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All orders divide the highest order?
Posted:
Aug 21, 1996 10:09 PM


U_m={natural numbers<=m and relatively prime to m} phi(m) = the number of elements in U_m if u is in U_m, then order(u)=n where n is the least natural number such that u^n=1 (mod m)
I know that all the orders in U_m divide phi(m)...but if U_m doesn't have an element with order phi(m), then a stronger statement can be made; for example, phi(8) is 4, but all the orders in U_8 are not only factors 4 but factors of 2. Note that 2 is the highest order of any element in U_8...Is this true in general, i.e., is it true that all the orders divide the highest order? Can anyone prove this?
Also, a related question: Does anyone know of a function f(m) that gives the highest order in U_m? I know f(m)=phi(m) exactly when m = 2, 4, p^n, or 2p^n, where p is any odd prime and n is any natural number...
Oh, one last question: I know it's hard in general to say what the order of u is in U_m, but are there any special elements for which it's easy to find the order? For example, for m>2, order(m1)=2. Also, I've noted that if you take u=the product of all elements of U_m, then order(u) divides 2. Going further, take u=the product of all elements of U_m which are less than m/2, and order(u) divides 4. What other statements like these can be made?



