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This is contained in the proof of Theorem 2.8.5 of Herstein's Abstract Algebra.

Suppose that $p$ and $q$ are primes with $p>q$ and $q \nmid p-1$. By Fermat's Theorem, $i^{p-1}\equiv 1(p)$. Furthermore, suppose that $0<i<p$ and $i^q \equiv 1(p)$.

Then it is claimed that $i \equiv 1(p)$.

I'm having trouble seeing how this is derived. What I have tried, $$i^{p-1}\equiv 1(p) \implies p \mid i^{p-1}-1,$$ $$i^q \equiv 1(p) \implies p \mid i^q-1.$$

Now, $p>q \implies p-1 \geq q \implies i^{p-1}-1 \geq i^q-1$. Thus, $$p \mid i^{p-1}-1-(i^q-1)=i^{p-1}-i^q= i^q(i^{p-1-q}-1).$$

Since $p \nmid i^q$, then we must have that $$i^{p-1-q} \equiv 1(p).$$ But I don't see how this implies that $i \equiv 1(p)$. Thanks in advance.

Arturo Magidin
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E.Y.D.
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    Duplicate $\ $ Hint: ${\rm ord}_p(i)\mid q,p!-!1 \Rightarrow {\rm ord}_p(i) |\mid (q,p!-!1) = 1.\ $ Will shortly be closed as a dupe. – Bill Dubuque Oct 19 '24 at 01:05
  • Your title left out the necessary condition that $i^q\equiv 1 \pmod p$. Obviously without it, the statement in the title is false. – fleablood Oct 19 '24 at 01:16
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    By the dupe, the order of $,i,$ divides coprimes $,q,p-1,$ so the order must be $\color{#c00}{\bf 1},,$ i.e. $,i^{\color{#c00}{\bf 1}}\equiv 1\pmod{!p}$ – Bill Dubuque Oct 19 '24 at 01:16
  • This order inference is ubiquitous, e.g. see Remark here and here and here. – Bill Dubuque Oct 19 '24 at 01:20
  • Good. Good. Good. – suckling pig Oct 19 '24 at 03:12
  • @BillDubuque Thank you for your hint and succesive comments, as well for referencing a similar question. I'm unfamiliar with the notation you use in the comment, although I can somewhat understand based on the original post and the linked information I find myself lost still. I'm trying to derive the result in the context of what Herstein has shown of modular arithmetic, not much beyond the basic definition really. Adding to what I wrote yesterday, I thought of writing $$p \mid i^{p-1-q}-1=(i-1)(i^{p-1-q-1}+i^{p-1-q-2}+...+1).$$ – E.Y.D. Oct 19 '24 at 20:58
  • With that it would come down to showing that $p \nmid (i^{p-1-q-1}+...+1)$, which presents similar difficulty to me. The author says "since $q \nmid p-1$, we conclude that $i \equiv 1(p)$." So presumably the fact that $q$ doesn't divide $p-1$ must be used to arrive at the conclusion, although I'm unsure as how to use it. I would appreciate help arriving at the result with this information only, thanks in advance. – E.Y.D. Oct 19 '24 at 21:03
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    Herstein is using the Order Theorem that I linked in the dupe. It is his Problem $2.4.31!:$ $,a^s = e,\Rightarrow, o(a)\mid s.,$ He already implicitly used it in this proof to infer $,o(a)=p\mid i^q-1.,$ So, as I said above, in the multiplicative group of $,\Bbb Z_p,,$ since $,i^q = 1 = i^{p-1},$ we infer $o(i)$ divides coprimes $,q,,p-1,$ so $,o(i)=\color{#c00}1,,$ i.e. $,i^{\color{#c00}1} = 1$ in $\Bbb Z_p,,$ i.e. $,i\equiv 1\pmod{p}.,$ He explicitly invoked that Problem earlier in his proof of Theorem 2.6.4 (Cauchy's Theorem), so he assumes it is known at this point. – Bill Dubuque Oct 19 '24 at 22:09
  • @BillDubuque, thanks again for your additional comment. With your last response I understand what I was initially missing and the proof is completely clear to me now. I knew it was something pretty evident but for some reason i was not seeing it at that point. Regardless, thank you for your answers and for providing links to additional information! – E.Y.D. Oct 22 '24 at 19:48

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