \[ \mathbb Q\subset\mathbb Q(\sqrt{-3}) \subset\mathbb Q(\sqrt{-3},\sqrt3)\subset K, \] with degrees \(2,2,3\) at successive steps. Here \(\sqrt{-3}=2\zeta+1=ib\), so the middle biquadratic field may also be written \(\mathbb Q(i,\sqrt3)\).