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The other posts linked on the closing of this question may seem to indirectly (or somewhere within the answer) provide an explanation. However, they don't answer this specific question at my level. Please note that I am asking 1. the definition of period, and 2. how that divides $p - 1$ as a direct consequence of Fermat's little theorem.

I found this in here, and as it is given, the period of the exponential function $f(x) \equiv a^x \pmod p$ divides $p -1$, presumably with $p$ being co-prime to $a \in \mathbb Z$ (?) by Fermat's little theorem.

I see that for an integer $a>0$ and for a prime $p$ the sequence $1,a,a^2, a^3, \dots \mod p$ is periodic, so for instance:

For $a=2$ and $p=7,$ the sequence would be $1,2,4,1,2,4,\dots$, so I presume the period is $3,$ which indeed divides $p - 1 = 6$. But I can't find a formal definition for period (tangentially similar to order, but not the same).

Also, by FLT, $a^p \equiv a \pmod p$ or equivalently (if $a$ and $p$ are coprime, $a^{p-1}\equiv1 \pmod p.$

But how do I prove that from the above it follows that period $ \mid\ {p-1}$?

JAP
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  • Lagrange's Theorem. – Arturo Magidin Jun 12 '23 at 04:49
  • Re: your edit. The definition of the order (least period) is aready given in the 2nd dupe, as is the application using Fermat and Euler's theorem. Please read more carefully. Everything that you need (standard arguments) are already in the linked dupes. – Bill Dubuque Jun 12 '23 at 13:59
  • Nothing is being "promoted" by duplication links. There are various answers given in the linked dupes, at various levels of generality, including all of the standard arguments. Again, if you don't understand thee standard arguments then please ask for elaboration in comments. This will help to improve the answers. – Bill Dubuque Jun 12 '23 at 14:05
  • Your result is a special case of Corollary$_1$ in the 2nd linked dupe, viz. by lil Fermat $,a^{\ell}\equiv 1$ for $\ell = p-1,$ so $\ell = p-1,$ is divisible by the order of $,a,,$ by the converse direction of the Corollary. The proofs are very simple (one-line of elementary number theory). What is not clear about that? – Bill Dubuque Jun 12 '23 at 14:21

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In the language of group theory, your question concerns the group $(\mathbb{Z}/p\mathbb{Z})^{\times}$, the group of units modulo a prime $p$. The "period" of $a^x$ is then simply the order of $a\pmod p$ in the group $(\mathbb{Z}/p\mathbb{Z})^{\times}$.

Now Lagrange's Theorem tells us that the order of an element divides the order of the group. Since the order of $(\mathbb{Z}/p\mathbb{Z})^{\times}$ is $p-1$, we have the desired result.

july
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  • Thank you, but the answer is vague precisely where I was hoping for help. For instance, $ord_p(a)$ is the minimum positive integer such that $a^{ord_p(a)} ≡ 1 \pmod p$. The general abstract algebra concept is not wrong. – JAP Jun 12 '23 at 05:22
  • Therefore the order of $a$ would not the same as the period. – JAP Jun 12 '23 at 11:32
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    Some authors make it precise by saying least period. Others define a period as any number after which the sequence repeats and the period as the least period. The order is the least period. – kodlu Jun 12 '23 at 11:37
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jun 12 '23 at 13:10
  • @JAP You can find answers to your questions in the linked threads. If you have questions about any of the answers there then you should post them in comments there so that the answers can be improved. – Bill Dubuque Jun 12 '23 at 13:55