Let \(R\) be a commutative integral domain with identity, and let \(G\) be a finite subgroup of the unit group \(R^*\). Prove that \(G\) is cyclic.
Solution
Let \[ K\coloneqq\Frac(R). \] Since \(R\) is an integral domain, the natural map \(R\hookrightarrow K\) is injective, so regard \(G\) as a finite subgroup of \(K^\times\).
1. If \(e\) is the exponent of \(G\), then \(G\) contains an element of order\(e\).
Write \[ e=\prod_{p}p^{a_p}. \] For each prime \(p\) dividing \(e\), the definition of the least common multiple implies that some \(g_p\in G\) has order divisible by \(p^{a_p}\). If the order of \(g_p\) is \[ p^{a_p}m_p \] with \(p\nmid m_p\), then \[ h_p\coloneqq g_p^{m_p} \] has order \(p^{a_p}\).
Because \(G\) is abelian, the elements \(h_p\) commute. Their orders are pairwise coprime, so \[ h\coloneqq\prod_p h_p \] has order \[ \prod_p p^{a_p}=e. \]
2. Every element of \(G\) is a root of\[ t^e-1\in K[t]. \]
By definition of the exponent, the order of every \(g\in G\) divides \(e\). Hence \[ g^e=1 \] for every \(g\in G\).
3. One has\[ \abs{G}\leq e. \]
By step 2, all \(\abs{G}\) distinct elements of \(G\) are roots in the field \(K\) of the nonzero polynomial \(t^e-1\), which has degree \(e\). A nonzero polynomial over a field has at most as many roots as its degree, so \(\abs{G}\leq e\).
4. One has\[ e=\abs{G}. \]
Step 1 gives an element \(h\in G\) of order \(e\). By Lagrange’s theorem, \[ e\mid\abs{G}, \] so \(e\leq\abs{G}\). Combined with step 3, this gives \(e=\abs{G}\).
5. The group \(G\) is cyclic.
The element \(h\) from step 1 has order \(e\), and step 4 gives \(e=\abs{G}\). Therefore \[ \abs{\langle h\rangle}=\abs{G}, \] so \(G=\langle h\rangle\).
6. Q.E.D.
Step 5 is the required conclusion.