Subgroups of nth root of unity over an arbitrary field

Question

Let $F$ be an arbitrary field and consider $K_n:=\{f\in F|f^n=1\}$, the nth root of unity in $F$ and I was asked to verify the following property:

Property I wanted to verify: Let $m=|K_n|$ then for every divisor $d|m$ there exists at most one subgroup of $K_n$ which has the cardinality of $d$.

I wanted to verify this because this characterises cyclic property of a finite Abelian group as I described here.

My progress thus far:

I figured there was a way using Theorem 1.1 in KConrad's note. This is because Thm 1.1 shows any finite subgroup of $F^{\times}$ are precisely roots to $X^l-1$ for some integer $l$ then the rest is clear enough.

However if I were to follow this method it would be clear straightaway that $K_n$ is cyclic and thus I thought this way was a bit of a waste and not intended for the original exercise.

Hence my question is if there exists some more straightforward way of showing $K_n$ satisfies the property I described above?

Many thanks in advance!

Answer 1

Your question is strange because you mentioned most of the ideas. If a subgroup $H_d\subset F^\times$ has $d$ elements then all its elements are roots of $x^d-1$, there are at most $d$ such roots (here we are using that $F$ is an integral domain) thus $H_d$ is precisely the roots of $x^d-1$. If $H_{p^k}$ exists (it exists in $\overline{F}$ iff $char(F)\ne p$) then any element of $H_{p^k}-H_{p^{k-1}}$ has order $p^k$ so $H_{p^k}$ is cyclic. If $H_n$ exists then it contains $ \prod_{p^k \| n} H_{p^k}$ which contains an element of order $n$ so $H_n$ is cyclic. Knowing that those groups are cyclic makes the reasoning more intuitive. There is no easier way to solve those questions.