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Is there any operation that makes a set of primes i.e. {2,3,5,7.... .} a group with identity 2?

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closed as off-topic by Clarinetist, G. Sassatelli, Silvia Ghinassi, RecklessReckoner, SchrodingersCat Jan 26 at 5:36

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Welcome to math.SE! This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – user37238 Jan 19 at 11:09
    
Ok I will take care to provide all necessary info regarding the question, next time. – user306288 Jan 19 at 11:22
up vote 3 down vote accepted

There are two ways to interpret your question.

  1. If you want the group to be a subgroup of $\mathbb{Z}$ with the usual addition then no. To see this, all we need to see is that $3$ has no inverse.

  2. If you just want the group to have as its underlying set the set of all primes, then yes. Since there are countably infinitely many primes, there is a bijection between the set of all primes and the set of integers, say $f$ where $f(2)=0$. Then we can define multiplication of two primes $a$ and $b$ by $$a\ast b=f^{-1}(f(a)+f(b))$$ it isn't hard to prove this will be a group isomorphic to $(\mathbb{Z},+)$ with the thing represented by the symbol $2$ as the identity.

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Specifically, the "obvious" $f$ is where $f(p)$ is the number of primes less than $p$. – Steve Jessop Jan 19 at 14:05
    
@SteveJessop maybe it isn't so obvious considering that your proposal for $f$ actually gives a bijection between the primes and the natural numbers, not the primes and the integers. – Sean English Jan 19 at 14:20
    
Good point, well made. Sticklers are demanding that groups actually come equipped with inverses now, are they? So it actually has to map to something like half that value if that value's even, and minus half the value if it's odd (round away from zero). Anyway I just meant to indicate that there should be no mystery about this bijection. It's not some high-falutin' existence theorem, we can write one down without too much trouble (and ideally less trouble than I had). – Steve Jessop Jan 19 at 15:17
    
@SteveJessop another common sense way to get the bijection would be to just compose your proposed bijection with the canonical bijection from $\mathbb{N}$ to $\mathbb{Z}$ given by the ordering $(0,1,-1,2,-2,...)$ of $\mathbb{Z}$. – Sean English Jan 19 at 15:21
    
If I consider the set {2,3,5} then applying operation given above on 3 nd 5 then 3*5=7 which is not the element of the set . Is there anyone who can help me ? – user306288 Jan 19 at 15:40

Hint: Take any bijection between the set of prime and Z which takes 2 to 0 and transport the structure with this bijection.

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Also works with any countable group, such as $\mathbb{Q}$, $\mathbb{Q}_0$ or any finite power of a finite or countable group. – Arnaud D. Jan 19 at 11:05

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