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The element $\zeta$ satisfies $T^2-\eta T+1=0$, so $[\mathbb{Q}(\zeta ):\mathbb{Q}(\eta )]\le 2$. Since $\eta =\zeta +\overline{\zeta }$ is real, $\mathbb{Q}(\eta )\subseteq \mathbb{R}$. A primitive root of order $m\ge 3$ is nonreal: the only real roots of unity are $\pm 1$, of orders at most two. Hence $\zeta \notin \mathbb{Q}(\eta )$ andEquivalently, nontrivial complex conjugation fixes precisely the maximal real subfield of a cyclotomic field in this degree-two extension.

The Cyclotomic character of a finite cyclotomic extension and the Chinese remainder theorem give the displayed Galois group. It has three order-two subgroups and three order-four subgroups; the latter are kernels of its three nonzero group homomorphisms to $C_2$. By the Fundamental theorem of Galois theory, its proper intermediate fields strictly above $\mathbb{Q}$ are the three quadratic fieldsand the three quartic fieldsThe first quartic has cyclic Galois group of order four and unique quadratic subfield $\mathbb{Q}(\sqrt{5})$; the second is a biquadratic extension, so they differ. The last is real whereas the others are not, by the maximal real subfield of a cyclotomic field formula. Also $\sqrt{-3}+\sqrt{5}$ generates the second quartic: if $s=\sqrt{-3}+\sqrt{5}$, then $\sqrt{-3}=(s-8/s)/2$ and $\sqrt{5}=(s+8/s)/2$.