Suppose that a and b belong to a group, a and b commute and |a| and |b| are relatively prime. prove that ....?

write |a|=m , |b|=n. let k=|ab|. 1=1^n= [(ab)^k]^n=a^(kn) b^(kn)=a^(kn) [b^n]^k=a^(kn) 1^k= a^(kn) a^(kn) = 1. |a|=m, m divides kn. since m , n relatively prime m divides k. switching things around (starting 1=1^m etc.) see n divides k. m , n divide k , m , n relatively prime, implies mn divides k. (ab)^(mn) = (a^m)^n (b^n)^m = 1^n 1^m =1 k divides mn. therefore k=mn

example: choose 2 , 6 in additive group z16. 2 has order 8, 6 has order 8 2+6 =8 has order 2. many other examples possible, e.g., when orders different share factor

|ab| = |a||b| . give example showing |ab| need not |a||b| when , b commute |a| , |b| not relatively pime ( don't use b = a^-1)

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