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Given $G= \{ z \in \mathbb{C} | \exists \ n \in \mathbb{Z}^{+} \text{such that} \ z^n=1\}$. Define a map $f : G \to G $ by $$f(z)=z^k$$

where $k>1$ is fixed and $k \in \mathbb{Z}^+$

My question : Prove that $f(z)$ is a onto group homomorphism ?

My attempt : No , take $z^{n/k} \in G$ now $f(z^{n/k})=z^n=1$ for all $z^{n/k} \in G$

$\implies$ $f$ become a constant i,e $f=1$

we know that constant function never give onto

therefore $f$ is not a onto homomorphism

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    $\begingroup$ Your solution is extremely unclear. How do you use $n$ here? In the definition of $G$, $n$ is just a variable-for every $z\in G$ there is some $n\in\mathbb{N}$ (which depends on $z$) such that $z^n=1$. It is not a constant number. $\endgroup$
    – Mark
    Feb 1, 2021 at 13:34
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    $\begingroup$ It is also not obvious what $z^{n/k}$ means if $k \not \mid n$. Also, nitpicking: a constant function can be onto, if the set is a singleton :p $\endgroup$
    – qualcuno
    Feb 1, 2021 at 13:39

2 Answers 2

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The problem comes from the fact that $n$ is not fixed ! For $z \in G$, a priori, the "corresponding $n$" could be anything.

A good idea to start would be to understand what the elements of $G$ look like. In fact, you can try to show (or may be you already know) that if $z^n = 1$, then you can write $z = e^{\frac{2ir\pi}n}$ for some $0 \leq r < n$.

This means that $G = \{e^{\frac{2ir\pi}n}\;|\; n \in \mathbb N^\ast , 0\leq r < n\}$. Now you can work out more precisely how your function acts on this set !

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From Numbra answer

Motive : to show $f$ is onto

$G = \{e^{\frac{2ir\pi}n}\;|\; n \in \mathbb N^\ast , 0\leq r < n\}$.

Now let $f $ is given by $e^{\frac{2ir\pi}n} \in G \implies a=e^{\frac{2ir\pi}{kn}}\in G$

according to question $f(a)=a^k=e^{\frac{2ir\pi}n}$

so $f$ is onto

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