- #1

Mr Davis 97

- 1,462

- 44

So, let, ##|(a,b)| = n##. Then ##(a,b)^n = [(a,1_B)(1_A, b)]^n = (a,1_B)^n(1_A, b)^n##, since ##(a,1_B)## and ##(1_A, b)## commute. Then ##(a,b)^n = (a^n,1_B)(1_A, b^n) = (a^n,b^n)##. From this, it seems pretty clear that ##n = lcm (|a|, |b|)##, however, I don't really how to rigorously show this, or whether what I've said is good enough.