Lemma 9.17.1. Let A be an abelian group of exponent dividing n such that \{ x \in A \mid dx = 0\} has cardinality at most d for all d | n. Then A is cyclic of order dividing n.
9.17 Roots of unity
Let F be a field. For an integer n \geq 1 we set
\mu _ n(F) = \{ \zeta \in F \mid \zeta ^ n = 1\}
This is called the group of nth roots of unity or nth roots of 1. It is an abelian group under multiplication with neutral element given by 1. Observe that in a field the number of roots of a polynomial of degree d is always at most d. Hence we see that |\mu _ n(F)| \leq n as it is defined by a polynomial equation of degree n. Of course every element of \mu _ n(F) has order dividing n. Moreover, the subgroups
\mu _ d(F) \subset \mu _ n(F),\quad d | n
each have at most d elements. This implies that \mu _ n(F) is cyclic.
Proof. The conditions imply that |A| \leq n, in particular A is finite. The structure of finite abelian groups shows that A = \mathbf{Z}/e_1\mathbf{Z} \oplus \ldots \oplus \mathbf{Z}/e_ r\mathbf{Z} for some integers 1 < e_1 | e_2 | \ldots | e_ r. This would imply that \{ x \in A \mid e_1 x = 0\} has cardinality e_1^ r. Hence r = 1. \square
Applying this to the field \mathbf{F}_ p we obtain the celebrated result that the group (\mathbf{Z}/p\mathbf{Z})^* is a cyclic group. More about this in the section on finite fields.
One more observation is often useful: If F has characteristic p > 0, then \mu _{p^ n}(F) = \{ 1\} . This is true because raising to the pth power is an injective map on fields of characteristic p as we have seen in the proof of Lemma 9.12.5. (Of course, it also follows from the statement of that lemma itself.)
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