my question was discussed in this thread:
Fundamental Theorem of Finite Abelian Groups and the mulitplicative group of a field.
I want to show that if G is a finite subgroup of the multiplicative group of an arbitrary field, then G is cyclic.
However, I don't really know what to do with the hint in the answer. It would be very nice if someone could go into more detail as to how to use that $x^k-1$ has at most $k$ roots in a field. What should $k$ be in this case? And how does the fundamental theorem come into play?
If $A$ is an Abelian group of order $n$, which is not cyclic, then it has exponent $m$ with $m<n$. That is, $m\ge1$ and $a^m=1$ for all $a\in A$. This follows from the structure theorem for finite Abelian groups.
If $A$ is a subgroup of a multiplicative group of a field, then this is impossible, since then $x^m-1=0$ has more than $m$ solutions.