Prove that every finite subgroup of the multiplicative group \(\mathbb C^\times\) is cyclic.
Solution
Let \(G\leq\CC^\times\) be finite, set \(n=\abs{G}\), and write \[ \mu_n = \left\{ z\in\CC^\times:z^n=1 \right\}, \qquad \zeta=e^{2\pi i/n}. \]
1. \(G\subseteq\mu_n\).
For every \(g\in G\), Lagrange’s theorem gives \[ g^n=1, \] because the order of \(g\) divides \(n=\abs{G}\). Hence every element of \(G\) lies in \(\mu_n\).
2. The group \(\mu_n\) has exactly \(n\) elements and\[ \mu_n=\gens{\zeta}. \]
The \(n\) numbers \[ 1,\zeta,\zeta^2,\ldots,\zeta^{n-1}. \] all satisfy \(z^n=1\). They are distinct: if \(0\leq j<k\leq n-1\) and \(\zeta^j=\zeta^k\), then \(e^{2\pi i(k-j)/n}=1\), so \(n\) divides \(k-j\), which is impossible because \(0<k-j<n\).
Conversely, if \(z^n=1\), then \(\abs{z}^n=1\), so \(\abs{z}=1\). Writing \(z=e^{i\theta}\), we have \(e^{in\theta}=1\), hence \(n\theta\in2\pi\ZZ\). Therefore \(z=\zeta^k\) for some integer \(k\). Thus these are exactly the elements of \(\mu_n\), and \(\mu_n=\gens{\zeta}\).
3. \(G=\mu_n\).
By step 1, \(G\subseteq\mu_n\). Step 2 gives \(\abs{\mu_n}=n=\abs{G}\). An inclusion between finite sets of equal cardinality is equality.
4. \(G\) is cyclic.
By step 3 and step 2, \[ G=\mu_n=\gens{\zeta}. \]
5. Q.E.D.
Step 4 proves the required conclusion.