 Solution

The multiplicative group of a field is abelian, so its finite subgroup $G$ is a finite abelian group. Let $d\mid |G|$, and suppose $H$ is a subgroup of order $d$. By Lagrange's theorem, the order of each $h\in H$ divides $d$, hence $h^d=1$. Thus $H$ is contained in the set of roots in $k$ of $t^d-1$.

This nonzero polynomial has degree $d$, so it has at most $d$ roots. Since the $d$ distinct elements of $H$ are already roots, they comprise the whole root set. Any second subgroup $K$ of order $d$ would also consist of $d$ roots of the same polynomial, forcing $K=H$. Therefore there is at most one subgroup of each order dividing $|G|$.

Part (c) now applies to giveThe root bound remains valid in positive characteristic, even if the formal derivative of $t^d-1$ vanishes: the argument counts distinct roots, so no separability assumption is needed.