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

Question

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.

Answer 1

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 $n$.

Answer 2

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

Answer 3

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.

Answer 4

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