Let \(F \subset L\) be fields such that \(L/F\) is a Galois field extension with Galois group equal to \(D_8 = \left< \sigma,\tau \mid \sigma^4 = \tau^2 = 1,~ \sigma\tau = \tau \sigma^3 \right>\). Show that there are fields \(F \subset E \subset K \subset L\) such that \(E/F\) and \(K/E\) are Galois field extensions, but \(K/F\) is not Galois.
Let \(L/F\) be Galois with \[ \operatorname{Gal}(L/F)=D_8=\langle\sigma,\tau\mid \sigma^4=\tau^2=1,\ \tau\sigma\tau=\sigma^{-1}\rangle. \] Show that there are fields \[ F\subset E\subset K\subset L \] such that \(E/F\) and \(K/E\) are Galois, but \(K/F\) is not Galois.
Solution
Let \[ H=\langle\sigma^2,\tau\rangle\cong C_2\times C_2, \qquad J=\langle\tau\rangle. \] The subgroup \(H\) has index \(2\) in \(D_8\), so \(H\trianglelefteq D_8\). Since \(H\) is abelian, \(J\trianglelefteq H\).
However, \(J\) is not normal in \(D_8\): for example, \[ \sigma\tau\sigma^{-1}=\sigma^2\tau\notin\langle\tau\rangle. \]
Define the fixed fields \[ E=L^H, \qquad K=L^J. \] Then \(F\subset E\subset K\subset L\). By the fundamental theorem of Galois theory, \[ E/F\text{ is Galois} \iff H\trianglelefteq D_8, \] so \(E/F\) is Galois. Likewise \[ K/E\text{ is Galois} \iff J\trianglelefteq H, \] so \(K/E\) is Galois. But \[ K/F\text{ is Galois} \iff J\trianglelefteq D_8, \] which is false.