Question: Let $p$ $\DeclareMathOperator{\ord}{ord}$ and $q$ are distinct odd primes and $n=pq$. Show that there is an integer not divisible by $p$ or $q$ such that $\ord_n$ of that integer is $\operatorname{lcm}(p-1,q-1)$.
I guess here, $\ord_n(x)$ means the order of $[x]$ in multiplicative group of units of $\Bbb{Z}_n$, denoted by $U_n$. So I reduced the problem to following:
Let $p$ and $q$ be two distinct odd primes, show that $U_{pq}$ has an element of order $\operatorname{lcm}(p-1,q-1)$.
The following is a well-known result in group theory:
If an abelian group $G$ has an element of order $m$ and an element of order $n$ then it has an element of order $\mathrm{lcm}(m,n)$ (It is an exercise in Topics In Algebra of Herstein)
In your question $U_n=U_{pq}$ has order $\varphi(pq)=(p-1)(q-1)$. Both $p,q$ have primitive roots. Let $a$ and $b$ are primitive roots modulo $p$ and $q$ respectively. Let $d=\gcd(p-1,q-1)$. Show that $a^{\frac{q-1}{d}}\pmod{pq}$ is an element of order $p-1$ in $U_n$ and $b^{\frac{p-1}{d}}\pmod{pq}$ is an element of order $q-1$ in $U_n$. Hence you can now conclude using the result stated at the beginning since $U_n$ is definitely abelian.
We have $$ U_{pq} \cong U_{p} \times U_{q} \cong C_{p-1} \times C_{q-1} \cong C_{\gcd(p-1,q-1)} \times C_{\operatorname{lcm}(p-1,q-1)} $$
The first isomorphism follows from the Chinese remainder theorem.
The second isomorphism follows from $U_p$ begin cyclic of order $p-1$ when $p$ is prime.
The third isomorphism is the interesting one and is a general fact. See this question.
