TITLE: Collecting proofs that finite multiplicative subgroups of fields are cyclic
QUESTION [87 upvotes]: I teach elementary number theory and discrete mathematics to students who come with no abstract algebra.  I have found proving the key theorem that finite multiplicative subgroups of fields are cyclic a pedagogical speedbump.  For example, Serre's proof in A Course in Arithmetic runs a full page, requires introducing Euler's $\phi$-function, and depends on a counting argument that might seem to beginners too clever or magical for a cornerstone result.
I'd like to have a collection of proofs of this fact, to compare their advantages,
to match their viewpoints to my various audiences, to contrast for my students, etc.
To get the ball rolling, here's the shortest argument I can think of (and if it's in the literature somewhere I'd love a reference).
Induction on the order of the subgroup. So suppose multiplicative 
subgroup $G$ of field $F$ has order $n$. If $n=p^k$ with $p$ prime and $G$ isn't 
cyclic, all  $p^k$ elements of $G$ satisfy $x^{p^{k-1}}-1=0$, impossible.
If $n=ab$, $gcd(a,b)=1$, then $(\cdot)^a:G\rightarrow G$ has a kernel $A$ of size at most $a$ and a range $B$ of size at most $b$ (since the $y\in B$ satisfy $y^b=1$), so $|A|=a$, $|B|=b$, and a product $xy$ of cyclic generators $x,y$ for $A,B$ respectively generates $G$. 

If you know published proofs distinctly different from either of these, please cite a source.  No need to spell out the details, but please mention a key feature to help avoid duplicates.  If you have your own favorite approach, please share it.

REPLY [2 votes]: Here's my version of a proof, which seems to me to be very elementary.  I'll prove the following assertion, which clearly applies to finite subgroups of the multiplicative group of a field:
Let $G$ be a finite abelian group such that, for every prime $p$, there are at most $p$ elements $g\in G$ satisfying $g^p=1$.  Then $G$ is cyclic.
I argue by induction on $n=|G|$.  So let $G$ be a non-trivial group satisfying the hypothesis. Then there exists an element of $G$ with order $>1$, and hence (by taking an appropriate power) an element of $G$ with order $p$ for some prime $p$.  The main observation I will use is that the $p$th power map on $G$ is a homomorphism, and therefore defines a $p$-to-$1$ surjective function $\phi\colon G\to G^p$ (as the hypothesis on $G$ implies that the kernel of $\phi$ must have order $p$.)
The subgroup $G^p$ has order $n/p<n$, and being a subgroup of $G$ also satisfies the hypothesis, so by induction is cyclic on some generator $a\in G^p$.  I will show the following: there exists $b\in G\smallsetminus G^p$ such that $b^p=a$.  Given this  it is straightforward to verify that $b$ has order $n$ (using that $G/G^p$ has prime order so is cyclic), whence $G$ is cyclic. 
There are two cases, depending on the behavior of the restriction of the $p$th power map to $G^p$, which gives a surjective homomorphism $\phi|_{G^p}\colon G^p\to G^{p^2}$.

$G^{p^2}=G^p$, so $\phi|_{G^p}\colon G^p\to G^p$ is a bijection.  Therefore the restriction $\phi|_{G\smallsetminus G^p}$ of the $p$th power map to the complement of $G^p$ is a  $(p-1)$-to-$1$ surjective function $G\smallsetminus G^p\to G^p$. In particular $a=b^p$ for some $b\in G\smallsetminus G^p$ as desired.
$G^{p^2}\neq G^p$.  Therefore $\phi|_{G^p}\colon G^p\to G^{p^2}$ is a surjective $p$-to-$1$ map. Since $\phi\colon G\to G^p$ is surjective and $p$-to-$1$, it follows that  the restriction $\phi|_{G\smallsetminus G^p}$ of the $p$th power map to the complement of $G^p$ is a $p$-to-$1$ surjective function $G\smallsetminus G^p\to G^p\smallsetminus G^{p^2}$.
The generator $a$ of $G^p$ cannot be in the proper subgroup $G^{p^2}$, so $a=b^p$ for some $b\in G\smallsetminus G^p$ as desired.