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Topic: All orders divide the highest order?
Replies: 1   Last Post: Aug 22, 1996 7:55 PM

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Lance Levine

Posts: 6
Registered: 12/12/04
All orders divide the highest order?
Posted: Aug 21, 1996 10:09 PM
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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(m-1)=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?







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