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

I can show that it's a homomorphism, but am a bit stuck after that

Question

Let $G=\left{z\in\mathbb{C}|z^n=1\text{ for some }n\in\mathbb{Z}^+\right}$. Prove that for any fixed integer $k > 1$ the map from G to itself defined by $z \mapsto z^k$ is a surjective homomorphism but not an isomorphism. 

I can show that it's a homomorphism, but am a bit stuck after that

anonymous · Answer

Let $G=\{z \in \mathbb{C} | z^n=1\text{ for some }n \in \mathbb{Z}^+  \}$, no idea why it doesn't show up in the question itself

sirm3d · Answer

isomorphism requires both injective and surjective homomorphism

anonymous · Answer

I know that but I wouldn't have a clue how to prove this is not injective :)

john_es · Answer

May be, you should consider two elements from the image space, and then try to find a pair of values from the domain space that have this image. For example, considering z=i^k, for some k's.

sirm3d · Answer

try prime and non-prime values of k.

anonymous · Answer

Showing that G is a group:
let $z_1, z_2 \in G$, then $z_1^m=1,z_2^n=1\text{ for some }n\in\mathbb{Z}^+$ and $(z_1z_2)^{mn}=1$ so G is closed under multiplication, $(z_1^{-1})^m=1$ and also closed under inverses so G is a group.

Homomorphism:
was quite easy since it follows from commutativity of ($\mathbb{C}$, *)
let $z_1, z_2\in G$, $k > 1$ then $\varphi(z_1z_2)=z_1^kz_2^k=\varphi(z_1)\varphi(z_2)$

Showing that it's surjective:
To show that it's surjective we need to show that for every $z \in G$, $\varphi(w)=z$ for some $w \in G$, let $w = \sqrt[k]{z}$, $\varphi(\sqrt[k]{z})=(\sqrt[k]{z})^k=z$, so all we need to show is that $w \in G$, so $w^n = 1$ for some positive integer, let this integer be kn where n is the order of z, then it's obvious $w^{kn} = 1$ and since kn is also a positive integer, $w \in G$.

Now to show it's not injective every complex number has k k-th roots, we know k > 1, so for every $z \in G$, there are at least 2 k-th roots, $r_1 \text{ and } r_2$, but $\varphi(r_1)=\varphi(r_2)$ so $\varphi$ is not injective and thus not an isomorphism.

anonymous · Answer

Completely forgot I could also take roots of numbers :D