What are the elements forming the subgroup $H = \langle a,b \rangle$ of $\Bbb C^\ast$ when $a = e^{2\pi i/5}$ and $b=e^{2\pi i/7}$? Is $H$ cyclic?

Question

What are the elements forming the subgroup $H = \langle a,b \rangle$ of $\Bbb C^\ast$ when $a = e^{2\pi i/5}$ and $b=e^{2\pi i/7}$? Is $H$ cyclic?

For the subgroup $H$ we have that $H=\{ae^{2\pi i/5} + be^{2\pi i/7} \mid a,b \in \Bbb Z\}$ but I don't know if it's supposed to get some better description for it?

$H$ is cyclic if I can generate it with single element, but I don't think it's possible to do this?

Answer 1

Hint:

1. $H=\langle a, b\rangle=\{a^mb^n \mid m, n\in\Bbb{Z}\}$

2. The order $O(a)$ of $a$ is $5$ and the order $O(b)$ of $b$ is $7$.

3. $\Bbb{C}^{\star}$ is an abelian group and $\gcd(O(a), O(b))=1$, hence $O(ab) =35$

4. Every subgroup of a cyclic group is cyclic.

5. $|G|<\infty$.Then $G=\langle g\rangle$ iff $|G|=O(g)$

$H=\langle e^{\frac{2\pi i}{35}}\rangle\cong \Bbb{Z}_{35}$