About a basic property of prime numbers.

Question

So I never really understood anything about prime numbers, other than the definition, and so I'm currently having problems with the following proposition:

let $p$ be a prime number and $n,k\in\mathbb{N}$. If $k|p^n$, then $p|k$.

I have no idea how to prove this, using just the most basic properties of prime numbers and divisibility. I'm thinking that maybe it's because of the fundamental theorem of arithmetic, because then one would have $p^n=kq$, for some $q\in\mathbb{Z}$, and using the prime factorization of $k$ and $q$ maybe one could get to the result, but I am not sure how to follow and formalize this.

Also, any recommendation/reference on the basic results on number theory needed in order to solve problems in elementary group theory would be appreciated, as this problem arrived when trying to prove a theorem on group theory.

Thanks in advance :))

Answer 1

Actually, the statement is false if $k=1$. You need to assume $k\geq 2$. In that case, by the fundamental theorem of arithmetics there is some prime number $q$ which divides $k$. Suppose $q\ne p$. Then $q|p^n$, and since $q$ is prime this implies $q|p$. But a prime number can't divide a different prime number, a contradiction. Hence $q=p$, and thus $p|k$.

Answer 2

You can write $p^n = q\cdot k$ for some $q \in \mathbb{N}$.

But then $k \in \{p,p^2,...,p^n\} $ because no other prime factor is present on the left side.

Equivalently to $p|k$, you can prove :

$$\exists t \in \mathbb{N} \setminus k = t\cdot p $$

but you know you can express $k$ as $p^r$ with $1\leq r \leq n $ so $t$ exists and it's equal to $p^{r-1}$.

I excluded $k=1$ because otherwise the statement is false, you can see that because $t$ would be equal to $\frac{1}{p}\notin \mathbb{N}$.