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anonymous
 3 years 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
anonymous
 3 years 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

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let \(G=\{z \in \mathbb{C}  z^n=1\text{ for some }n \in \mathbb{Z}^+ \}\), no idea why it doesn't show up in the question itself

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0isomorphism requires both injective and surjective homomorphism

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I know that but I wouldn't have a clue how to prove this is not injective :)

John_ES
 3 years ago
Best ResponseYou've already chosen the best response.1May 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0try prime and nonprime values of k.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Showing 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 kth roots, we know k > 1, so for every \(z \in G\), there are at least 2 kth 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
 3 years ago
Best ResponseYou've already chosen the best response.0Completely forgot I could also take roots of numbers :D
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