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

This is called the *group of $n$th roots of unity* or *$n$th 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

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 $p$th 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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