# Prove that $f(z)$ is a onto group homomorphism?

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

• 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.
– Mark
Feb 1, 2021 at 13:34
• 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 Feb 1, 2021 at 13:39

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 !

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