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I saw this proof in "Proofs from The Book" by Aigner and Ziegler. The proof is starts with supposing that primes, $\mathbb{P}$, is finite. Pick the greates prime number $p$. Consider the Mersenne number $2^p -1$. Now, assume that $q$ is a prime factor of $2^p -1$.

So since $q \mid 2^p-1$ we can say that $2^p \equiv 1 \pmod q$. Then the order of $2$ in the multiplicative group $\mathbb{Z}_q$ is $p$.

My question is, how can we say that the order of $2$ is $p$? Why is that the order of $2$ is exactly $p$ and not a multiple of $p$?

(The rest of the proof: "Order of any element divides the order of the group, so we conclude that $ \begin{align*} p \mid q-1 \implies p<q \end{align*}$").

Bill Dubuque
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Ninja
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  • Note the proper use of \mid and \pmod, as in my edit to this question. $\qquad$ – Michael Hardy May 17 '16 at 21:27
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    If $2^p\equiv 1\pmod q$, that would mean the order of $2$ in the group $\mathbb Z_q$ is a divisor of $p$, not a multiple of $p$. Or in other words, $p$ is a multiple of the order. $\qquad$ – Michael Hardy May 17 '16 at 21:29
  • Instead of assuming there are only finitely many prime numbers and deducing a contradiction, what if we just use the argument above to show that for every prime number $p$, all prime factors of the Mersenne number $2^p - 1$ are bigger than $p$? That can be shown without the assumption that only finitely many primes exist, and it entails that for every prime there is some larger prime, and from that it follows that there are infinitely many primes. $\qquad$ – Michael Hardy May 17 '16 at 21:37
  • Of course it's order must be $p$. I finally see it, my bad. – Ninja May 17 '16 at 21:57
  • See the Corollary in the linked dupe. – Bill Dubuque Jun 10 '24 at 14:22

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The order is the smallest possible exponent. Clearly it has to divide $p$ and is not equal to $1$, and multiples of $p$ are larger than $p$ itself. Therefore the order is $p$.

Nathan Weckwerth
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