**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.6 Exercise 1.6.19**

Let $G = \{ z \in \mathbb{C} \ |\ z^n = 1\ \mathrm{for\ some}\ n \in \mathbb{Z}^+ \}$. Prove that for any fixed integer $k > 1$ the map $\varphi : G \rightarrow G$ given by $z \mapsto z^k$ is a surjective group homomorphism but is not an isomorphism.

Solution: First, we note that $G$ is in fact a group under complex multiplication: $1 \in G$, if $z^n = w^m = 1$ then $(zw)^{mn} = 1$, and if $z^n = 1$ then $(z^{-1})^n = 1$.

Now with $k > 1$ and $z,w \in G$, we have $$\varphi(zw) = (zw)^k = z^k w^k = \varphi(z) \varphi(w),$$ so that $\varphi$ is a group homomorphism.

Now let $k > 1$ and $z \in G$. Then $z^n = 1$ for some $n \in \mathbb{Z}^+$. Note that $\sqrt[k]{z^n}$ exists in $\mathbb{C}$, and $$\varphi(\sqrt[k]{z^n}) = z^n = 1.$$ So every element of $G$ has a preimage under $\varphi$, so that $\varphi$ is surjective.

Note however that every complex number has $k$ distinct $k^\mathrm{th}$ roots; for all $k > 1$ and $z \in G$, let $\zeta_1, \zeta_2$ be two of these roots. Then $\zeta_1 \neq \zeta_2$ but $\varphi(\zeta_1) = \varphi(\zeta_2)$. So $\varphi$ is not injective for any $k > 1$.