= Solution
Let $G$ be a finite subgroup of $K^\times$. It is abelian since multiplication in the <field> commutes. The structure theorem therefore writes $G$ as a product of cyclic groups with invariant factors $d_1\mid\cdots\mid d_s$, omitting trivial factors. Every element satisfies $g^{d_s}=1$, so $G$ consists of roots of $t^{d_s}-1$. The root bound from part (ii) gives
$$
|G|\leq d_s,
$$
while the decomposition gives $|G|=\prod_jd_j\geq d_s$. Equality forces every factor except the last to be trivial. Hence \b[$G$ is cyclic]. The trivial subgroup is cyclic as well. This is the <finite multiplicative subgroup of a field is cyclic> theorem, proved here using the preceding two parts.
The infinite case is false. In $\mathbb C^\times$, the group of all <roots of unity> is infinite, but every element has finite order. If it were cyclic, its generator would have finite order and the whole group would be finite, a contradiction.
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