Suppose the field \(K\) is a Galois extension of the field \(F\), such that the Galois group of \(K\) over \(F\) is isomorphic to the 8 element dihedral group \(D_8\).
Determine the number of fields \(L\) such that \(F \subset L \subset K\), \(L \neq F, L \neq K\), and indicate the dimension of \(L\) over \(F\) in each case.
Refine your answer to (a) by indicating how many fields in each dimension are Galois over \(F\).
Solution
There are eight strict intermediate fields: three of degree two and five of degree four over \(F\). All three quadratic fields and exactly one of the five quartic fields are Galois over \(F\).
1. We may write\[ G=\operatorname{Gal}(K/F) =\{1,r,r^2,r^3,s,rs,r^2s,r^3s\}, \qquad r^4=s^2=1,\quad srs=r^{-1}. \]The subgroup-to-field correspondence has\([K^H:F]=8/|H|\), and \(K^H/F\) is Galois exactly when\(H\lhd G\).
Choose an isomorphism with the symmetries of a square, taking \(r\) to a quarter-turn and \(s\) to a reflection. The displayed elements and relations describe these eight symmetries. Since \(K/F\) is Galois, its fixed field \(K^G\) is \(F\). The fixed-field theorem for a finite group of field automorphisms gives \([K:F]=|G|=8\), so the finite Galois correspondence applies with the asserted degree and normality properties [@DF04].
2. The nontrivial proper subgroups of \(G\) are exactly\[ \langle r^2\rangle,\quad \langle r^j s\rangle\ (j=0,1,2,3),\quad \langle r\rangle,\quad \langle r^2,s\rangle,\quad \langle r^2,rs\rangle. \]The first five have order two and the last three order four.
Let \(R=\langle r\rangle\), a cyclic subgroup of order four. Its only subgroups are \(\{1\}\), \(\langle r^2\rangle\), and \(R\): containing \(r\) or \(r^3\) forces all of \(R\), and any other subgroup is contained in \(\{1,r^2\}\).
For any \(H\leq G\), put \(T=H\cap R\). If \(H\subseteq R\), this already lists all possibilities. Otherwise choose a reflection \(t=r^j s\in H\). Every reflection has order two, by \(sr^j=r^{-j}s\). The product of two reflections is a rotation. Hence each element of \(H\) outside \(R\) lies in the coset \(tT\), and conversely \(tT\subseteq H\). Therefore \[ H=T\mathbin{\sqcup}tT. \]
If \(T=\{1\}\), this gives \(H=\{1,r^j s\}\), one of four distinct order-two subgroups. If \(T=\langle r^2\rangle\), it gives \(H=\{1,r^2,r^j s,r^{j+2}s\}\). This set is a subgroup: \(r^2\) commutes with the order-two element \(r^j s\), so their four products form a group. The set depends precisely on the parity of \(j\), giving the two displayed subgroups of order four. Finally, \(T=R\) gives \(H=G\). Together with the subgroups contained in \(R\), these cases exhaust every \(H\leq G\). Removing \(\{1\}\) and \(G\) leaves exactly the eight subgroups stated above, without repetitions.
3. The strict intermediate fields and Galois counts are
| Degree over \(F\) | Fixed fields | Number | Number Galois over \(F\) |
|---|---|---|---|
| \(2\) | \(K^{\langle r\rangle}\), \(K^{\langle r^2,s\rangle}\), \(K^{\langle r^2,rs\rangle}\) | \(3\) | \(3\) |
| \(4\) | \(K^{\langle r^2\rangle}\) and \(K^{\langle r^j s\rangle}\) for \(j=0,1,2,3\) | \(5\) | \(1\) |
The subgroup orders in step 2 and the degree formula in step 1 give exactly the stated fields and degrees. Different subgroups have different fixed fields, so none of these entries is duplicated. The subgroup \(G\) corresponds to \(F\) and the subgroup \(\{1\}\) to \(K\).
All three subgroups of order four have index two and are normal: both the left and right cosets of any element outside an index-two subgroup are its complement. The subgroup \(\langle r^2\rangle\) is central, since \(r\) commutes with \(r^2\) and \(sr^2s=r^{-2}=r^2\). It too is normal. No reflection subgroup is normal, because \[ r(r^j s)r^{-1}=r^{j+2}s\ne r^j s. \] Conjugation takes its unique nonidentity element outside that subgroup. The normality criterion in step 1 therefore gives precisely the last column of the table.