Let $G= \{ z \in \mathbb{C} \ \ | \ \ az^{n}=b \}, a \in \mathbb{C^{*}}, b \in \mathbb{C}$.

Question

Let $G= \{ z \in \mathbb{C} \ \ | \ \ az^{n}=b \}, a \in \mathbb{C^{*}}, b \in \mathbb{C}$.

Find $a$ and $b$ such that $G$ is a subgroup of $(\mathbb{C^{*}},\cdot)$ where $\cdot$ is the regular multiplication.

The answer should be $a=b$.

Answer 1

$G$ is a subgroup thus it contains $1$, $a1^n=b$ implies $a=b$. Conversely, $a=b=\neq 0$, $az^n=b$ is equivalent to $z^n=1$, $G$ is the group of $n$ roots of the unity.

Answer 2

What do you need for $G$ to be a subgroup of $(\Bbb{C}^*, \cdot)$?

Well:

1) $1 \in G$

2) For any $g, h \in G$ you have $g \cdot h \in G$

3) For every $g \in G$, $g^{-1} \in G$.

Can you conclude with this that $a=b$?