# Are there Groups of Strictly Primes

## Motivation

Since Euclid's proof of the infinitude of the primes, the structure and properties of primes has always fascinated mathematicians. This lead to great work in their properties and distribution, such as the Prime Number Theorem. However, much of the study of the structure of primes is done via Analytic Number Theory where much of the work is done using the tools of Analysis.

I am aware of Algebraic Number Theory (it is my goal to have this be my field of expertise so forgive me if perhaps this question comes from my ignorance), but I have not seen groups comprised strictly of primes. Rather, one sees the structure of primes$-$or even number fields in general rather$-$studied through use of Algebraic Geometry, algebraic number fields, Iwasawa Theory, group cohomologies, et cetera.

However, these employ notions far "broader" such as fields (larger special rings that add more structure than just the notion of a group). One doesn't see a group consisting strictly of prime numbers.

## My Attempts

For example, in my research into absolute primes (also known as permutation primes), cyclic primes, and palindromic primes, often I am forced to long calculations with big $O$'s and large scale modular arithmetic (some of the more basic tools of Analytic Number Theory). It would be nice to form a group out of these primes and work with them from a group perspective instead. For those that don't know, absolute primes are primes whose digit permutations are also prime and cyclic primes are those whose cyclic permutations of digits remain prime. Even with this added structure of permutations acting on primes, I have never been able to find an operation on the set of such primes that creates a group$-$nevertheless a ring (in any nontrivial way that focuses on the primes at hand and does not end up reflecting more the structure of $\mathbb{Q}$, $\mathbb{Z}$, or $\mathbb{R}$).

When I took a step back and tried to make a group out of the primes in general, any structure I attempted to create failed either with closure under the operation or under inverses. Consulting the literature has yielded nothing. For example, attempting to create a group of primes under ordinary multiplication lacks closure. Moreover, would would inverses look like? Then adding "primes" of the of the form $\frac{1}{p}$ really doesn't give one anything useful. Switching or even adding ordinary addition creates $\mathbb{Q}$ which of course is nothing new. Obviously, I have found people studying primes using groups, rings, and fields. However, the groups, rings, or fields they employ do not consist "primarily" of primes. Using such objects seems like a great way to approach primes.

As Alexey Sosinsky put it,

"The notion of a "group", viewed only 30 years ago as the epitome of sophistication, is today one of the mathematical concepts most widely used in physics, chemistry, biochemistry, and mathematics itself."

## Question

To avoid an unanswerable question or one that would require years of research, my question is this: are there examples of groups consisting of only primes in the literature?

This could mean a set $G=\{p \mid p \text{ prime}\}$ under some natural operation or a natural operation on a set like this $G=\{\frac{1}{p}\mid p \text{ prime}\}$. Examples of rings or fields out of "mostly" primes would also be great. To clear up the possible lack of rigor of "mostly primes", say the group, ring, or field must have only finitely many elements that are not of prime form. Meaning they are primes, something of the form $\frac{1}{p}$ where $p$ is prime along with the ordinary primes, et cetera. (basically, they "look" like a prime in some form as in the sets $G$ above). If you can produce such an example in the literature, please be explicit in its construction or provide a citation. Thank you.

EDIT: As suggested by the comments and Marie, the group operation should arrive naturally and certainly nontrivially.

• It is a nice question you asked there. I have always been fascinated actually by how the hell did people managed to put a group structure on an elliptic curve. I mean, I know the definitions and some of the links between them, I just have no idea what went in the brains of the people who created these things in the first place... you seem to be trying to be one of these guys with crazy ideas :D +1 Oct 26, 2013 at 0:04
• Take an arbitrary set of prime numbers, and there's guaranteed to be at least one group $G$ with the same cardinality. Define operations on the set of prime numbers so that it becomes a group isomorphic to $G$. You're going to have to be more specific about the operations if you want to rule out trivial cases like these. Oct 26, 2013 at 0:51
• Well, you could always just let $\operatorname{Sym}(\Omega)$ (the symmetric group of countable degree) act on $\{p\in\mathbb{N}:p\text{ prime}\}$, or similarly make it into a ring by a faithful action of $\mathbb{Z}$ or $\mathbb{Q}$. This has nothing to do with "prime-ness," though. Oct 26, 2013 at 0:51
• Also, though this isn't exactly what you asked for, you might find it interesting to look at the product of all $p$-adic integers, as it is a group defined using all the primes. Oct 26, 2013 at 1:01
• @mathematics2x2life Dirichlet's theorem on primes in arithmetic progressions is an important result! :) Oct 26, 2013 at 1:54

As written your question seems to be underdetermined: you want a group law on the set $$\mathcal{P}$$ of prime numbers. As has already been pointed out, no problem: every nonempty set admits a group law, and for a countably infinite set this is especially easy to do (and does not require the Axiom of Choice): simply choose a bijection to your favorite countably infinite group and transport the group law using this bijection. Thus you can make a group with underlying set $$\mathcal{P}$$ which is: cyclic, finitely generated abelian but not cyclic, infinitely generated abelian, nilpotent of any desired nilpotency class, solvable but not nilpotent of any desired solvability class, simple,...If I am not mistaken the set of nonisomorphic group laws on $$\mathcal{P}$$ has the cardinality of the continuum.

