Let a be in a group G with |a| = m. If n is relatively prime to m, show that a = b^n for some b in G.
My Proof so far:
Now the order of a is m, so I have a^m = e, the identity. n is relatively prime to m, so I have ng + r = m for some integers g and r, with m > r > n.
umm... what now?
Well, everything goes well to that point Now by Euclidean division, there are integers g and r, such that gn+rm=1. So , that is .n is relatively prime to m, so I have
Ps. Topsq:
Suppose for some i<n, and prove you chose a bad startThen show that one of these elements must be your "b" for a given n.
Let the order the order of be . And consider the cyclic subgroup . Now the order of this cyclic group is . Thus, the generators of this group are where is relatively prime to . In particular, by hypothesis. That means . Now is a premutation of thus is found among and hence . Let for some . And the proof is complete.