Every finite multiplicative subgroup of a field is cyclic

Let F be a field and G \leq F^\times be a finite multiplicative subgroup. Prove that G is cyclic.

We prove this first in the case that G is a p-group, say |G| = p^k, by induction on k. Given G, we let \psi(d) denote the number of elements of G of order d.

Suppose |G| = p; then certainly G \cong Z_p is cyclic and \psi(d) = \varphi(d) for each d dividing p.

Suppose that every subgroup of order p^t is cyclic for some k \geq t \geq 1 and that for all such t, \psi(p^t) = \varphi(p^t). Let G be a subgroup of order p^{k+1}. Note that the elements of G are precisely the roots of x^{p^{k+1}} - 1 in F. Similarly, the roots of x^{p^k} - 1 are precisely those elements of G whose order divides p^k. Suppose \alpha and \beta are two such roots; that is, \alpha^{p^k} = 1 and \beta^{p^k} = 1. Now (\beta^{-1})^{p^k} = 1, so that (\alpha\beta^{-1})^{p^k} = 1. Since 1^{p^k} = 1, the set of roots of x^{p^k} - 1 (i.e. the set of elements of G whose order divides p^k) form a subgroup of F^\times of order p^k. By the induction hypothesis, this group is cyclic, and \psi(p^t) = \varphi(p^t) for each 1 \leq t \leq k. Note also that for each such t, \psi(p^t) with respect to G is equal to \psi(p^t) with respect to this subgroup. Since \sum_{t=1}^{k+1} \psi(p^t) = p^t = \sum_{t=1}^{k+1} \varphi(p^t), we have \psi(p^{k+1}) = \varphi(p^{k+1}). In particular, G contains an element of order p^{k+1}, and so is cyclic.

Thus every finite p-subgroup of F^\times is cyclic. Now let G be an arbitrary finite subgroup; since G is abelian, it is the direct product of its p-subgroups. Since these are cyclic, G is cyclic.

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