 Solution

Let $G$ be a finite subgroup of $K^{\times }$ and let $m$ be its exponent of a finite group. A finite abelian group contains an element whose order is its exponent: combine elements of maximal prime-power order across the primary components. Hence $m\le |G|$.

Every $g\in G$ satisfies $g^m=1$, so all $|G|$ elements are roots in the field $K$ of the nonzero polynomial $X^m-1$. The Lagrange root bound over a field gives $|G|\le m$. Therefore $|G|=m$, and the element of order $m$ generates $G$. This proves that a finite subgroup of a field multiplicative group is cyclic.