On a question of capitulation (Q361244)

Let \(q \equiv 3 \bmod 4\) be a prime number, and set \(k = {\mathbb Q}(\sqrt{-q}\,)\). The compositum \(L\) of \(k\) and the maximal real subfield \(F = {\mathbb Q}\big(\sqrt{2 + \sqrt{2}}\,\big)\) of the field of \(16\)-th roots of unity has the cyclic quartic subfield \(K = {\mathbb Q}\big(\sqrt{-q(2 + \sqrt{2}\,)}\,\big)\) over which \(L\) is unramified. The authors show that if \(q \equiv 7 \bmod 16\), then the maximal unramified \(2\)-extension of \(K\) has degree \(8\) and the quaternion group as a Galois group.