I need the Proof:
a, b are two elements of a group G such that ab=ba.
If o(a)=m & o(b)=n and m,n are relatively prime to each other,then
o(ab)=mn.
It's obvious that .
To show that mn is the smallest such number, note that the order of ab must be a multiple of both a and b. Depending on how strictly your teacher grades, you may want to explicitly justify this statement. Since m and n are relatively prime the smallest such number is their LCM, which is mn.
Yes, and that's what the problem essentially is about!
Suppose (where the last equality is justified because commute). Use the division algorithm to write , with and . Then . Now raise this expression to the power , eliminating from the left side; you obtain . Now we know has the same order as (prove it, using ), and therefore we must have , which implies . Similarily we must have . Therefore is divisible both by and therefore divisible by .