- Prove that the multiplicative group of units of a finite field must be cyclic, and so is generated by a single element.
Solution
1. Let \(F\) be a finite field and put \(G=F^\times\).Then \(G\) is a finite abelian group.
Multiplication in a field is commutative, and every nonzero element is a unit.
2. Let\[ m=\operatorname{lcm}\{\operatorname{ord}(g):g\in G\}, \]the exponent of \(G\).There exists an element \(h\in G\) of order exactly \(m\).
Write \[ m=\prod_{\ell}\ell^{a_\ell} \] over the finitely many primes dividing \(m\). For each such prime \(\ell\), by the definition of the least common multiple there is some \(g_\ell\in G\) whose order has \(\ell\)-adic valuation \(a_\ell\). If \(\operatorname{ord}(g_\ell)=\ell^{a_\ell}r_\ell\) with \(\gcd(\ell,r_\ell)=1\), then \[ h_\ell=g_\ell^{r_\ell} \] has order \(\ell^{a_\ell}\). Because \(G\) is abelian and the orders of the \(h_\ell\) are pairwise coprime, the product \[ h=\prod_\ell h_\ell \] has order \[ \prod_\ell \ell^{a_\ell}=m. \]
3. Every element of \(G\) is a root of the polynomial \(x^m-1\in F[x]\).
For \(g\in G\), its order divides \(m\) by the definition of \(m\), so \(g^m=1\).
4. Hence \(|G|\le m\).
A nonzero polynomial of degree \(m\) over a field has at most \(m\) roots. By 3, all \(|G|\) elements of \(G\) are roots of \(x^m-1\).
5. Since \(h\in G\) has order \(m\), one also has \(m\le |G|\).Therefore \(m=|G|\).
The cyclic subgroup \(\langle h\rangle\) has exactly \(m\) elements and is contained in \(G\).
6. Thus \(\langle h\rangle=G\), so the multiplicative group of a finite field is cyclic.
By 5, the subgroup \(\langle h\rangle\) and the whole group \(G\) have the same finite cardinality.