= 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 $m\leq|G|$.
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 $|G|\leq m$. Therefore $|G|=m$, and the element of order $m$ generates $G$. This proves that a <finite subgroup of a field multiplicative group is cyclic>.
Back to article page
