How do I correctly express of even simplify this operation (number theory)?

The first important observation is:

what does this do to the set of natural numbers?

Let p=2, q=3. Then basically, given any number X not divisible by 2 or 3, you have

f[pq] : 2^j 3^k X → 2^k 3^j X

It swaps them. You see, you're writing messy prime factorizations, but who CARES about the primes? The big old product notation is cumbersome. Let's try:

n = 2^(k_2) 3^(k_3) ... p^(k_p) ... q^(k_q) ...

Instead let's let the coordinate {n} =

(k_2 , k_3 , .. , k_p , ... , k_q , .... )

That's an infinite list of non-negative integer coordinates. We allow 0s, that way the coordinates always match up with the same primes. For a few examples, try:

{15} = ( 0 , 1 , 1 , 0 , 0 , 0 .... )

{24} = ( 3 , 1 , 0 , 0 , 0 , 0 .... )

So what does f[pq] become? Simple. You swap the coordinates for p and q! It's that easy.

It doesn't really matter if the coordinates are for "2,3,5,7,..." or for "1,2,3,4,..." - they're just coordinates, one-by-one. So don't worry about which way you're thinking of it.

So f[pq] becomes the transposition:

http://en.wikipedia.org/wiki/Transposition_%28math...

of coordinates p and q, in the infinite symmetric group:

http://en.wikipedia.org/wiki/Symmetric_group

So the set of all possible f[pq], taken with functional composition, is the exact same (speaking algebraically), as the infinite symmetric group (on the set of natural numbers, or primes, it's the same in terms of laying out the coordinates). That's the set of all possible permutations on the integer coordinates.

So yes, this has been studied, quite a bit. You've constructed it from a very different context involving exponent-swapping, but it is essentially permutations that you're looking at. The context you constructed it in, though, that's interesting too.

This infinite symmetric group you have is acting on the set of natural numbers, by swapping its exponents, in what's called a group action:

http://en.wikipedia.org/wiki/Group_action

However, this group action doesn't lead to too much different, because it acts on the group the way it normally acts. Nothing special happens when you apply it to the exponents of a prime factorization, because you still wind up swapping numbers around the same way the permutation group swaps itself around.