Let $\mathbb{C}^\times$ the multiplicative group of complex numbers different of zero. Let $H$ the subgroup $\mathbb{C}^\times$ of generated by $\{i, e^{\frac{2i \pi}{5}}, -1\}$. Find the order of $H$.
I found that the neutral element is one $1$, the inverse of $i$ is $-i$, the inverse of $e^{\frac{2i \pi}{5}}$ is $e^{\frac{-2i \pi}{5}}$, the inverse of $-1$ is itself and other unnecessary items; so $O(H)= 6$. Am I wrong here? Is anyone can help me at this point?
The generators can be written as $e^{\pi i/2},e^{2\pi i/5},e^{\pi i}$. Then the values these can generate through multiplication are exactly the values of the form $$e^{2\pi i(m/4 + n/5)}$$ for integral $m$ and $n$ (the $-1$ adds nothing because it can itself be generated from $i$).
These in turn are exactly the distinct values $$e^{2\pi i k/20}$$ for integral $k$, of which there are twenty.
Hint:
Is $a$ and $b$ have orders $r$ and $s$, and $\langle a\rangle\cap\langle b\rangle=\{1\}$, $ab$ has order $\operatorname{lcm}(r,s)$.
