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A while ago I asked the question "Is there a group-theoretic proof that multiplicative groups of integers modulo a prime are cyclic?". Instead of an answer I got a few comments from competent users (including Keith Conrad, whose paper I mentioned in the question) telling me that the answer was, in short, 'no'.

Lately the question has been popping into my mind anew, and thus I thought about adding a bounty to the post, but I have a few concerns:

  1. Is it OK to ask a question that may have no answer? I'm fine with a few of my reputation points going to waste if there is indeed no proof of the kind. I worry instead that making a question that may have no answer goes against what the site is intended to be.

  2. Would it be discourteous (which I genuinely do not intend to be) to the users who left the comments? The comments are, I should mention, quite reasonable. As an example Conrad's comment reads

...the multiplicative group of integers modulo m is generally not cyclic, so proving it is when m is a prime number (or an odd prime power) will need to use something that distinguishes those choices of m from others, and a very basic one is that the multiplicative group of integers modulo a prime is a field, which is not a purely group-theoretic issue.

This is sensible, and the fact that whenever m is prime we get a field is a natural direction on which to build an idea as to why these groups are cyclic. I yet remain with a small hope that a group-theoretic proof could exist. Can I, with all being said, still add the bounty?

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  • $\begingroup$ FYI you won’t get the rep back if nobody answers. $\endgroup$
    – Ekadh Singh - Reinstate Monica
    Oct 19, 2021 at 16:33
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    $\begingroup$ @EkadhSingh-ReinstateMonica as I mentioned I'm OK with that (: $\endgroup$
    – Sam
    Oct 19, 2021 at 16:34
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    $\begingroup$ There is nothing discourteous about offering a bounty to encourage effort toward your goal. The Comments provided may help to focus thoughts about circumventing the apparent obstacle, or they may prove justified. Either way I'd hope no one involved takes offense. $\endgroup$
    – hardmath
    Oct 19, 2021 at 21:26
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    $\begingroup$ "Is it OK to ask a question that may have no answer?" Absolutely imo, afterall proving a question doesn't have an answer requires maths after all! $\endgroup$
    – Adil Mohammed
    Oct 20, 2021 at 7:34
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    $\begingroup$ Godel proved you can't always know if a question has an answer so... $\endgroup$
    – Jerry Guern
    Oct 20, 2021 at 10:19
  • $\begingroup$ Thanks for the comments. I think I'll add the bounty then. $\endgroup$
    – Sam
    Oct 22, 2021 at 12:38
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    $\begingroup$ Many times a problem remained unsolved until a new concept was introduced. Usually the person asking finds an answer for themself if no one else answers. $\endgroup$
    – Тyma Gaidash
    Oct 23, 2021 at 2:38
  • $\begingroup$ Have you asked someone how it might be possible to rigorously prove that group-theoretic axioms alone cannot prove this result? That is, converting the following comment of KCd "More seriously, the unit group of $\mathbb Z/(m)$ is generally not cyclic, so proving it is when $m$ is a prime number (or an odd prime power) will need to use something that distinguishes those choices of $m$ from others, and a very basic one is that $\mathbb Z/(p)$ is a field, which is not a purely group-theoretic issue" into something rigorous would be a fitting answer to the question as well. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Oct 23, 2021 at 15:01
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    $\begingroup$ I had nothing about axioms of logic in mind in my comment to the OP's post. All I meant is that there is no known proof and I explained why it makes sense that a proof of the kind the OP was seeking is unlikely. Nothing more, nothing less. There are proofs of the cyclicity based on group-theoretic properties (like $(\mathbf Z/p\mathbf Z)^\times$ having at most $n$ elements of order dividing $n$ for each $n$), but verifying such a property for $(\mathbf Z/p\mathbf Z)^\times$ so far always relies at some point on something more, like the field structure of $\mathbf Z/p\mathbf Z$. $\endgroup$
    – KCd
    Oct 24, 2021 at 10:38

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