Given $k \in \mathbb{N}, G_k = \{z \in \mathbb{C} |z^k =1 \} $. Probe that if $n$ and $m$ are coprime, the function $f: G_n \times G_m \rightarrow G_{mn}, f(\alpha, \beta) =\alpha\beta$ is bijective.
I tried thinking on a $w_1,w_2 \in \mathbb{C}$ such:
$$w_1 = e^{{i2k\pi}/n}, k\in \mathbb{N}$$ $$w_2 = e^{{i2k ' \pi}/m}, k'\in \mathbb{N} $$
then $$ f(w_1,w_2) = e^{{i2k\pi}/n} e^{{i2k'\pi}/m} = e^ {{i2\pi (mk+nk')}/nm} $$ then if I get to prove that $(mk+nk')/nm$ is bijective, $f(\alpha,\beta)$ it's also bijective, right?
I'm getting a little confused whith the fact that $e^{i*someting}$ is not bijective, though
Since $G_n \times G_m$ and $G_{mn}$ both have $mn$ elements, it is enough to prove that $f$ is injective.
Suppose $\alpha\beta=\alpha'\beta'$. Then $\alpha^{-1}\alpha'=\beta^{-1}\beta'$. Now, $\alpha^{-1}\alpha'\in G_m$ and $\beta^{-1}\beta'\in G_n$ and so $\alpha^{-1}\alpha'=\beta^{-1}\beta'\in G_m \cap G_n$. Since $\gcd(m,n)=1$, we have $G_m \cap G_n=\{1\}$. Thus $\alpha^{-1}\alpha'=\beta^{-1}\beta'=1$ and $f$ is injective.