1
$\begingroup$

I'm learning group theory and I'm trying to consider the "symmetry" of a certain group of natural numbers:

Here's the idea, all natural numbers are comprised of multiples of primes. So a subset would be those comprised of 2 unique primes.

Consider the function:

f(x,y) = xp * yq

where p and q are chosen unique prime numbers and x and y are natural numbers.

can the inputs x and y, with the function itself being thought of as "the operator", (since x*p and y*q is also a condition of the operator *)?

QUESTION 1: My main question is, can the set of (x,y) -> f(x,y) = xp * yq be thought of as a group? If so, what is the inverse function and identity (x,y)?

Does this break down because a function might not necessarily be considered a "binary operation"?

I have a feeling this is much simpler than what I'm making it out to be, my idea is to see what kind of symmetry or other symmetry-like structure(s) that prime numbers have on the generation of the natural numbers.

QUESTION 2: If this isn't a group in group theory, what kind of abstract algebraic structures ought I be looking at?

QUESTION 3: As a side note, can a finite set of 1 - n | (1, 2, 3, n) have a prime number set of "generators"? Are these primes technically group generators, or are they something else entirely different, perhaps named a "set generator"?

$\endgroup$
8
  • $\begingroup$ For Q1: are you asking if the set $\mathbb N$ with the operation $(x,y) \mapsto f(x,y)$ satisfies the group axioms? For Q3: I'm afraid that I don't understand your question at all. Perhaps you can clarify a few things for me -- what do you mean by "a finite set of $1-n$|(1,2,3,n)"? What do you mean by "generators" in this context? $\endgroup$ Aug 22, 2014 at 19:48
  • $\begingroup$ For Q1 I'm only asking if the subset of N where say p,q = 2,5 and f(1,3) = 1*(2)*3*(5), so does this f satisfy the group axioms? $\endgroup$ Aug 22, 2014 at 20:01
  • $\begingroup$ As for Q3, I know that in group theory a generator is basically data that can have operations on it that generate the group set, so in a manner of speaking, prime numbers with multiplication "generate" the set of all natural numbers. Forgive me for being so loose with definitions, I'm trying to wrap my head around some concrete groups that I find interesting. So I was wondering that if Natural numbers have any type group making function, would prime numbers be the actual group generators of that set? $\endgroup$ Aug 22, 2014 at 20:03
  • $\begingroup$ I'm still a little confused on Q1: to define a group, you must specify an underlying set $S$, and define a function from $S \times S \to S$ (binary operation). If you take $\mathbb N$ as the set, the map $(n,m) \mapsto n \cdot p \cdot m \cdot q$ is indeed a binary op. But the group axioms aren't satisfied: there is no identity element. Note that $f(n,m)>n$ for all $m \in \mathbb N$, so it is impossible that there exists an element $e \in \mathbb N$ so that $f(n,e)=n$. To addres Q2, the natural numbers with this operation do form a semi-group. $\endgroup$ Aug 22, 2014 at 20:22
  • $\begingroup$ @MorganO Ok, I was defining the set to be "all the answers" to the function f(x,y) and making the binary operation to be the same function, I think that's partly where the group breaks down, and yes I was not finding an identity element and hoped there would be one. Thank you for specifying that it forms a semi-group, I'll take a look at what that may imply for the function. $\endgroup$ Aug 22, 2014 at 20:31

1 Answer 1

1
$\begingroup$

So thinking of groups as having something to do with numbers isn't really a great idea. Yes, the integers are an example of a group, but they're a very degenerate / misleading one in a number of ways.

The definition of a group abstracts the idea of a group of symmetries: a group element is a symmetry (or "automorphism") of some other object. A symmetry is something you can do to an object without breaking it. The group axioms basically say that:

  1. (Associativity) If you do three things in a row to an object there are two ways to think of that -- $(ab)c$ or $a(bc)$ -- and it doesn't matter which way you think about it.
  2. (Identity) One thing you can do to an object without breaking it is just to leave it alone.
  3. (Inverses) If you can do something to an object without breaking it, that also means you can undo that thing.

Let's talk about why / how the integers form a group: consider the directed graph

$$ \cdots \rightarrow \circ \rightarrow \circ \rightarrow \circ \rightarrow \circ \rightarrow \cdots $$

and note that its symmetries consist of translating each vertex the same number of slots to the left or right. That is the sense in which the integers form a group: they represent translational symmetries. Trying to do any kind of number theory that doesn't have a natural interpretation in terms of translations is going to require you to work with something richer than the additive group of integers; in particular, you're probably going to want to use the ring structure.


Now, if you want to consider the set of numbers that are divisible by, say, 5 and 7 but not by any other prime, we could think about what kind of properties it has. Obviously you can't divide, because for instance 5/7 isn't an integer. You can't add the elements together because 5 + 7 = 12 is divisible by 2 and 3. You can, however, multiply two such numbers together and get another one.

So the set $\{5^a 7^b \; | \; a, b \geq 0 \}$ together with the integer multiplication operation is what's called a monoid -- a set with an associative binary operation and an identity for that operation. I don't know that that tells you anything about that set that you didn't already know, though.


You've asked what the primes are algebraically. They have to do with multiplication, so they have to do with the ring of integers, not just the group of integers. In particular, they're generators of the "prime ideals" of $\mathbb{Z}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.