However the first paragraph does not take into account that the elements of the countably infinite set are prime numbers in any way: any countably infinite set would work as well. You acknowledge that you want more than this, namely a natural group law on $$\mathcal{P}$$. But what do you mean by this? You don't say, so your question cannot really be answered in its current form.

There is a natural algebraic structure on $$\mathcal{P}$$, but it is not a group law. Rather, the natural structure is that $$\mathcal{P}$$ forms a basis for the commutative monoid $$(\mathbb{Z}^+,\cdot)$$ of positive integers under multiplication. This is precisely a restatement in more algebraic language of the uniqueness of prime factorization, and it has as a consequence that the monoid $$(\mathbb{Z}^+,\cdot)$$ is isomorphic to a free commutative monoid of countably infinite rank, namely $$\bigoplus_{i=1}^{\infty} (\mathbb{N},+)$$. (These considerations are worked out in more detail in the exercise G1) from the first problem set of an under/graduate number theory course I taught at UGA in 2007 and 2009: see here.)

In summary, the whole point of the prime numbers is that they are not closed under their natural operation. Your question is a bit like asking for a way of combining two chemical elements to get a third chemical element. Such operations can be defined, but are in a clear chemical sense unnatural: rather we combine elements to get compounds. That's the point of elements.

There are however natural group laws on quotient sets of the primes, especially if we generalize $$\mathcal{P}$$ to the set of prime ideals in a Dedekind ring $$R$$. Namely, let $$\mathcal{I}(R)$$ be the monoid of all nonzero ideals of $$R$$, and let $$\operatorname{Cl} R$$ be the ideal class group of $$R$$. Then we have natural maps

$$\mathcal{P} \stackrel{\iota}{\rightarrow} \mathcal{I}(R) \stackrel{q}{\rightarrow} \operatorname{Cl} R$$.

Now once again the map $$\iota$$ is an injection which realizes $$\mathcal{P}$$ as a basis for the free commutative monoid $$\mathcal{I}(R)$$: this is the natural algebraic structure on the primes! The map $$q$$ is a surjective monoid homomorphism. Denote by $$\Phi$$ the composite map $$\mathfrak{p} \mapsto [\mathfrak{p}]$$ which sends a prime ideal to its ideal class. Since $$\mathcal{P}$$ is a set of generators for $$\mathcal{I}(R)$$ and $$q$$ is surjective, $$\Phi(\mathcal{P})$$ is certainly a set of generators for $$\operatorname{Cl} R$$ (which is a group, but we get a set of generators even as a monoid -- i.e., without having to take inverses). One can ask whether or not $$\Phi$$ is actually surjective -- i.e., when every ideal class can be represented by a prime ideal. In this paper of mine I call $$R$$ replete when $$\Phi$$ is surjective. When this holds, the class group $$\operatorname{Cl} R$$ is a group formed out of the images of the prime numbers. Moreover, we can define something which is "almost" a group law on $$\mathcal{P}$$: we try to define $$\mathfrak{p}_1 \cdot \mathfrak{p}_2 = \mathfrak{p_3}$$ if $$[\mathfrak{p_1}][\mathfrak{p_2}] = [\mathfrak{p_3}]$$, but it doesn't quite work: $$\mathfrak{p}_3$$ is not well-defined but only its ideal class: i.e., if $$\mathfrak{p}_4$$ is another prime ideal such that there are $$\alpha,\beta \in R \setminus \{0\}$$ with $$\alpha \mathfrak{p}_3 = \beta \mathfrak{p}_4$$, then also $$\mathfrak{p}_1 \cdot \mathfrak{p}_2 = \mathfrak{p}_4$$. So it is a group law on the quotient of $$\mathcal{P}$$ by the equivalence relation of equality up to a principal ideal: it is still natural and important in the study of $$R$$ and $$\operatorname{Cl} R$$. When $$R$$ is not replete, the relationship between $$\mathcal{P}$$ and $$\operatorname{Cl} R$$ is more complicated.

