Let \(G\) be a finite subgroup of the multiplicative group of a field. We shall show that it is cyclic.

In fact, we will show something far stronger: if \(G\) is a group with at most \(n\) solutions to \(x^n = 1\) (i.e., at most \(n\) elements of order dividing \(n\)) for each \(n\), then every finite subgroup of \(G\) is cyclic. (Note that we do not even presume abelianness here, though we will get it as a consequence). A field’s multiplicative group of course satisfies the stated property by the fact that a polynomial of degree \(n\) can have at most \(n\) distinct roots.

Proof: Let \(G\) be finite and have at most \(n\) elements of order dividing \(n\), for each \(n\); we shall show that \(G\) is cyclic.

Let \(g\) be any element in \(G\), of order \(n\); note that the powers of \(g\) will then provide \(n\) many values of order dividing \(n\). This is the maximum allowed by our presumption. Thus, these are ALL the values of order dividing \(n\), and among them, we find that precisely \(\phi(n)\) are of order precisely \(n\) [where \(\phi\) refers to the Euler totient function; see also Möbius inversion].

Thus, there are either zero or \(\phi(n)\) many values of order \(n\), for each \(n\). Can there be zero values of order \(\vert G \vert\)? No, for summing up the upper bound \(\phi(n)\) over all proper divisors of \(\vert G \vert\) 1, we get \(\vert G \vert - \phi( \vert G \vert )\); since this is less than \(\vert G \vert\), these proper divisor orders cannot account for all the elements of the group, and thus there must be at least one (indeed, precisely \(\phi( \vert G \vert )\)) many elements of order \(\vert G \vert\), i.e. generating the entire group, completing our proof.

Footnotes:

  1. Remember, by Lagrange’s theorem, every element’s order must divide \(\vert G \vert\). ↩