# Primitive roots of unity in $\mathbb{Z}/p$

Can anyone help me with this problem?

Let $p$ be a prime number. Prove that if the field $\mathbb{Z}/p$ has a primitive $n^{th}$ root of unity, then $n \mid (p-1).$

Any sources or books for reference? Any hints also appreciated.

## 4 Answers

The order of the group of units is $p-1$. The order of any element divides the order of the group.

Detail: Suppose that $a$ is a primitive $n$-th root of unity. Then $a^n\equiv 1\pmod{p}$, and there is no positive interger $m\lt n$ such that $a^m\equiv 1\pmod{p}$.

Let $p-1=qn +r$, where $0\le r\le n-1$. Because $a^{p-1}\equiv 1\pmod{p}$ (Fermat's Theorem), we have $$1\equiv a^{p-1}=a^{qn}a^r=(a^n)^q a^r\equiv a^r\pmod{p}.$$ Thus $a^r\equiv 1\pmod{p}$. If $r\ne 0$, this contradicts the fact that $a$ is a primitive $n$-th root of unity, meaning that $a$ has order $n$.

Hint: What is the order of a primitive $n^{th}$ root of unity? What does Lagrange's theorem say?

Any book that talks about finite fields should state the theorem that the multiplicative group of non-zero elements is cyclic.

Any book on elementary number theory should state the theorem that the multiplicative group of units modulo $p$ (for $p$ prime) is a cyclic group.

• This honestly looks a bit off to me. Neither of these (perfectly standard and good to know, of course) facts are either necessary or sufficient to answer the question at hand. Rather one needs to know (1) that every nonzero element of $\mathbb{Z}/p\mathbb{Z}$ is a unit -- this is usually deduced from Euclid's Lemma -- and (2) Lagrange's Theorem, or at least its special case that the order of an element of a finite commutative group divides the order of the group (Lagrange's Little Theorem). – Pete L. Clark Apr 12 '13 at 16:19

Hints:

$$\Bbb F_p:=\Bbb Z/p\Bbb Z\implies |\Bbb F_p^*|=p-1$$

$$w^n=1\pmod p\implies w\in\Bbb F_p^*$$

== In any finite group, the order of any element divides the order of the group