The Chinese remainder theorem for groups

Homomorphism. $\varphi(\bar k+\bar\ell)=(\overline{k+\ell}\bmod m,\overline{k+\ell}\bmod n)=\varphi(\bar k)+\varphi(\bar\ell)$.

Injectivity. If $\varphi(\bar k)=(\bar0,\bar0)$, then $m\mid k$ and $n\mid k$. Since $\gcd(m,n)=1$, $mn\mid k$, so $\bar k=\bar0$.

Surjectivity. Both sides have $mn$ elements, and injectivity plus equal cardinalities gives surjectivity.

Conclusion. $\mathbb{Z}/mn\mathbb{Z}\cong\mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/n\mathbb{Z}$, so the cyclic group of order $mn$ splits as a direct product of coprime cyclic factors.