primitive n-th roots of unity Show that the primitive n-th roots of unity have the form $e^{2ki\pi/n}$ for $k,n$ coprime for $0\leq k\leq n$.
Since all primitive n-th roots of unity are n-th roots of unity by definition they all have that form, the question is, how to show $k$ and $n$ are coprime. 
 A: Primitivity means that no positive power of $\zeta=e^{2\pi i k/n}$ less than $n$ will achieve unity. If $k$ is not coprime to $n$ and $\gcd(k,n)=m$, then observe $\zeta^{(n/m)}=e^{2\pi i (k/m)}=1$ but $n/m<n$ if $m>1$.
A: More generally, if $G$ is a cyclic group of order $n$ and $g$ is one of its generators, then the order of $g^k$ is $n/(k,n)$. In particular, the generators are $g^k$ with $(k,n)=1$. 
This is easy to prove: $t$ is the order of $g^k$ iff $t$ is the smallest number such that $1=(g^k)^t=g^{kt}$ iff $kt$ is the smallest multiple of $k$ that is also a multiple of $n$ iff $kt=\operatorname{lcm}(k,n) = kn/(k,n)$ iff $t=n/(k,n)$.
A: Just to help your intuition in support of @lhf's and @anon's answers...
Are you aware of the formula (for positive integers $n$)
$$
n = \sum_{0<d|n}\phi(d)
$$
where $\phi(n)$ is Euler's totient function? It's basically giving you a breakdown of how many $n$th roots of unity there are which are primitive $d$th roots of unity for each $d$ dividing $n$ (there are $\phi(d)$).
For example, of the six sixth roots of unity, one has order $1$, one order $2$, two have order $3$, and two have order $6$: $\phi(1)=\phi(2)=1$ & $\phi(3)=\phi(6)=2$. Try drawing them on the unit circle.
So with this intuition, we need to show that the order $d$ of $\zeta=e^{2\pi ik/n}$ is $n$ iff $k$ & $n$ are relatively prime, i.e. iff their gcd $g=(k,n)$ is $1$. However, note that for $t=n/g$, $\zeta^t=e^{2\pi ik/g}=1$ since $g|k$, so that $g>1$ implies that the $ord(\zeta)|t<n$, i.e. that $\zeta$ is not a primitive $n$th root but rather a primitive $t$th root of unity, where $t$ is a proper divisor of $n$.
