For $n ≥ 2$ we let $G_n ⊂ Z_n$ denote the subset of all integers mod n which are invertible $\mod n$. Let $m, n \in \mathbb{Z}$ $m, n ≥ 2$ and $(m, n) = 1$. Define a mapping $f : Z_m × Z_n → Z_{mn}$ as follows: set $f([a]_m, [b]_n) = [x]_n$ where $x$ is the unique solution $(\mod mn)$ given by the Chinese Remainder Theorem to the simultaneous linear congruences $$[x]_m = [a]_m$$ $$[x]_n = [b]_n.$$ Show that $f$ is a bijective map. Show that $f(G_m × G_n) ⊂ G_{mn}$. Show that $f : G_m × G_n −→ G_{mn}$ is bijective.
My attemps: pick $([a]_m, [b]_n),([a']_m, [b']_n) \in \mathbb{Z}_n \times \mathbb{Z}_m$ so that $f([a]_m, [b]_n) = [x]_n$ and $f([a']_m, [b']_n) = [x']_n$. If $[x]_n = [x']_n$, then by the definition of the map, we have that $[a]_m = [a']_m$ and $[b]_n = [b']_n$. Hence, $f$ is injective.
Now, take any $y \in \mathbb{Z}_{mn}$. Need to find a point $( \times , \times)$ in $\mathbb{Z}_n \times \mathbb{Z}_m$ so that $f( \times , \times) = y = [x]_n$. By construction, we can just take $( \times , \times) = ([a]_m,[b]_m)$. Hence, $f$ is surjective and therefore bijective.
Now, take $x \in f( G_m \times G_n) \implies x = [x]_n$ where $[x]_n$ is unique solution $\mod mn$ to $$[x]_m = [a]_m \in G_m$$ $$ [x]_n = [b]_n \in G_n$$
Therefore, $[x]_{mn} = [ab]_{mn} \in G_{mn}$. So, $x$ is unique solution to previous equation and hence $x$ lies in $G_{mn}$. therefore, $f(G_m × G_n) ⊂ G_{mn}$.
Is this correct? Also, is there a faster way to prove the last statement of the problem? can someone help me? I appreciate it, thanks
