Prove that oa1 a2 lcmoa1 oa2Solutionif ab ba then in gener

Prove that o((a1, a2)) = lcm(o(a1), o(a2))

Solution

if ab = ba, then in general we have

(ab)^n = a^n b^n

We also note that
a^n * b^n = e
means that
a^n = b^-n

that is, the above is only true when a^n is in <b>, which means that a^n = b^-n = b^n = e.

With that in mind, we deduce that
(a b)^n = e
iff
a^n * b^n = e
iff
a^n = e  
and  
b^n = e

This, in turn, is only true when o(a)|n and o(b)|n. The lowest n for which this is true is the least common multiple. The statement follows.