Divisors of a multiple of a prime (or any integer)

Question

I have been studying group theory for some time now, and I have noted that quite a few theorems/proofs considering finite groups rely on results from number theory, a branch of mathematics of which I have very little knowledge.

One question that I would like to have answered concerns the divisors of a product of integers.

Let m and n be integers. Is it true (and can you prove/disprove) that the divisors of mn are either divisors of m, divisors of n or common multiples of m and n?

In particular, is it true that the divisors of mp with p a prime number are either multiples of p or divisors of m?

Notice that $6 \cdot 15 = 90$ has divisor $9$, which is neither divisor of $6$ or $15$, nor common multiple, nor common divisor of both. — Sil, Apr 30 '20 at 10:38
By the fundamental theorem of arithmetic every integer has a unique factorization into prime numbers — Peter Melech, Apr 30 '20 at 10:39
Welcome to Mathematics Stack Exchange. Euclid's lemma says if $p$ is prime and divides $ab$ then $p$ divides $a$ or $b$, but that may not be true if $p$ isn't prime — J. W. Tanner, Apr 30 '20 at 10:41
Your last conjecture is correct: if $p$ is prime then any divisor of $mp$ is either a multiple of $p$ or a divisor of $m$. This follows quickly from unique factorization. — lulu, Apr 30 '20 at 10:55
An Introductory Course in Elementary Number Theory see chapter "2.3 The Fundamental Theorem of Arithmetic" for more information on what the previous comments already mentioned. — Vepir, Apr 30 '20 at 13:15
What is true is that a divisor of a product is a product of divisors. This is equivalent to uniqueness of prime factorizations (and many other equivalent notions) — Bill Dubuque, Apr 30 '20 at 14:28

Divisors of a multiple of a prime (or any integer)

0 Answers0