TITLE: Proving that the order of $(a,b)$ is the lcm of $|a|,|b|$
QUESTION [5 upvotes]: I want to solve the following exercise from Dummit & Foote's Abstract Algebra text:

Prove that the elements $(a,1)$ and $(1,b)$ of $A \times B$ commute and deduce that the order of $(a,b)$ is the least common multiple of $|a|$ and $|b|$.

The first part is easy: $$(a,1)(1,b)=(a \star 1,1 \diamond b)=(a,b)=(1 \star a,b \diamond 1)=(1,b)(a,1) $$
The second part isn't that much harder to prove. The thing is I can't see how a proof can be deduced by the mere fact that $(a,1)$ and $(1,b)$ commute. Is such a proof possible?
Thanks.

REPLY [3 votes]: Hint $\ \ 1 = (a,b)^k = (a,1)^k (1,b)^k\iff (a,1)^k = (1,b)^{-k}\in \langle(a,1)\rangle\cap \langle (1,b)\rangle = \{(1,1)\}$  
$\qquad\  \iff (a,1)^k = (1,1) = (1,b)^k\iff|a|,\,|b|\mid k\iff {\rm lcm}(|a|,|b|)\mid k$