## Every finite multiplicative subgroup of a field is cyclic

Let be a field and be a finite multiplicative subgroup. Prove that is cyclic.

We prove this first in the case that is a -group, say , by induction on . Given , we let denote the number of elements of of order .

Suppose ; then certainly is cyclic and for each dividing .

Suppose that every subgroup of order is cyclic for some and that for all such , . Let be a subgroup of order . Note that the elements of are precisely the roots of in . Similarly, the roots of are precisely those elements of whose order divides . Suppose and are two such roots; that is, and . Now , so that . Since , the set of roots of (i.e. the set of elements of whose order divides ) form a subgroup of of order . By the induction hypothesis, this group is cyclic, and for each . Note also that for each such , with respect to is equal to with respect to this subgroup. Since , we have . In particular, contains an element of order , and so is cyclic.

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

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