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
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.
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 .