Let $m=p_1p_2...p_r$ be a product of distinct primes, and let $U_m$ be the group of units of $\mathbb{Z}/m\mathbb{Z}$. Let $e$ be the least common multiple of $p_1-1, p_2-1, ..., p_r-1$.
(i) Show that the order of every element $a$ of $U_m$ divides $e$.
(ii) Explain how to use the Chinese Remainder Theorem to find an element of $U_m$ of order e.
What I wrote for (i) :
Let $a \in U_m$ and $n$ be the order of $a$ and $n$ divides $e$. By division algorithm we can write $e=nq+r$ with $q,r \in \mathbb{N}$ and $0 \leq r<m$.
We have $a^e=1$. But $a^e=a^{nq+r}=(a^n)^q * a^r$.
Since $n$ is the order of $a$, $a^n=1$. Hence, we get $a^r=1$.
Since $0 \leq r <n$ and $a^r=1$, $r$ must be zero because the order of $a$ is $n$.
$r=0$, thus $e=nq$ and finally $n$ divides $e$.
Is that correct? Any idea for (ii)?