Note, $R_n$ is the multiplicative group mod$n$ of all numbers $k$ relatively prime with $n$, where $1\leq k <n$.
I don't know how to proceed here. Hints only please.
None of the suggestions below make sense to me. Would the following be a suitable candidate for an isomorphism between $R_{mn}$ and $R_m \times R_n$? Define $\phi :R_m\times R_n\rightarrow R_{mn}:(k,j)\mapsto kj$mod($mn$)
Here is a roadmap:
If $A$ and $B$ are rings and $R=A\times B$, then $R^\times = A^\times \times B^\times$.
The natural ring homomorphism $\mathbb Z \to \mathbb Z/m \times \mathbb Z/n $ has kernel $mn\mathbb Z$.
The induced ring homomorphism $\mathbb Z/mn \to \mathbb Z/m \times \mathbb Z/n $ is surjective (by comparing sizes).
$\mathbb Z/mn \cong \mathbb Z/m \times \mathbb Z/n $. (This is the Chinese remainder theorem.)
If you haven't seen rings and quotients, you may proceed as follows. Define $f: R_{mn} \to R_m \times R_n$ by $f(x \bmod mn) = (x \bmod m, x \bmod n)$ and prove:
$f$ is well defined.
$f$ is a group homomorphism.
$f$ is injective.
$f$ is surjective. The easy way is to compare sizes, if you know that $\phi(mn)=\phi(m)\phi(n)$. Otherwise, you need to use that $1=mu+nv$ for some $u,v \in \mathbb Z$ and then $(a \bmod m,b \bmod n)=f(mub+nva \bmod mn)$.
