Do we know that $\mathbb{C}^{*}$ is an abelian group? Let $U= \{ z \in \mathbb{C}: \exists n \in \mathbb{N} \text{ such that } z^n=1\}$.
Show that $(U, \cdot)$ is a subgroup of $\mathbb{C}^{*}=\mathbb{C}-\{ 0\}$
Let $a,b \in U$.
So,$\exists n,m \in \mathbb{N} \text{ such that } a^n=1, b^m=1$.
We have to check if $a \cdot b' \in U$.
We suppose $k=m \cdot n$
Can I get from the relation $(a \cdot b')^k$ to this : $a^k \cdot (b^{-1})^k$?
Do we know that $\mathbb{C}^{*}$ is an abelian group?

The multiplication of $\Bbb C^*$ is the same multiplication of $\Bbb C$ as a field, which is commutative, therefore the group is abelian.
