I came across this result: $\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}/n_1\mathbb{Z} \times \mathbb{Z}/n_2\mathbb{Z}$ implies that $\left(\mathbb{Z}/n\mathbb{Z}\right)^{\times} \cong \left(\mathbb{Z}/n_1\mathbb{Z}\right)^{\times} \times \left(\mathbb{Z}/n_2\mathbb{Z}\right)^{\times}$
where $\left(\mathbb{Z}/n\mathbb{Z}\right)^{\times}$ denotes the multiplicative inverse group of $\mathbb{Z}/n\mathbb{Z}$.
So, I've been thinking that since the Chinese Remainder Theorem is about a ring isomorphism, namely $\phi : \alpha[n] \mapsto (\alpha[n_1],\alpha[n_2])$, I was thinking that $\phi(\alpha[n] \beta[n]) = \phi(\alpha[n])\phi(\beta[n])$ has to do something about the above result.
Here, $\alpha[n],\beta[n]$ denote some classes in $\mathbb{Z}/n\mathbb{Z}$.
Many thanks in advance.
PS: $n_1, n_2$ are relatively prime and $n=n_1 n_2$
First, every ring isomorphism $R\cong S$ induces a group isomorphism $U(R)\cong U(S)$. Here I write $U(R)=R^{\times}$ for the group of units of $R$. Secondly $U(S_1\times S_2)\cong U(S_1)\times U(S_2)$, see here:
Structure of a group of units of a ring composed of the direct product of two subrings
Now apply this to $R=\Bbb Z/n$, $S_1=\Bbb Z/n_1$, $S_2=\Bbb Z/n_2$. By the CRT we have $R\cong S_1\times S_2$ for coprime integers $n_1$ and $n_2$. Hence $$ U(n)\cong U(n_1)\times U(n_2), $$ with $U(n)=(\Bbb Z/n)^{\times}$.