## Meepi one year ago 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

1. Meepi

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

2. sirm3d

isomorphism requires both injective and surjective homomorphism

3. Meepi

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

4. John_ES

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.

5. sirm3d

try prime and non-prime values of k.

6. Meepi

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.

7. Meepi

Completely forgot I could also take roots of numbers :D