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I'm on the preliminary basics of Primitive Roots. My textbook states this theorem (with no explanation) that is apparently essential to the concept of Primitive Roots.

Suppose $p$ is a prime and $\operatorname{ord}_p(a) = d$. Then for each natural number $i$ with $\gcd(i,d)=1$, $\operatorname{ord}_p(a^i) = d$

Could I receive some help please as to what information this theorem is trying to tell me and the purpose of this information? I think it may be saying that we can raise the base of the order (the "$a$" in the phrase $\operatorname{ord}_p(a) = d$) to a power $i$ without affecting the validity of the statement. But I honestly have no idea for how to verify if my interpretation is correct.

Bernard
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Bobby B
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  • Are you asking for understanding what the theorem says, or how to prove it? As you seem to already know, it is key to understanding primitive roots to understand how taking powers of an element $a$ of order $d$ affects the order of the resulting power $a^i$. – hardmath Mar 29 '21 at 21:56
  • Note $\operatorname{ord}_p(a) = d$ means that $a^k \equiv 1 \pmod{p} \iff d \mid k$. – John Omielan Mar 29 '21 at 21:58
  • I don't care about proofs. I only wish to know the meaning and purpose of a theorem. So are you saying that applying the exponent $i$ onto $a$ has no affect on the order of the modulo? That is, $a^d \equiv 1 \pmod{p}$ and $(a^i)^d \equiv 1 \pmod{p}$ are both valid? – Bobby B Mar 29 '21 at 22:03

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A generalisation of this theorem might help understand better.

Remember the order of $a\bmod p$ is the order of the multiplicative group generated by the congruence class $a\bmod p$.

Now , one may wonder what is the order of a power of a, and the (easy) result is the following: $$\operatorname{ord}_p a^k=\frac{\operatorname{ord}_p a}{\gcd\bigl(k,\operatorname{ord}_p a\bigr)}$$ Therefore, if $k$ and $\operatorname{ord}_pa$ are coprime, $a\bmod p$ and $a^k\bmod p$ have the same order.

Bernard
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  • You can also say that whenever $k$ and ord$_p a$ are coprime, $a^k$ is a generator of the subgroup generated by $a$ – user289143 Mar 29 '21 at 22:14
  • This is confusing. Before I ask about your main point, could you expand on the phrase, "the order of $a\bmod p$ is the order of the multiplicative group generated by the congruence class $a\bmod p$" .If I remember correctly, a congruence class is the collection of possible numbers for $x$ satisfy $x \equiv a \pmod{n}$. So what do you mean by the "order of the multiplicative group"? – Bobby B Mar 29 '21 at 22:19
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    The set of congruences classes $\mathbf Z/p\mathbf Z$ is a field when $p$ is prime, and as in all fields, the non-zero elements $(\mathbf Z/p\mathbf Z)^*$ are a multiplicative group and it can be shown that this multiplicative group is cyclic. Is that clearer? – Bernard Mar 29 '21 at 22:32
  • That's even worse. I understand only the most basic and elementary concepts about Number Theorey. I think your knowledge of this subject is way too far above mine for us to properly communicate. – Bobby B Mar 29 '21 at 22:46
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    In fact, this is a particular case of the characterization of generators of a cyclic group (https://math.stackexchange.com/q/2155137). – Jean Marie Mar 29 '21 at 23:12
  • @BobbyB: Didn't you ever have a course on the basics of group theory? – Bernard Mar 30 '21 at 08:28
  • No, I'm studying Applied Math and group theory isn't a requirement of my degree. My current Number Theory course is skewed towards understanding the practical application of cryptography, but I struggle a lot with knowing how to form the pure math proofs it demands. – Bobby B Mar 31 '21 at 19:22
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    You should try to learn some basics in group theory, in my opinion. Group theory is fundamental in many domains of mathematics, and even in cryptography, there are parts such as cryptography via elliptic curves (which have a group structure, and are from the domain of algebraic geometry). – Bernard Mar 31 '21 at 19:30