I have to find all the finite stable subsets of $(\mathbb{Z}, \cdot)$. I know that a subset $M$ is a stable subset of $(\mathbb{Z}, \cdot)$ iff $\forall$ $x, y \in M$ we have $x \cdot y \in M$.
Intuitively, I though that the only integers that might be "capable of stability" are $A = \{-1,0, 1\}$. So then I quickly checked all of $A$'s subsets and reached the conclusion that the finite stable subsets of $(\mathbb{Z}, \cdot)$ are the following:
$$\bigg\{ \{ 0 \}, \{ 1 \}, \{ 0, 1 \}, \{ -1, 1 \}, \{ -1, 0, 1 \} \bigg\}$$
Is my solution correct? Did I miss something? Or did I get the problem's statement completely wrong?
This is correct. It shouldn't be hard to check that these sets all work, so all that should be left is to justify your intuition that nothing besides $-1$, $0$, and $1$ are capable of stability.
To do this rigorously, we want to start with a subset $M$ that is stable, and show it can't contain some $n$ not in $\{-1,0,1\}$. Say it does contain some $n$. The "capable of stability" condition is essentially the property that $M$ must have finitely many elements, so you want to find infinitely many elements in $M$ that you can generate from a single $n\not\in\{-1,0,1\}$.
All you know at the beginning is that $n\in M$, and you want to find more elements. Since $M$ is closed under multiplication, you can start multiplying elements of $M$ together; the only product you can make from the start is $n\cdot n=n^2$, but now $\{n,n^2\}\subset M$, and you can find more numbers that are supposed to be in $M$. In particular, $n\cdot n^2=n^3\in M$, ...