Prove that the order of an element in the group $N$ is the lcm(order of the element in $N$'s factors $p$ and $q$)

Question

How would you prove that

$$\operatorname{ord}_N(\alpha) = \operatorname{lcm}(\operatorname{ord}_p(\alpha),\operatorname{ord}_q(\alpha))$$

where $N=pq$ ($p$ and $q$ are distinct primes) and $\alpha \in \mathbb{Z}^*_N$

I've got this:

The order of an element $\alpha$ of a group is the smallest positive integer $m$ such that $\alpha^m = e$ where $e$ denotes the identity element.

And I guess that the right side has to be the $\operatorname{lcm}()$ of the orders from $p$ and $q$ because they are relatively prime to each other. But I can't put it together, any help would be appreciated!

Answer 1

Hint. There are natural maps $\mathbb{Z}^*_N\to\mathbb{Z}^*_p$ and $\mathbb{Z}^*_N\to\mathbb{Z}^*_q$ given by reduction modulo $p$ and reduction modulo $q$. This gives you a homomorphism $\mathbb{Z}^*_N\to \mathbb{Z}^*_p\times\mathbb{Z}^*_q$.

What is the kernel of the map into the product? What is the order of an element $(x,y)$ in the product?

Answer 2

Hint $\rm\ \ pq\:|\:a^n\!-\!1\!\!\overset{(p,q)=1\!\!}\iff p,q\:|\:a^n\!-\!1\!$ $\rm \iff\! ord_p a, ord_q a\:|\:n\!\iff\! lcm(ord_p a, ord_q a)\:|\: n$