This question concerns an extension \(K\) of \(\mathbb Q\) such that \([K:\mathbb Q]=8\). Assume that \(K/\mathbb Q\) is Galois and let \(G=\Gal(K/\mathbb Q)\). Furthermore, assume that \(G\) is non-abelian.
Prove that \(K\) has a unique subfield \(F\) such that \(F/\mathbb Q\) is Galois and \([F:\mathbb Q]=4\).
Prove that \(F\) has the form \(F=\mathbb Q(\sqrt{d_1},\sqrt{d_2})\) where \(d_1,d_2\) are non-zero integers.
Suppose that \(G\) is the quaternionic group. Prove that \(d_1\) and \(d_2\) are positive integers.
Solution
1. Since \(G\) is a nonabelian group of order \(8\), we have\[ G\cong D_8\quad\text{or}\quad G\cong Q_8. \]In either case \(G\) has a unique normal subgroup \(H\) of order \(2\), namely \(Z(G)\).
The classification of groups of order \(8\) gives the three abelian groups together with \(D_8\) and \(Q_8\). Since \(G\) is nonabelian, only the latter two occur.
For \(D_8=\langle r,s\mid r^4=s^2=1,\ srs=r^{-1}\rangle\), the unique normal subgroup of order \(2\) is \(\langle r^2\rangle=Z(D_8)\); the reflection subgroups of order \(2\) are conjugate and hence not normal. For \(Q_8\), the unique element of order \(2\) is \(-1\), so \(\{\pm1\}=Z(Q_8)\) is the unique subgroup of order \(2\).
2. There is a unique subfield \(F\subset K\) with\[ [F:\mathbb Q]=4 \]and \(F/\mathbb Q\) Galois.
By the Galois correspondence, subfields \(F\) with \([F:\mathbb Q]=4\) correspond to subgroups \(H\le G\) of order \(2\). Such an \(F/\mathbb Q\) is Galois exactly when \(H\trianglelefteq G\). By 1 there is exactly one such normal subgroup, namely \(H=Z(G)\). Thus \[ F=K^{Z(G)} \] is the unique quartic Galois subfield.
3. The Galois group of \(F/\mathbb Q\) is\[ \operatorname{Gal}(F/\mathbb Q)\cong C_2\times C_2. \]
Since \(F=K^{Z(G)}\), Galois theory gives \[ \operatorname{Gal}(F/\mathbb Q)\cong G/Z(G). \] For both \(D_8\) and \(Q_8\), the quotient by the center is the Klein four group.
4. There exist nonzero integers \(d_1,d_2\) such that\[ F=\mathbb Q(\sqrt{d_1},\sqrt{d_2}). \]
By 3, the group \(\operatorname{Gal}(F/\mathbb Q)\cong C_2^2\) has three distinct subgroups of order \(2\). Their fixed fields are three distinct quadratic extensions of \(\mathbb Q\). Choose two of them, say \[ E_1=\mathbb Q(\sqrt{d_1}), \qquad E_2=\mathbb Q(\sqrt{d_2}), \] where \(d_1,d_2\in\mathbb Z\setminus\{0\}\) may be taken squarefree. Since \(E_1\ne E_2\), their compositum has degree \[ [E_1E_2:\mathbb Q]=4. \] As \(E_1E_2\subseteq F\) and \([F:\mathbb Q]=4\), we get \[ F=E_1E_2=\mathbb Q(\sqrt{d_1},\sqrt{d_2}). \]
5. Suppose now that \(G\cong Q_8\). Then \(F\) is contained in \(\mathbb R\).
Fix the given embedding \(K\subseteq\mathbb C\). Complex conjugation restricts to an element \[ c\in\operatorname{Gal}(K/\mathbb Q)=Q_8 \] of order at most \(2\). If \(c=1\), then every element of \(K\), hence every element of \(F\), is real.
If \(c\ne1\), then \(c\) must be the unique element of order \(2\) in \(Q_8\), namely the central involution \(-1\). Hence \[ \langle c\rangle=Z(Q_8), \] and therefore \[ F=K^{Z(Q_8)}=K^{\langle c\rangle}. \] But the fixed field of complex conjugation consists precisely of the real elements of \(K\). Thus again \(F\subseteq\mathbb R\).
6. In the quaternionic case, \(d_1\) and \(d_2\) may be chosen positive.
By 5, each quadratic subfield \(E_i=\mathbb Q(\sqrt{d_i})\subseteq F\) is contained in \(\mathbb R\). A quadratic field \(\mathbb Q(\sqrt d)\) with squarefree integer \(d\) is real exactly when \(d>0\). Hence the squarefree representatives \(d_1,d_2\) in 4 can be chosen positive.