Let \(F\) be a field. Prove that every finite subgroup of the multiplicative group \(F^\times\) is cyclic.
Solution
Let \(G\leq F^\times\) be finite. Choose \[ g\in G \] whose order \[ m\coloneqq\operatorname{ord}(g) \] is maximal among the orders of elements of \(G\).
1. If two commuting elements \(x,y\) have finite coprime orders \(r,s\),then\[ \operatorname{ord}(xy)=rs. \]
Since \(x\) and \(y\) commute, \[ (xy)^{rs}=x^{rs}y^{rs}=1. \] If \((xy)^k=1\), then \[ x^k=y^{-k}. \] Raising to the \(s\)th power gives \[ x^{ks}=1, \] so \(r\mid ks\). Since \(\gcd(r,s)=1\), one has \(r\mid k\). Similarly, raising \(x^k=y^{-k}\) to the \(r\)th power gives \(s\mid k\). Thus \(rs\mid k\). Therefore the least positive such \(k\) is \(rs\).
2. Every \(h\in G\) has order dividing \(m\).
Let \[ n\coloneqq\operatorname{ord}(h). \] Suppose, for contradiction, that \(n\nmid m\). Then there is a prime \(p\) such that \[ p^a\mid m,\quad p^{a+1}\nmid m, \] and \[ p^b\mid n,\quad p^{b+1}\nmid n \] with \(b>a\).
Write \[ m=p^a r, \qquad n=p^b s, \] where \(p\nmid r,s\). The element \[ x\coloneqq g^{p^a} \] has order \(r\), while \[ y\coloneqq h^s \] has order \(p^b\). Their orders are coprime. Since \(G\subset F^\times\) is abelian, step 1 gives \[ \operatorname{ord}(xy) = rp^b. \] Because \(b>a\), \[ rp^b>rp^a=m, \] contradicting the maximality of \(m\). Therefore \(n\mid m\).
3. Every element of \(G\) is a root of the polynomial\[ X^m-1\in F[X]. \]
For \(h\in G\), step 2 gives \[ \operatorname{ord}(h)\mid m. \] Hence \[ h^m=1, \] so \(h\) is a root of \(X^m-1\).
4. One has\[ \abs{G}\leq m. \]
By step 3, all distinct elements of \(G\) are roots of the nonzero degree-\(m\) polynomial \(X^m-1\). A nonzero polynomial of degree \(m\) over a field has at most \(m\) roots.
5. One also has\[ \abs{G}\geq m. \]
The cyclic subgroup \[ \langle g\rangle \] has exactly \(m\) elements and is contained in \(G\).
6. Therefore\[ \boxed{G=\langle g\rangle} \]and \(G\) is cyclic.
Steps 4 and 5 give \[ \abs{G}=m. \] But \(\langle g\rangle\subseteq G\) already has \(m\) elements, so the two finite sets are equal.
7. Q.E.D.
Step 6 proves the assertion for an arbitrary finite subgroup \(G\leq F^\times\).