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How to find multiplicative orders of all elements in F 13?

I am working on some Finite fields and I was referring to some online class material. Is there any way to find this?

kingmakerking
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  • This question appears to be off-topic because it doesn't involve programming. – ntoskrnl Mar 07 '14 at 12:42
  • @user2888239 I did not claim that it is programming, I am more concerned about the maths part – kingmakerking Mar 07 '14 at 13:12
  • [This FAQ](http://stackoverflow.com/help/on-topic) covers what is considered on-topic on Stack Overflow. This question might be better suited at [Math.SE](http://math.stackexchange.com/). – ntoskrnl Mar 07 '14 at 13:18

1 Answers1

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The non-zero elements in F 13 form a multiplicative group of order 12. You can represent them by the numbers 1, 2, 3, ..., 12. Algebra tells you that the group is cyclic. It turns out that 2 is a generator. Knowing the order of an element g in a group G it is straight forward to determine the order of any element on the form g^i. You can use this to determine the orders of all the elements.

A different method is to directly use the definition of the order of an element. That is for each element you calculate g, g^2, g^3, g^4, ... The smallest number d for which g^d = 1 is the order of that element. Given the small size of the group F 13* this is quite doable.

user515430
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  • I figured out that.I figured out that I can do with the following python code. for i in range(1,50): if((pow(2,i))%13==1): print i ............... I did the same for 2,3,4....12 – kingmakerking Mar 07 '14 at 10:29
  • as per your explaination, Mul_ord(1) = 1, Mul_ord(2) = 12, Mul_ord(3) = 3, Mul_ord(4) = 6, Mul_ord(5) = 4, Mul_ord(6) = 12, Mul_ord(7) = 12,Mul_ord(8) = 4, Mul_ord(9) = 3, Mul_ord(10) = 6, Mul_ord(11) = 12, Mul_ord(12) = 2 Is it correct? – kingmakerking Mar 07 '14 at 15:48
  • Your numbers are correct. You can double-check using that ord(g^i) = ord(g)/GCD(ord(g), i) combined with the fact that 2 is a generator. – user515430 Mar 07 '14 at 16:36