So, the set of $n^{th}$ roots of unity are defined as:
$C_{n} = \{e^\frac{2 \pi k i}{n} \mid k \in \{0,...,(n-1)\}\}$
And the group $S^1$ is defined to be:
$S^1 = \{z \in \mathbb{C} \mid \lvert z \rvert = 1 \} = \{ e^{i\theta} \mid \theta \in \mathbb{R} \}$
Now, to show that $C_n \lt S^1$ We need to demonstrate closure of products and inverses in $C_n$. I have (nominally) done this, though I'm a bit hung up on fulfilling certain criteria. Allow me to write out below and then I will elaborate. Let $c_{1},c_{2} \in C_n$
- $c_{1}c_{2} = e^\frac{2 \pi k_{1} i}{n}e^\frac{2 \pi k_{2} i}{n} = e^\frac{2 \pi (k_{1} + k_{2}) i}{n} \in C_n$
- $c_{1}^{-1} = e^\frac{-2 \pi k i}{n} = e^\frac{2 \pi (-k) i}{n} \in C_n$
How exactly can I show that these products and inverses fulfill the requirements to be an element of $C_n$? Namely, that $(k_1 + k_2)$ and $(-k_1) \in \{0,...,(n-1)\}$. Now I know that $e^{i\theta}$ where in this case $\theta = \frac{2\pi k}{n}$ is a periodic function, though I would like to be able to show when $k' \geq n$ or $k' \lt 0$ that $e^{\frac{2\pi k' i}{n}}$ is equivalent to $e^{\frac{2\pi k i}{n}}$ where $k \in \{0,...,(n-1)\}$ i.e., one of the roots of unity.
Are you familiar with modular arithmetic?
$k_{1}+k_{2}= kn+r $ for some $0\leq r\leq n-1$ and $k$ is some integer.
Hence $$c_{1}c_{2} = e^\frac{2 \pi k_{1} i}{n}e^\frac{2 \pi k_{2} i}{n} = e^\frac{2 \pi (k_{1} + k_{2}) i}{n} = e^{\frac{2o\pi(k_{1}+k_{2})}{n}}=e^{\frac{2i\pi(kn+r)}{n}}=e^{2i\pi k}\cdot e^{\frac{2\pi i r}{n}}=e^{\frac{2\pi i r}{n}}\in C_{n} $$.
and $-k = n-k\pmod n$ .
Hence $c_{1}^{-1} = e^\frac{-2 \pi k i}{n} = e^\frac{2 \pi (-k) i}{n}=e^{\frac{2\pi i(n-k)}{n}} \in C_n$.
I suggest you get familiar with the group $\Bbb{Z}_{n}$ as soon as possible to better understand all of these.