Let a and b belong to a group. If |a| and |b| are relatively prime, show that <a> intersect <b> = {e}.

Answer 1

let c be in ^ . then there exist x,y such that c = ax = by.
let u = x^-1, v=y^-1
cu = a, cv = b contradicting the relative primeness of a,b

Answer 2

Let a^j = b^k, for j, k integers.

Then there exists an integer m which divides o(a) -- and hence is relatively prime to o(b) -- such that 1=a^(jm) = b^(km).

But since m is relatively prime to o(b), and we just showed that the o(b) divides km, it follows that the o(b) divides k.

Thus, b^k = 1

QED.