Do you know what the quaternion group is? How many elements are there of each order? Suppose I have a field extension of the rationals with Galois group the quaternion group. How many quadratic extensions does it contain? Can any of them be imaginary?
Solution
Write \[ Q_8=\{\pm1,\pm i,\pm j,\pm k\}, \qquad i^2=j^2=k^2=ijk=-1. \]
1. The element orders in \(Q_8\) are:\[ 1\text{ element of order }1, \quad 1\text{ element of order }2, \quad 6\text{ elements of order }4. \]
The identity has order \(1\). The element \(-1\) has order \(2\). Each of \(\pm i,\pm j,\pm k\) squares to \(-1\) and therefore has order \(4\).
2. The group \(Q_8\) has exactly three subgroups of order \(4\):\[ \langle i\rangle, \qquad \langle j\rangle, \qquad \langle k\rangle. \]
Every subgroup of order \(4\) contains an element of order \(4\), hence is the cyclic subgroup generated by one of \(i,j,k\) up to sign. These are distinct.
3. If \(L/\QQ\) is Galois with Galois group \(Q_8\), then \(L\) contains exactly three quadratic subfields.
By the Galois correspondence, quadratic intermediate fields correspond to subgroups of index \(2\), equivalently subgroups of order \(4\). Apply 2.
4. All three quadratic subfields are real.
Fix an embedding \(L\subseteq\CC\). If \(L\) is totally real, there is nothing to prove. Otherwise complex conjugation restricts to a nontrivial element \[ c\in\operatorname{Gal}(L/\QQ) \] of order \(2\). By 1, \(Q_8\) has a unique element of order \(2\), namely \(-1\), so \(c=-1\) in the group-theoretic notation.
Let \(E\) be any quadratic subfield, and put \[ H=\operatorname{Gal}(L/E). \] Then \(|H|=4\). Every subgroup of order \(4\) in \(Q_8\) contains the unique involution \(-1\), so \(c\in H\). Thus complex conjugation fixes \(E\) pointwise. Therefore every element of \(E\) is real, and \(E\) is a real quadratic field.