# Primitive roots of the unity in $\mathbb C$

Let $$\omega$$ be a primitive $$n-$$th root of unity.

(i) Show that its powers $$\omega^k$$, for $$k ∈ {1, \ldots, n}$$, are all different;

(ii) Deduce that they are precisely all the $$n-$$th roots of unity.

I know that the powers have to be different as otherwise the order of $$ω$$ would be less than $$n$$ so $$\omega$$ wouldn't be a primitive $$n-$$th root of unity but I don't know how to prove this rigorously.

• your idea is good: if $\omega^k=\omega^j$ then $\omega^{k \mod n}=\omega^{j \mod n}$ so $k \mod n=j \mod n$ so if $j,k\in\{1,\dots,n\}$ then $j=k$ Nov 6, 2019 at 22:25
• why does ω^k(mod n)=ω^j(mod n) imply that k(mod n) = j(mod n) ?
– user720013
Nov 6, 2019 at 22:35
• $\omega^a=\omega^b\implies\omega^{a-b}=1$ but since $\omega^n=1$ but $\omega^k\ne1$ for $0<k\le n$ we have $a\equiv b\mod n$ Nov 6, 2019 at 22:38
• oh ok, it took me a bit to understand this as I haven't been taught it yet but it makes sense now
– user720013
Nov 6, 2019 at 22:50

## 1 Answer

$$\omega \in \Bbb C \tag 1$$

such that

$$\omega^n = 1, \tag 2$$

but

$$\omega^k \ne 1, \; 1 \le k < n; \tag 3$$

if

$$\exists 1 \le j \ne l \le n, \; \omega^j = \omega^l, \tag 4$$

we may without loss of generality assume

$$j < l; \tag 5$$

then from (4),

$$\omega^{l - j} = 1; \tag 6$$

but

$$0 < l - j < n, \tag 7$$

in contradiction to (3); thus (4) is false and the $$\omega^k$$, $$1 \le k \le n$$, are all distinct.

We note that each $$\omega^k$$, $$1 \le k \le n$$, is an $$n$$-th root of unity, since

$$(\omega^k)^n = \omega^{kn} = \omega^{nk} = (\omega^n)^k = 1^k = 1. \tag 8$$

In accord with (2) and (8), we see that each $$\omega^k$$ is a root of

$$x^n - 1 \in \Bbb Q[x] \subsetneq \Bbb C[x]; \tag 9$$

now since

$$\deg (x^n - 1) = n, \tag{10}$$

this polynomial has at most $$n$$ distinct roots; so since

$$\vert \{\omega^1 = \omega, \omega^2, \ldots, \omega^{n - 1}, \omega^n = 1\} \vert = n, \tag{11}$$

every $$n$$-th root of unity is in this set; there are no others.

• You have explained this really well thank you!!
– user720013
Nov 7, 2019 at 15:11
• You are most welcome, my friend. And thanks for the "acceptance"! Cheers! Nov 7, 2019 at 15:25