Need help in showing that $z\mapsto z^k$ is a surjective homomorphism.

Question

Let $$G=\{z\in \Bbb{C}\mid z^n=1 \quad\text{for some }n \in \Bbb{Z}^{+}\}.$$ Show that for any fixed integer $k>1$, the map $z\mapsto z^k$ from $G$ to itself is a surjective homomorphism but not an isomorphism.

Proving that the map is a homomorphism is easy - for any $z_1,z_2 \in G$ we have $\varphi(z_1z_2)=(z_1z_2)^k=z_1^kz_2^k=\varphi(z_1)\varphi(z_2)$. Proving that it is not an isomorphism is also straightforward - since $\operatorname{ker}\varphi=\{z\in G \mid z^k=1\}$ which is not $\{1\}$, we can conclude that the mapping is not an isomorphism.

I just need some hint to show that the mapping is surjective. Thanks in advance!

Answer 1

Every element $z$ in $G$ can be written in the follow form $$z={\rm exp}\left(\frac{2\pi i a}{n}\right)$$ for some $a,n \in \mathbb N, n\neq 0$, so a preimage is $${\rm exp}\left(\frac{2\pi i a}{nk}\right).$$