= Solution
If $\theta=0$, the <splitting field> is $K$ and its <Galois group> is trivial, hence cyclic. Otherwise choose $\alpha$ with $\alpha^n=\theta$. The roots are $\alpha\zeta_n^j$, so the <splitting field> is $K(\alpha)$ because $\zeta_n\in K$. The roots are distinct since the characteristic does not divide $n$, making this a <Galois extension>. Each automorphism satisfies $\sigma(\alpha)=c_\sigma\alpha$ with $c_\sigma\in\langle\zeta_n\rangle$. Because that root-of-unity group is fixed by all automorphisms, $c_{\sigma\tau}=c_\sigma c_\tau$. Thus $\sigma\mapsto c_\sigma$ is an injective homomorphism into a cyclic group. \b[The splitting-field <Galois group> is cyclic, of order dividing $n$.]
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