# divisibility of a product of (unique) primes by another product of (unique) primes

Assume that A is a set of prime numbers (without duplicates). Let P_A be the product of all elements of A. Assume that B is another set of prime numbers (again without duplicates) and let P_B be their corresponding product. If P_A divides P_B, that is, the reminder of P_B/P_A is zero, could we assume that set A is a subset of set B? That is, all primes in A are also present in B? For example, A = {2, 3, 5}, B = {2, 3, 5, 7}, P_A = 30, and P_B = 210.

I suspect that this might not hold in general, but I haven't studied properties of prime factors. For example, if A = {a, b, c, d} and B = {a, c, e, f, g}, it might be (?) that the division (efg)/(bd) has zero reminder although that A is not a subset of B. But does this happen for any possible combinations of primes up to a given prime number N, and is there a bound for such an N?

Thank you very much for your time.

For example, if A = {a, b, c, d} and B = {a, c, e, f, g}, it might be (?) that the division (efg)/(bd) has zero reminder although that A is not a subset of B. But does this happen for any possible combinations of primes up to a given prime number N, and is there a bound for such an N?

If $e,f,g,b,d$ are prime, then $bd$ will not divide $efg$.

I recommend you read up on Divisor and Fundamental theorem of arithmetic.

What you're asking basically boils down to the following question:

Let there be a set of primes $A$, and a product of primes $P_{A}$. Is there a prime number $q$ that divides $P_A$, where $q$ is not in $A$?

The answer is no, because of the fundamental theorem of arithmetic.

For $P_A$ to divide $P_B$, all factors of $P_A$ must be factors of $P_B$. So basically, for $P_A$ to divide $P_B$, each element in $A$ must be in $B$. Thus $A\subset B$.