What would be the simplest way to prove that every element in $F_{128}$ is a primitive root except zero $(0)$ and the identity.

Well, clearly 0 can not be a primitive root, and i also know that $F_128$ has got 126 primitive roots since, $p - 1$, being fairly large when i choose $2^7$ immediately yields this result. now, do i have to use brute force by checking all the 126 elements and showing that they are primitive roots or is there a better way. If there is, can someone show it to me, because i could not figure it out - let alone find it. Thanks.


1 Answer 1


Note that $\Bbb F_{128}^\times$ is a group with 127 elements. Since 127 is prime, every one of the 126 non-trivial elements generates the whole group.

  • $\begingroup$ ok, so $0$ is trivial and that easily puts it out of the way, but what guarantees, that the other elements generate the whole group. I dont understand how the total number of elements in the group being a prime, guarantees that the other $p - 1$ elements generate the group. or does that only apply to this case? $\endgroup$
    – guthik
    May 10, 2015 at 21:22
  • $\begingroup$ @user1825142 The order of an element in a group must divide the order of the group, by Lagrange's theorem. Since the order is prime, all non-identity elements in the multiplicative group of nonzero elements (ie, elements other than $1$) have order $127$. That is to say, they are primitive roots. $\endgroup$
    – Rolf Hoyer
    May 10, 2015 at 21:28
  • $\begingroup$ Thanks a lot, that makes it much clearer. :) $\endgroup$
    – guthik
    May 10, 2015 at 21:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .