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For $K=\mathbb{Q}(\zeta )$, every Galois automorphism sends the primitive root $\zeta$ to another primitive $m$th root of unity, uniquely ${\zeta }^a$ for $a\in (\mathbb{Z}/m\mathbb{Z}{)}^{\times }$. The Cyclotomic character of a finite cyclotomic extension isIf ${\chi }_m(\sigma )=a$ and ${\chi }_m(\tau )=b$, then $(\sigma \tau )(\zeta )=\sigma ({\zeta }^b)={\zeta }^{ab}$, proving the group homomorphism property. Its kernel is trivial: an automorphism fixing $\zeta$ fixes the field it generates. Thus the character is injective. Here its domain is the finite cyclotomic Galois group; the corresponding map from an absolute Galois group need not be injective.

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 $\sqrt{5}=(s+8/s)/2$.