A sum of powers of primitive roots of unity

For the primitive roots of unity $\omega_n = e^{i2\pi/n}$ I want to prove that $$\sum_{k=0}^{n-1} \omega_n^{lk} = 0$$ if $n$ doesn't divide $l$.

I have already proven the well-known result $$\sum_{k=0}^{n-1} \omega_n^{k} = 0$$ so I only need to show that if I raise the $\omega_n$s to the power of $l$, I get every power of $w_n$ exactly once. This is a simple algebraic statement but I don't see how to prove it.

• What you're trying to show is wrong : if $n=4$ and $l=2$, the elements $\omega_4^{2k}$ for $k=0,1,2,3$ are $1,-1,1,-1$. – Arnaud D. Jun 20 '16 at 17:21
• Ah, thank you for catching this! – Marc Jun 20 '16 at 17:42
• I believe the @ArnaudD. comment was intended to go under the "Fly by night" answer. One would ordinarily interpret "what you're trying to show" as the statement in the original question $\sum_{k=0}^{n-1} \omega_n^{lk}=0$, in which case it looks like the Arnaud comment is claiming $1 + -1 + 1 + -1 \neq 0$ (though I think it is really claiming $\{-1, 1, -1\}$ is not a permutation of $\{e^{i4\pi/4}, e^{i6\pi/4}, e^{i 8 \pi/4}\}$. – Michael Jun 20 '16 at 17:50
• Thanks for posting this clarification. I understood what he meant but it is a bit ambiguous. – Marc Jun 20 '16 at 17:56
• Yes sorry, my comment was about the sentence "I only need to show that if I raise the $\omega_n$s to the power of $l$, I get every power of $\omega_n$ exactly once.", but not about the original question. I should have been more precise. – Arnaud D. Jun 20 '16 at 18:43

Say that sum is $s$. Show that $\omega_n^l s=s$ and that $\omega_n^l\ne 1$.
Hint: Show that $(\omega_n^{\ell 1},\ldots,\omega_n^{\ell (n-1)})$ gives a permutation of $(\omega_n^{1},\ldots,\omega_n^{n-1})$ when $\ell$ and $n$ are coprime.
• In fact the sum is $0$ assuming just that $n$ does not divide $l$. – David C. Ullrich Jun 20 '16 at 17:24
• @Marc No, it's not the same. He's assuming that $n$ and $l$ are coprime, not just that $n$ does not divide $l$. Did you see the comment to your question? The thing about this being a permutation of that is false under your hypotheses... – David C. Ullrich Jun 20 '16 at 17:40
• @Marc If $F$ is a finite set and $f:F\to F$ is surjective or injective then $f$ is bijective. – David C. Ullrich Jun 20 '16 at 18:01