Finite multiplicative subgroup of a field is cyclic
= Finite multiplicative subgroup of a field is cyclic
Every finite <subgroup> $H$ of the multiplicative group of a <field> is <cyclic group>[cyclic]. If $m$ is the exponent of $H$, commutativity lets one combine elements whose orders realize the prime-power factors of $m$, producing an element of order $m$. Every element of $H$ is a root of $X^m-1$, so the <Lagrange root bound over a field> gives $|H|\leq m$. Since $m\leq|H|$, equality holds and that element generates $H$.