Proving 'If (2^n)-1 is Prime, Then n is Prime' Using Groups

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The discussion centers on the proof of the statement "If (2^n)-1 is prime, then n is prime" using group theory. The user attempts to establish a connection between the cyclic nature of the group {1, 2, ..., p-1} and the order of its elements, concluding that n must be a multiple of p-1. However, a critical correction is made regarding the order of elements, emphasizing that while the group is cyclic, not every element has order p-1, as demonstrated with the example of p=7. This highlights the need for a deeper understanding of group properties in relation to prime numbers.

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Bleys
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I was trying to prove the statement "If (2^n)-1 is prime then n is prime". I've already seen the proof using factorisation of the difference of integers and getting a contradiction, but I was trying to use groups instead. I was wondering if it's possible, since I keep getting stuck.
So far I've got:
If (2^n)-1=p where p is prime then 2^{n}\equiv1 mod(p). The group {1,2,...,p-1} is cyclic and every element has order p-1. So n = k(p-1) for some positive integer k. But doesn't this mean n is not prime? which is wrong I know. Could someone point me in the right direction or a theorem that might be useful?
 
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Interesting problem... I'm not yet sure of a proof, but your statement

Bleys said:
The group {1,2,...,p-1} is cyclic and every element has order p-1.

is wrong. All you know is that every element has an order that divides p-1. (E.g. for p=7 the element 2 has order 3).
 
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