From: "ali omer" <ali_1122334455@...>

I appreciate your concern.

Thank you very much.I found by guess that the order of x mod(p^k)=the order of x mod(p)

multiplied by p^{k-1} but I can't prove it and I don't know why it should be multiple of

it's order mod(p).

-------------------------------------------

Counterexample: Order(3,11) = 5, Order(3,11^4) = 605 = 5·11^2 <> 5·11^3 =

Order(3,11)·11^(4-1)

What happens here is that Phi(p) = p-1 and Phi(p^k) = p^(k-1)·(p-1), and it's true that

the order of the element mod n must divide Phi(n), so for small values of p you can expect

a lot of "simple" behaviours (it's the law of small numbers), like a lot of orders being

p^(k-1)·e, where e is the order mod p.

Regards. Jose Brox