Let $G$ be the group of non-zero complex numbers under multiplication and let $H=\{z \in\mathbb C :z^n=1\}$. Find the cosets of $H$ in $G$.

Question

What I think is that the cosets are going to be $zH$ where $z^q=1$, where $q$ is an irrational number.

Proving the cosets are distinct: Let $z^{q_1}H =z^{q_2}H$ then $z^{q_1 -q_2} \in H$ which would probably mean that $q_1-q_2=n$. I need some help.

Answer 1

First, $\mathbb{C}\setminus \{0\} = S \times \mathbb{R}_+$ where $S =\{z: |z|=1\}$ and $\mathbb{R}_+$ is the multiplicative group of positive real numbers. This can be seen by the polar/trigonometric form of the complex number $z \mapsto (e^{i\arg{z}},|z|)$

Now, we can prove that $S/H \cong \mathbb{R}/\mathbb{Q}$, where $\mathbb{R}$ and $\mathbb{Q}$ are the additive groups.

Let $\phi \colon \mathbb{R} \to S/H$ be the group homomorphism described by $\theta \mapsto [e^{2\pi i\theta}]$. If $\theta \in \ker \phi$ then $e^{2n\pi i \theta}=1$, which means that $ \theta = \frac{k}{n} $ for some $k$ and $n$. That is, $\mathbb{Q} \supset \ker\phi$. Since $\mathbb{Q} \subset \ker \phi$ we have the result by Isomorphism Theorems.

So $\mathbb{C}\setminus \{0\}/H \cong \mathbb{R}_+ \times \mathbb{R}/\mathbb{Q}$

Answer 2

The notation leads me to believe that what is meant is "the group $\mathbb C \backslash \{ 0 \}$ of all non-zero complex numbers under multiplication" and that $n$ is meant to be either "non-negative integer" or "positive integer." If $n=0$ then $H=\mathbb C \backslash \{ 0 \}$ and there is only one coset, $\mathbb C \backslash \{ 0 \}.$ If $n=1,$ then $H= \{1\}$ and the cosets are all the $\{z\}$ for $z \in \mathbb C \backslash \{ 0 \}$. If $n>1$ then the cosets are all the $$\{z \text {cis}\frac{2\pi 1}{n},...,z \text {cis}\frac{2\pi n}{n}\}$$ for $z \in \mathbb C \backslash \{ 0 \}$.