 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) giveswhile the decomposition gives $|G|={\prod }_j{d}_j\ge d_s$. Equality forces every factor except the last to be trivial. Hence $G$ 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.