If $m = n_1 n_2 \cdots n_k $ where $\gcd(n_i~,n_j)=1 ~~ \forall i \neq j$, then prove that:
$$U(m) = U_{m/n_1} (m) \times U_{m/n_1} (m) \times \cdots\times U_{m/n_k}$$
where $\times$ refers to the internal direct product and when $r$ divides $m$, $U_r(m) = \{x ~~\in ~~ U(m) \mid x \bmod r = 1\}$. We already know that $U_r(m)$ is a subgroup of $U(m)$
Attempt: If $\exists$ a group $G$ such that $G =HK \mid H \cap K = \{e\}$ and where $H,K$ are normal subgroups, then we say that $G = H \times K$ where $\times$ refers to the internal cross product.
So, we have to show that , as a precursor to the requirements of internal direct product, the following things :
(i) $U(m) = U(n_1 n_2 \cdots n_k) = U_{m/n_1}(m)~~ U_{m/n_1} (m) \cdots U_{m/n_k}(m)$
(where if $A_1, A_2, \ldots , A_n$ are groups, then $A_1~A_2 \cdots A_n = a_1~a_2\cdots a_n ~~\forall a_1 \in A_1 ,\forall a_2 \in A_2,\ldots \forall a_n \in A_n$ )
(ii) $U_{m/n_1}(m), U_{m/n_1} (m), \cdots,U_{m/n_k}(m)$ are normal subgroups of $U(m)$
(iii) $(~U_{m/n_1}(m) U_{m/n_1} (m)\cdots U_{m/n_j}(m) ) \cap U_{m/n_{j+1}}(m) = \{e\} $
One conclusion which we can draw is that for any $x~\in ~U(m)~~: x U_{m/n_j}(m) x^{-1} = x x^{-1}U_{m/n_j}(m) \subseteq U_{m/n_j}(m) $
Hence, all $U_{m/n_j}(m) $ are normal subgroups.
I have little idea how to prove parts (i) and (iii) of the requirements. Help will be appreciated. Thank you