Question on a homomorphism of a set G.

I'm having difficulty showing the given a map, say $\phi(z)=z^k$, is surjective. This question is from D & F section 1.6 - #19

Let $G$ =$\{z \in \mathbb C|z^n=1 \text{ for some } n \in \mathbb Z^+\}$. Prove that for any fixed integer $k>1$ the map from $G$ to itself defined by $z \to z^k$ is a surjective homomorphism, but is not an isomorphism.

To show $\phi(z)=z^k$ is a homomorphism let $z_a=e^\frac{2\pi ia}{n}$ and $z_b=e^\frac{2\pi ib}{n}$ where $a,b\in \mathbb Z^+$ and $a\neq b$ then $\phi(z_az_b)=(e^\frac{2\pi ia}{n}$$\cdot$$e^\frac{2\pi ib}{n}$)=$e^\frac{2\pi ik(a+b)}{n}$=$e^\frac{2\pi ika}{n}$$\cdot$$e^\frac{2\pi ikb}{n}$=$\phi(z_a)\phi(z_b)$.

I had what I thought was a proof for $\phi$ being surjective until I thought about this specific example. Here's the example: Suppose $z_a$ is a root of unity then so is a power $z^t_a$. My thoughts on $G$ are that it consists of all integer powers of the "basic" roots of unity for a partiuclar $n$. Let $n=3$ and $k=3$, then $\phi(e^\frac{2t\pi ik_i}{3})=(e^\frac{2t\pi ik_i}{3})^3 =e^{2t\pi ik_i}= 1$ for any integer $t\in \mathbb Z$ and integer $k_i$ such that $0\leq$$k_i$$\leq$ $2$. So we have $\phi[G]=\{1\}$, $\phi$ is not surjective in this instance.

What have I misunderstood about the question? Are there any glaring errors I'm making? Is $n$ supposed to be a particular value and not just any integer value?

Your help is appreciated.

G = $\{e^{i2\pi\over n}: n\in \mathbb Z^{+}\}$ so for $g,h\in G, (gh)^{k} = g^{k}h^{k}$ so this is easily seen to be a homomorphism.

Let $g\in G$ be given by $e^{i2\pi \over m}$. Then $w = e^{i2\pi \over mk} \in G$ and $w^{k} = e^{i2\pi \over m}$ so g is surjective.

This map is not injective because $e^{i2\pi}$ and $e^{i2\pi \over k}$ both get mapped to 1.