We want to proove that Euler's $\phi$ function is multiplicative. We have to show that $\forall m,n \in \mathbb{Z^+}, $ if $\gcd(m,n)=1$ then $\phi(mn)=\phi(m)\phi(n)$.
We know that $\phi (n):=|U(\mathbb{Z}_n)|$. Also, $ \mathbb{Z}_{nm} \cong \mathbb{Z}_n \times \mathbb{Z}_m, \forall m,n \in \mathbb{Z^+} $ with $\gcd(m,n)=1$. (isomorphic rings)
So in these two isomorphic rings the no zero unit elements in $\mathbb{Z}_{nm}$, will be the "same" in $\mathbb{Z}_n \times \mathbb{Z}_m $. So, $|U(\mathbb{Z}_{nm})| = |U(\mathbb{Z}_n)|| U( \mathbb{Z}_m) | \implies \phi(mn)=\phi(m)\phi(n) $. Is this right?
Furthermore, can we say that $U(\mathbb{Z}_{nm}) \cong U(\mathbb{Z}_n) \times U(\mathbb{Z}_m)$ (multiplicative groups)? And can we prove it finding a bijection without Chinese remainder theorem?
(I know that this question already exists but the answer I am looking for is different and abstract).
Thank you