Show that $\left(\Bbb{C}^*,\cdot\right)$ is not cyclic

Question

My question is,

Show that $\left(\Bbb{C}^*,\cdot\right)$ is not cyclic.

I'm trying to show that $\Bbb{C}^*$ cannot be generated by an element in $\Bbb{C}^*$, but I cannot complete the proof.

Answer 1

Each cyclic group is countable. The group $(\mathbb{C}^\ast, \cdot)$ is not countable, so it is not cyclic.

Answer 2

I will give you some hints to solve your problem:

1) If the generator $g$ of $\mathbb{C}^*$ has norm greater or smaller than 1, how many powers of $g$ have norm equal to 1?

2) If the generator of $\mathbb{C}^*$ has norm equal to 1, can one of its powers be of norm different from 1?

Now it's easy to conclude. You can answer easy to the above questions using tha De Moivre formula for powers of a complex number or by looking at the exponential formula of the generator you go to consider.

Answer 3

Suppose $(\mathbb{C}^*,\cdot)$ is cylclic. Now $\langle2,i\rangle$ is a subgroup of $(\mathbb{C}^*,\cdot)$, but it is cyclic since it is a subgroup of a cyclic group and therefore $\langle2,i\rangle=\langle a\rangle$ for some $a\in\mathbb{C}^*$. Now $a^n=2$ and $a^m=i$ for some $n,m\in\mathbb{Z}-\{0\}$, which imples $$a^{nm}=2^n=i^m\implies 2^n=|i^m|=1$$ But $2^n\neq 1$ since $n\neq 0$ and it is a contradiction.