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.
Best Answer
Let $n = |G|$ and let $m$ be the l.c.m. of the orders of the cyclic factors of $G$. Then $x^m = 1$ for all $x \in G$; since we are in a field this equation has at most $m$ roots, which shows that $m \geq n$. It follows that $m = n$ and $G$ is cyclic.
Of course here one uses the classification of finite abelian groups as product of cyclic groups, which you may want to avoid.