How to show that $ord_n(x)=lcm(ord_p(x), ord_q(x))$ for primes p,q such that $n=pq$

Question

I am trying to show

$ord_n(x)=lcm(ord_p(x), ord_q(x))$ for primes p,q such that $n=pq$

I tried a few different things and found some similar questions but I failed to understand them. Please help me?

I thought of simply using the definition of ord and lcm and their commutative properties to prove it but I get $lcm(\text{ord}_p(x),\text{ord}_q(x))|\text{ord}_n(x)$ and not sure how this would prove the above.

Another proof that I fail to understand:

This follows because $x^j \equiv 1 (mod n) \iff x^j \equiv 1 mod p$ and $x^j \equiv 1 mod q$ but it is not immediately obvious why that is the case. Because the above is the definition of order

Similar questions I read but failed to understand:

The last proof you say you don’t understand is just the Chinese Remainder Theorem, which I think is the most straightforward way of showing what you want. — Diego Artacho, Apr 24 '22 at 17:58
Can you expand please. How is that the case? i.e. I get why the statements are true but I do not see how they connect to LCM — user2789433, Apr 24 '22 at 18:02

score 1 · Accepted Answer · answered Apr 24 '22 at 18:04

Notice that $$x^k\equiv 1\pmod{n}\implies x^k\equiv 1\pmod{pq}\implies x^k\equiv 1\pmod{p}, 1\pmod{q}$$ By the Chinese Remainder Theorem (since $p,q$ are relatively prime). Thus, by basic properties of the order, $ord_p(x), ord_q(x)|k\implies lcm(ord_p(x), ord_q(x))|k$ by the definition of lcm. Hence it suffices to show that $k=lcm(ord_p(x), ord_q(x))$ is a solution to $x^k\equiv 1\pmod{pq}$. This is true because $x^k\equiv 1\pmod{p}, 1\pmod{q}$, and the result follows by the chinese remainder theorem.

How to show that $ord_n(x)=lcm(ord_p(x), ord_q(x))$ for primes p,q such that $n=pq$

1 Answers1