When is $$R$$ replete? When $$R = \mathbb{Z}_K$$ is the ring of integers of a number field, then the repleteness of $$R$$ is a weak form of the Chebotarev Density Theorem (applied to the Hilbert class field of $$K$$), which says moreover that the fibers of the map $$\Phi$$ all have the same density. In my paper I observe that when $$k$$ is a field, $$E_{/k}$$ is an elliptic curve, and $$k[E^{\circ}]$$ is the coordinate ring of the affine curve $$E^{\circ} = E \setminus \{O\}$$, then $$k[E^{\circ}]$$ is (a Dedekind domain, as is well known) and is at least very close to being replete: it satisfies a condition that I called weakly replete which is good enough for the applications. But when is $$k[E^{\circ}]$$ actually replete? I give only partial information about this in Theorem 14 of my paper:

$$\bullet$$ If $$k$$ is alebraically closed, $$k[E^{\circ}]$$ is not replete.
$$\bullet$$ If $$k$$ does not have characteristic $$2$$ and $$k[E^{\circ}]$$ is not replete, then let $$y^2 = P_3(x)$$ be a Weierstrass equation for $$E$$. Then the $$x$$--coordinate map $$E^{\circ}(k) \rightarrow k$$ is surjective.

I don't say so in the paper, but the second part implies that $$k[E^{\circ}]$$ is replete whenever $$k$$ is a Hilbertian field of characteristic different from $$2$$: in particular when $$k$$ is a number field. And now that I look back on it, the characteristic not equal to $$2$$ hypothesis just looks like a bit of laziness on my part.

This discussion was a bit technical, but since you say you are interested in number fields and elliptic curves I thought you might be interested. In fact, giving a precise necessary and sufficient condition on $$E$$ and $$k$$ that makes $$k[E^{\circ}]$$ replete remains an open question. I am pretty sure that no one is working on this..

You can put a group structure on any nonempty set (this is equivalent to the axiom of choice). The question should be: does there exist a "natural" group structure on the set of all primes? In all humility, I don't believe that such a thing exists.

However, perhaps you'll be happy to hear this:

Let $\ell$ be a prime number, and let $\mathbf Z_\ell$ denote the ring of $\ell$-adic integers. It is a compact topological commutative ring. Let $\mathbb P$ denote the set of all prime numbers.

Theorem (Chebotarev - Dirichlet): the set $\mathbb P-\{\ell\}$ is dense in $\mathbf Z_\ell^\times$.

I do believe, however, that the above theorem comes as close as possible to answering your question (this is my nice way of answering "no"). Primes give rise to conjugacy classes in Galois groups via the "Frobenius", so abelian Galois groups are a natural place to look if one would like to put a group structure on the set of all primes. The example above is the most natural one which appears in this context (because $\mathbf Z_\ell^\times$ appears as the Galois group of the infinite cyclotomic extension $\mathbf Q(\mu_{\ell^\infty})/\mathbf Q$).

• This is nevertheless a good answer though you are correct in saying it's not quite what I'm looking for. I do believe given the structure and fundamental nature of primes that there must be some group associated with them. But as Erdös said, "...mankind is not ready for such questions.". Though I admit, the question is ill phrased and difficult to do so any better (it took me many rewrites). I had considered possible Galois representation approaches given my professors research. In any case, though not a direct answer, this is certainly a possible good place to start! Thank you! Oct 26, 2013 at 1:59
• Dear @mathematics2x2life: There are plenty of groups 'associated' with the primes; your question is whether the primes themselves form a group in any meaningful way, and I certainly don't believe that they do. So, it is my belief against yours, but the burden of proof is on you. As for mankind not being ready for your question: you'll have to forgive me if I find this claim a bit presumptuous. Regards, Oct 26, 2013 at 2:05
• I would agree there are many groups associated with the primes. Indeed, many of them are very useful. Though we have a disagreement in terms of 'groups' of primes, that's the beautiful thing about Science and Mathematics, now it's up to me (or others) to show it or you (or others) to show it can't. My only reasoning for believing it is that it would be beautiful if it were true and much of Mathematics tends to be. Erdös said that of the Collatz Conjecture, if that is 'beyond' mankind for the moment then finding the perfect group structure for primes (affirmative or negative is beyond us). Oct 26, 2013 at 2:15
• Dear @PedroTamaroff, thank you. I thought that "equivalent with" was equivalent to "equivalent to". Non-native English speakers unite! Oct 26, 2013 at 2:19
• "You can put a group structure on any set (this is equivalent to the axiom of choice)." There is exactly one set which does not admit a group structure! Dec 23, 2013 at 21:43