Let $k$ be a positive integer, $S$ is a set of $k$ nonzero complex numbers, sastifying $a,b\in S\Rightarrow ab\in S$. Show $\forall\ a\in S, a^k=1$, and find the sum of elements of $S$.
Clearly, $a,a^2,\dots,a^{k+1}\in S$, and hence for some $i<j, a^i=a^j, a^{j-i}=1$. Then what to do?
Let $S=\{s_1,\dots,s_k\}$, and $a\in S$. We have $as_i\in S$ for all $i=1,\dots,k$. Then $\{as_1,\dots,as_k\}\subseteq S$, and since both sets have $k$ elements it follows that $\{as_1,\dots,as_k\}=S$. The product of their elements is the same, so $a^ks_1\cdots s_k=s_1\cdots s_k$ and thus $a^k=1$. Moreover, $a(s_1+\dots+s_k)=s_1+\dots+s_k$ and therefore $s_1+\dots+s_k=0$. (Here I assumed that $k\ge2$ and thus one can choose an $a\in S$ such that $a\neq1$.)
By your argument, we know that $S$ is a group (it is worth mentioning that this is the group of the $n^{th}$ roots of unity), and we also know that for any $x \in S$ we have $\lvert x \rvert = 1$. Let $a \in S$ such that $arg(a) = min_{x \in S} \{ arg(x) \}$ where we define $arg(z) \in [0,2 \pi)$. We will prove that $X = S$ where $ X = \{ a, a^2, ..., a^k \}$. Clearly $X \subset S$.If there is $b \in S \setminus X$ then consider $J = max_{i \in \{ 1,...,k \} } \{ arg(b \cdot a^{-i}) > 0 \}$. Then $arg(b \cdot a^{-J}) < arg(a)$ (because if that's not the case then $arg(b \cdot a^{-J-1}) > 0$ which is a contradiction because J was the maximum of that set), but this is contradictory with the choice of $a$. Then we conclude that $X = S$, so we now know that \begin{equation*} S= \{ a,a^2,...,a^k \} \end{equation*} so because $S$ has k elements we know that if $i,j \in \{1,...,k \}$ and $i \neq j$ then $a^i \neq a^j$, so the fact that $1 \in S$ implies that $a^k = 1$. Now we prove that for any $b \in S$ we have $b^k = 1$.Let $\theta \in [0,2 \pi)$ be such that $a = e^{i \theta}$. Then $\theta = \frac{2 \pi}{k}$. Then if $b \in S$ we have $b = a^{j \frac{2 \pi}{k}}$ for some $j \in \{1,...,k \}$ so that \begin{equation*} b^k = a^{2 j \pi} = 1 \end{equation*} Now for the sum of the elements. If $S = {1}$ it's trivial. If there is $a \in S$ such that $a \neq 1$ then because $S$ is a group we know that $a S = S$ where $aS = \{ a \cdot s : s \in S \}$. Then if $T$ is the sum of the elements of S we have \begin{equation*} a T = T \implies T(a - 1) = 0 \implies T = 0 \end{equation*} because $a \neq 1$.
