# Extension of map $z\mapsto z^p$

Let $p$ be a prime and $G$ be the group of p-power roots of 1 in $\mathbb{C}$. Prove that $z\mapsto z^p$ is a surjective homomorphism and deduce that $G$ is isomorphic to a proper quotient of itself.

I have done the first part of the proof, and in fact it is here: Intuitive explanation for a map $z \to z^p$. What I am having trouble with is the "deduce..." part. I'm not sure what is meant by "a proper quotient of itself," and I don't know where the isomorphism will come from since we only have a surjective homomorphism. Am I supposed to show that the function from the proof is also one-to-one?

Any surjective group homomorphism $f : G\to H$ induces an isomorphism $G/\ker(f)\to H$ (this is simply the first isomorphism theorem). Thus, your surjective map \begin{align*} G&\to G\\ z&\mapsto z^p \end{align*} induces an isomorphism $G/\ker(z\mapsto z^p)\cong G$.
The above answers where the isomorphism comes from, let's cover what it means by "proper quotient." A proper quotient of a group $G$ is a quotient by a normal nontrivial subgroup $N.$ That is, the quotient map $\pi : G\to G/N$ is surjective (as always) but not injective. So, what you need to show is not that $z\mapsto z^p$ is one-to-one, but rather that it is not one-to-one! This is equivalent to $\ker(z\mapsto z^p)$ being a nontrivial subgroup of $G,$ and this isn't too hard to show - you don't need to determine the kernel entirely, you only need to exhibit one nontrivial element contained in it!