Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $a=\text{ord}(x,m) : $ $a$ is the minimum value for which $x^a \equiv 1\pmod{m} $.

By inspection it appears that $$\text{ord}(x,p^b) = p^{b-1} \cdot \text{ord}(x,p)$$ where $x,p,b,$ belong to $\mathbb{Z}^+$, $x>1$, $p$ is odd prime.

  1. Is this assertion actually true?
  2. If so, what is a simple proof of it?

I may have a "proof" of my own, but even if it is correct (which I doubt) it seems much too complicated.

share|cite|improve this question
Note it is obviously not true if $x=1$, for example. What is true is that $o(x,p^b)=p^k o(x,p)$ for some $0\leq k < b$. – Thomas Andrews Jun 10 '12 at 18:48

This is not quite true. Take a primitive root $g$ mod $p$. If what you're saying was true, we'd have $ord(g,p^2) = p \cdot ord(g,p) = p(p-1)$, and $g$ would be a primitive root $g$ mod $p^2$. However, this is not true in general. What is true that either $g$ or $g+p$ is a primitive root $g$ mod $p^2$.

The smallest counter-example I've found along this line is $ord(14,29) = 28 = ord(14,29^2)$.

The smallest counter-example in general is $ord(3,11) = 5 = ord(3,11^2)$.

share|cite|improve this answer

This is probably what you seek (T4.4 in William J. LeVeque, Fundamentals of Number Theory)

Suppose prime $\rm\:p\nmid a,\:$ and $\rm\:t =$ order of $\rm\:a\ (mod\ p).\:$ Suppose $\rm\:p^k\:|\:a^t\!-\!1\:$ but $\rm\:p^{k+1}\nmid a^t\! -\! 1.\:$ Then if $\rm\:p\! >\! 2\:$ or $\rm\:k\! >\! 1,\:$ the order of $\rm\:a\ (mod\ p^n)\:$ is $\rm\:t\:$ for $\rm\:n \le k,\:$ and $\rm\:t p^{n-k}$ for $\rm\:n \ge k.\:$

share|cite|improve this answer
Mr Dubuque: I have gotten the book you recommended. I don't understand what the 'p-component' is. What is this double '||' mean? It looks like divide '|' but not quite. I understand p^k | a^t - 1 but not p^k || a^t - 1. – Nick Jun 17 '12 at 18:10
@Nick $\rm\:p^k\,||\,n\:$ means that $\rm\:k\:$ is the highest power of $\rm\:p\:$ dividing $\rm\:n,\:$ i.e. $\rm\:p^k\:|\:n\:$ but $\rm\:p^{k+1}\nmid n.\:$ That was LeVeque's wording. I've changed it to be clearer. – Bill Dubuque Jun 17 '12 at 23:46
No problem. You are clearly someone of exceptional knowledge and I am honored you would take time for someone like me (and the many others I am sure you have helped) – Nick Jun 17 '12 at 23:58

This isn't true as written because if $p=2$ then we have that $ord(3,2^4) = 4$ since $$ 3^4 \equiv 1 \pmod{16} $$ but $2^3 \cdot ord(3,2) = 8$ and $4 \neq 8$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.