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After taking Modern Algebra I, I was wondering to find a group property among primes. It doesn't make sense. Does it?

Anyways here is my first instance

For any $p,q\in \mathscr{P}(prime)$ define a multiplication $*$ on $\mathscr{P}$ by $$p*q=\biggl \lfloor \frac{p\cdot q}{2}\biggl\rfloor $$ Clearly, $\mathscr{P}$ is not closed under $*$. But there are primes which behaves well under $*$. For those primes $(\mathscr{P},*)$ is a group with identity element $2$.

So, under what operation prime form a group?

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    $\begingroup$ By the process known as transporting the structure. Let $f$ be a bijection from the set of primes to the set of integers. Define $$p\star q=f^{-1}(f(p)+f(q)).$$ I doubt there is an operation defined by any elementary functions. For if there were, it would be big news. $\endgroup$ Commented Mar 28, 2015 at 8:47
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    $\begingroup$ Here is such operation: $p*q=p$. However, it violates the "single-identity-element" characteristic required in order to define it as a group. $\endgroup$ Commented Mar 28, 2015 at 9:12
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    $\begingroup$ @JyrkiLahtonen Transport of structure is the term I'm familiar with. $\endgroup$ Commented Mar 28, 2015 at 9:41
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    $\begingroup$ A remark on your first structure: Any prime $p \geq 3$ can't really have an inverse $p'$ s.th. $p*p' = 2$. That limits your "subset of primes where $*$ is well behaved" to $\{2\}$ which is the trivial group. I'm not sure what would happen if you only require monoid (no inverse), semigroup (no inverse, no unit) or magma (no inverse, no unit, not associative) structure. But I guess you would still end up with $\{2\}$ as the only sensible subset. $\endgroup$ Commented Mar 28, 2015 at 10:46
  • $\begingroup$ I disagree with your statement that under $*$ the well-behaved primes form a group. $\endgroup$
    – Debug
    Commented Oct 26, 2021 at 17:11

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Primes are naturally* thought of as the free generators of a group (the multiplicative group of invertible rational numbers $\Bbb Q^\times$ modulo torsion, i.e. up to sign), not themselves a group. Indeed the freeness aspect means they're as far away as possible from being closed under the operation of multiplication. More precisely it means they satisfy no multiplicative relations, which in essence is equivalent to the uniqueness of prime factorizations. And of course multiplication is the operation used to define primes in the first place, so any other kind of natural operation to be considered must somehow derive from multiplication one way or another.

I don't see any such group operation; do you? The operation $p*q:=\lfloor\frac{pq}{2}\rfloor$ seems pretty meaningless to me. What significance is it supposed to have? Or is it just randomly picked?

As mentioned in the comment, you can turn any countable set $X$ into a group isomorphic to an equal-size group $G$ by simply invoking a set-theoretic bijection $X\leftrightarrow G$. Then you simply need to relabel the elements of $G$'s multiplication table with the corresponding elements of $X$ to get the multiplication table of $X$ with its newfound group structure. This kind of construction (called transport of structure) is completely blind to any features that $X$ may have beyond just being a barren set, so there's no way we could consider this a "group structure on $X$" if the object $X$ already has special established meaning.

There is another type of structure we could identify primes as - a topological space, or more algebraically, an affine scheme. Algebraic geometry is the right context to understand what this perspective is all about, but it's quite abstract and sophisticated.

*The fundamental theorem of arithmetic (existence and uniqueness of prime factorizations for integers) is taken for granted too often as an obvious fact. It isn't as obvious as you might think; there are number rings which do not have prime factorizations for all elements and it's hard to point to any obvious feature on the surface that tells us when/why they are factorial or not.

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