20180911, 20:42  #1 
Mar 2016
11×23 Posts 
order of a function =/= p1
A peaceful night for all persons,
i am looking for an algebraic structure which has not an order as p1 or p+1 What about (r 1)(x) = f(x,y) mod p where r is element N and p prime. (0 r )(y) Would be nice to get a short link about this topic. Greetings from the algebraic structures, there seems to be more than expected Bernhard Last fiddled with by bhelmes on 20180911 at 20:43 
20180912, 15:22  #2 
Aug 2006
2^{2}×5×293 Posts 
I'm so confused by this question.
You're looking for an algebraic structure A, that is, a set S and a collection O1, O2, O3, ..., On of operators on S. You then want A to have an order other than p1 or p+1, but what do you mean by "order" and what is p? It sounds like p is some fixed prime number and your set is {0, 1, ..., p1}, with your operations being some "natural" operation or operations mod p. If your operations are unary, each element of S has an order; perhaps you mean the exponent of the group so induced. Since this divides the order of the group, that is, p, it can't be equal p+1 and can equal p1 only if p = 2. Otherwise, please clarify what you mean. 
20180912, 18:10  #3 
Sep 2002
Database er0rr
C97_{16} Posts 
I too was confused. I think Bernhard means f(x,y) = [1,r;0,1]*[x,y]~ in parispeak. HTH.

20180912, 20:11  #4 
Aug 2006
1011011100100_{2} Posts 

20180912, 20:22  #5 
Mar 2016
11·23 Posts 
A peaceful and pleasant night for everyone,
if i have a function f(x) mod p, where p is a prime and x is element 0 ... p1 and the function should have as result a natural number from 0 to p1 then there will be a cycle of function terms. f(x) : Np > Np The amount of results of f(x) is limited, therefore if i make a recursion, there has to be a repetition of f(x). i take an x0, calculate x1=f(x0) mod p, calculate x2=f(x1) mod p and so on I thought that the expression "order" describes the huge of the cylic structure. For example, you could take a 2x2matrix A and a vector (x,y) and consider the function f(x,y)=A(x,y) mod p where the mod p is taken for every x and y The question was, if A=[r,1;r,0] so that the eigenwert is r, what will be the huge of the cyclic structure ? I will try to improve my mathematical english, but it is not easy for me to explain an idea and translate it exactly in another language. Greetings from the recursive functions Bernhard 
20180913, 16:23  #6 
Aug 2006
2^{2}×5×293 Posts 
I don't see a useful way of answering that in general, it depends on the value of p, r, and the starting values of the column vector. For p = 2, r = 0 you get order 1. For p = 2, r = 1 you get 1 if you start with [2;2] and 3 otherwise, which I will denote
[3 3] [3 1]. For p = 3, r = 0 you get order 1; for p = 3, r = 1 you get [8 8 8] [8 8 8] [8 8 1]; for p = 3, r = 2 you get [3 1 3] [1 3 3] [3 3 1]. For p = 5, r = 0 you get order 1; for p = 5, r = 1 you get [20 4 20 20 20] [20 20 20 4 20] [ 4 20 20 20 20] [20 20 4 20 20] [20 20 20 20 1]; for p = 5, r = 2 you get [24 24 24 24 24] [24 24 24 24 24] [24 24 24 24 24] [24 24 24 24 24] [24 24 24 24 1]; for p = 5, r = 3 you get [4 4 1 4 4] [1 4 4 4 4] [4 4 4 1 4] [4 1 4 4 4] [4 4 4 4 1]; for p = 5, r = 4 you get [3 3 3 3 3] [3 3 3 3 3] [3 3 3 3 3] [3 3 3 3 3] [3 3 3 3 1]. Code for generating these: Code:
findCycle(ff:closure,startAt,flag=0)={ my(power=1,len=1,tortoise=startAt,hare=ff(startAt),mu); while (tortoise != hare, if (power == len, tortoise = hare; power <<= 1; len = 0 ); hare = ff(hare); len++ ); tortoise=hare=startAt; for (i=1,len, hare = ff(hare) ); while(tortoise != hare, tortoise=ff(tortoise); hare=ff(hare); mu++ ); if(flag, print("mu = "mu)); len }; addhelp(findCycle, "findCycle(ff, startAt): Finds the length of the first cycle that startAt, ff(startAt), ff(ff(startAt)), ... enters into. Prints the prefix length (steps taken before the cycle begins)."); f(p,r,startAt)=my(A=[r,1;r,0]);findCycle(v>A*v%p,startAt); g(p,r)=matrix(p,p,x,y,f(p,r,[x;y])); 
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