Let a and b belong to a group. if /a/ and /b/ are relatively prime, show tht <a> n <b> = {e}?

Answer 1

Simple enough: Let x be any element of ∩ . Then |x| divides both |a| and |b|. Since |a| and |b| are relatively prime, this means |x| must be 1, so x = e. So ∩ = {e}.

The key is knowing that if x ∈ then |x| divides |a|. To see this, suppose x = a^k, |x| = j and |a| = n. Then x^n = (a^k)^n = (a^n)^k = e.

Let n = j.q + r where q and r are integers such that 0 ≤ r < j. Then x^(jq + r) = e, so
(x^j)^q . x^r = e
<=> x^r = e since x^j = e.
But x^r = e and 0 < r < j contradicts j being the order of x. So we must have r = 0. So n = jq, i.e. j divides n.

Answer 2

WOW......