If n is a positive integer, then $\displaystyle \sigma(n)$ will denote the cyclic group of order n.
If $\displaystyle gcd(m,n)=1$, prove that $\displaystyle \sigma(mn)\cong\sigma(m)\times\sigma(n)$.
There is a known result which says if $\displaystyle C$ is cyclic group of order $\displaystyle n$ then $\displaystyle C$ is isomorphic to $\displaystyle \mathbb{Z}_n$ (or in your notation $\displaystyle \sigma (n)$). To show that $\displaystyle \mathbb{Z}_m \times \mathbb{Z}_n$ is isomorphic to $\displaystyle \mathbb{Z}_{nm}$ it is sufficient to show $\displaystyle | \mathbb{Z}_m \times \mathbb{Z}_m | = nm$ (which is immediate) and also that $\displaystyle \mathbb{Z}_m\times \mathbb{Z}_n$ is cyclic. Thus, we need to find a generator. Show that $\displaystyle ([1]_m,[1]_n)$ has order $\displaystyle nm$ and therefore it must generate the whole group. This will show the group is cyclic and complete the proof.