I have been trying to prove this:
Prove that if n, m are coprime, then
$f: G_n \times G_m \rightarrow G_{n.m}$
$f (t,s)\rightarrow t.s$
is a bijective function.
I started by attempting to check if it is injective. I didn't know which way to do that, so I tried to show it using properties. So far, I've shown that $G_n \cap G_m=G_ {(n:m)}=G_1=1$ and that $G_n\neq G_m$, which intuitively made me think it is injective.
Then, I noticed that if I take any $t\in G_2$ and any $s\in G_3$ and plug them in the function, I get $t.s \in G_6$, which is obvious, but made me realize that I could write any $G_i$ like this, as long as it is the product of two primes, which would show it is also surjective.
How could I write this properly? I believe that my attempts are still very intuitive and would appreciate some help to prove it more rigorously.
EDIT: $G_k= \{z \in \Bbb C : z^k = 1\}$
Hint Prove that $f$ is a group homomorphism, and then find $ker(f)$.