4

I am trying to work out the multiplicative order of each non zero element in $F_7$.

Lets say I am looking at the number $3$. I know its order is $6$.

Instead of having to work out the powers of three and work out the remainder when divided by $7$, is their an easier way to calculate the order especially when the numbers get bigger?

xxxxxxxxx
  • 13,302
Al jabra
  • 2,331
  • 2
    This is in practice the problem of finding discrete logarithms... I don't think there is an easy way to compute the order of an element. – Crostul Mar 16 '16 at 18:00

1 Answers1

3

The easiest way to find order of an element of $F_p$ by hand, is calculate remainder of powers of your number for all prime divisors of $p-1$, and then by use of them to calculate order of divisors of $p-1$ by the way that I explain below by help of an example:

For example you want to calculate order of $13 \in F_{31}$ :

$13*13=169 \equiv 14 \pmod {31}$
$13^3=13*13^2\equiv 13*14=182 \equiv 27 \pmod{31}$
Now $13^5=13^2*13^3 \equiv 14*27\equiv 6 \pmod {31}$

$13^{10}=13^5*13^5 \equiv 6*6 \equiv 5 \pmod {31}$
$13^{15}=13^5*13^{10} \equiv 6*5 = 30 \pmod {31}$

And so order of $13$ in $F_{31}$ is $30$.

  • 3
    This is the idea. It works for all small fields. In a larger $\Bbb{F}_q$ finding the factorization of $q-1$ may prove to be the bottleneck. Admittedly $q$ needs to be ridiculously large for that to be the case. But in crypto applications we may have $q\approx 2^{1000}$ or so. – Jyrki Lahtonen Mar 16 '16 at 19:47
  • 1
    @JyrkiLahtonen I don't think that this is the idea. One is not interested in the powers of the numbers for all prime divisors of p-1 but is interested in the powers of the number for all (p-1)/q, where q is a prime divisor of p-1. If these powers are all different to 1 then p-1 is the order of the number. So in this post the (31-1)/5=6 is not checked, so ist is not proved that the order of 13 is 30. – miracle173 Apr 24 '21 at 08:33
  • @miracle173 It is the idea, but you are right in that the poster forgot to check the sixth power. Most likely I thought that they were first checking the sixth power as opposed to the fifth. In other words, I didn't read carefully enough and simply assumed that the poster is competent. Finding the primes $q$ in your comment is exactly the bottleneck I referred to. This was 5 years ago, so I don't remember how it went. If you are inclined to post a better answer, be my guest! – Jyrki Lahtonen Apr 24 '21 at 09:12