= 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 give
$$
\boxed{G\text{ is cyclic}.}
$$
The 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.
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