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 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
 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

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Meepi
 one year 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

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.0isomorphism requires both injective and surjective homomorphism

Meepi
 one year 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
 one year 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.

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.0try prime and nonprime values of k.

Meepi
 one year 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.

Meepi
 one year ago
Best ResponseYou've already chosen the best response.0Completely forgot I could also take roots of numbers :D
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