Is there a way to produce cartesian products by prime factorization of a number?

Question

Not sure if I am able to put together the question there correctly, but I would like to explain with an example.

Suppose I have two Sets containing 10^3 elements. The number of elements consisting in the Cartesian products of these two set will be (10^3)^2.

Now, Prime Factorization of 10^3 = (2^3)*(5^3)

Can I form the same set containing (10^3)^2 elements by breaking and computing Cartesian product of 2^3 and 5^3 individually ?

Answer 1

The short answer is "no". Let's look at a smaller example. Let $A=\{1,2,3,4,5,6\}$ Then $A\times A$ has $36$ elements. Can we form this as a product of a $4$-element set and a $9$-element set? Say $$A\times A=B\times C\tag1$$ where $|B|=4,\ |C|=9$. There are only $4$ different first components of the elements on the right-hands, but $6$ different ones on the left-hand side, so $(1)$ cannot hold.

Answer 2

Not directly but isomorphically you can.

Let $\mathbb Z_{k} = \{0,.......,k-1\}$ and

Let $\phi_{m,k}: \mathbb Z_{m}\times \mathbb Z_k\to \mathbb Z_{mk}$ via $\phi(a,b) = ak + b$ be a bijection and we can say

$\mathbb Z_{m}\times \mathbb Z_k \approx \mathbb Z_{mk}$ because the bijection exists.

Then $\mathbb Z_{10^3}\times \mathbb Z_{10^3} \approx \mathbb Z_{2^6}\times \mathbb Z_{5^6}$ via

$\phi^{-1}_{2^6,5^6}\circ\phi_{10^3,10^3}:\mathbb Z_{10^3}\times \mathbb Z_{10^3} \to \mathbb Z_{2^6}\times \mathbb Z_{5^6}$

.....

Or perhaps in other words:

Define the relationship $R$ between $\mathbb Z_{10^3}\times \mathbb Z_{10^3}$ and $\mathbb Z_{2^6}\times \mathbb Z_{5^6}$ as $(a,b) R (c,d)$ if and only if $10^3a + b = 2^6c + d$. This relationship is a bijection.

So if we are willing to accept, say, that $(123,45)_{10^3,10^3} = 123*10^3 + 45 = 123045 = 5^6*7+ 13670= (7,13670)_{2^6, 5^6}$ then we can say $\mathbb Z_{10^3}\times \mathbb Z_{10^3} \approx \mathbb Z_{2^6}\times \mathbb Z_{5^6}$