Show that this Set contains all multiples of a whole number

Question

Let S be a nonempty subset of $\mathbb{Z}$. Suppose S satisfies the following constraints

If $x,y \in S$ then $x+y \in S $

If $x \in S$ and $y \in \mathbb{Z}$ then $xy \in S$

Show that S is the set of all integer multiples of m for some $m \in \mathbb{N}\cup{\{0\}} $

So far I can think of a few things that might be useful, this might be a proof by cases where we split it up into the case where $ S\cap\mathbb{N} =\emptyset$ and $ S\cap\mathbb{N} \neq\emptyset$ which is when $m=0,m\neq 0$ respectively. But how do I use the constraints to come to that conclusion? I don't need a full solution but a hint would be nice. Thanks!

Answer 1

Let $m$ be the minimal positive in $S$. Then by the second rule $km\in S$ for each $k\in\Bbb Z$.

Answer 2

Hint:

First do the trivial case $S=\{0\}$. Then move onto the non-trivial case.

Let $T \subseteq S$, where $T=\{a \in S \, | \, a > 0\}$. Now use well-ordering principle to claim that $T$ has a smallest element. Call it $m$, use that to show that $S=m\mathbb{Z}$.