I want to show that if $\gcd (n_i, n_j)=1$ for $i\not= j$ and $m=n_1 \cdots n_k$,then $$ U(m) = U_{\frac{m}{n_1}}(m) \times \cdots \times U_{\frac{m}{n_k}}(m), $$ where $\times$ is the symbol for internal direct product, $U(m) = \{x\in \mathrm{Z}^+ | \gcd (x,m)=1\}$, and $U_{s}(m)=\{x\in U(m) \mid x \pmod{s}= 1\}$.
My attempt:
Since $U(m) \cong U(n_1) \oplus \cdots \oplus U(n_k)$ and $U_{\frac{m}{n_j}}(m) \cong U(n_j)$ (where $\oplus$ is external direct product), it follows that $$ U(m) \cong U_{\frac{m}{n_1}}(m) \oplus \cdots \oplus U_{\frac{m}{n_k}}(m). $$
I’m also aware that because $U(m)$ is Abelian, all its subgroups — including the $U_{\frac{m}{n_j}}(m)$ — are normal. I also know that the $U_{\frac{m}{n_j}}(m)$ only share $1$ in common. I’ve been trying to use these pieces of information.
I think this is very close to finishing, but I’m not sure how to go from here. Any help would be appreciated.
Edit I tried to show one of the conditions for $U(m)$ being the internal direct product of the previously mentioned subgroups. Namely, that $U(m)=U_{\frac{m}{n_1}}(m)\cdots U_{\frac{m}{n_k}}(m)$.
My attempt:
Since $U(m) \cong U_{\frac{m}{n_1}}(m) \oplus \cdots \oplus U_{\frac{m}{n_k}}(m)$, it follows that all the elements in $U(m)$ are of the form $\mathrm{lcm}(|u_1|,\cdots ,|u_k|)$ where $u_i\in U_{\frac{m}{n_i}}(m)$. Also, $U(m)$ is certainly Abelian, which means for all $x_j\in U(m)$, $|x_1\cdots x_k|$ divides $|x_1|\cdots |x_k|$. When considering these two facts, LaGrange’s Theorem, and that the $n_i$’s are pairwise coprime, any $x\in U(m)$ must be a product of $u_i$’s.
I’d like to know if this is wrong in any way.