# Prove that there exists $n_0 \in \mathbb{Z}$ such that $I = \{kn_0 | k \in \mathbb{Z}\}$ given that $I$ satisfies the following properties [duplicate]

Assume that the set $$I \subseteq \mathbb{Z}$$ satisfies the following properties:

(a) There exists $$n \in I$$ such that $$n \not = 0$$

(b) If $$m, n \in I$$, then $$m + n \in I$$

(c) If $$m \in I$$ and $$a \in \mathbb{Z}$$, then $$am \in I$$

Prove that there exists $$n_0 \in \mathbb{Z}$$ such that $$I = \{kn_0 | k \in \mathbb{Z}\}$$

Not too sure if I am correct but I am thinking if $$n_0$$ is actually $$gcd(m, n)$$, then $$I$$ would be the set of all integer combinations of $$m$$ and $$n$$. So essentially this question is to relate the set of all integer linear combinations to the set of all integer multiples of the $$gcd(m, n)$$.

You are right in guessing that whenever $$m, n \in I$$, then $$\gcd(m,n) \in I$$ by the Lemma of Bézout. We will use this in the following proof.
Let $$n_0$$ be the smallest non-zero positive integer in $$I$$ (if no such $$n_0$$ exists, then $$I = \{0\}$$ and $$n_0 = 0$$). Then, by (b), we get that $$\{kn_0 : k \in \mathbb Z\} \subset I$$. For the other inclusion, suppose that $$m \in I$$. If $$m = 0$$ then $$m = 0\cdot n_0$$. If not, assume $$m$$ is positive by replacing $$m$$ by $$-m$$ otherwise. Then $$0<\gcd(m,n_0) \leq n_0$$, and by the Lemma of Bézout, $$\gcd(m,n_0) \in I$$. But now, since $$n_0$$ is minimal, we have that $$\gcd(m, n_0) = n_0$$. In other words, $$m$$ is a multiple of zero. This yields $$I \subset \{kn_0:k \in \mathbb Z\}$$.
For additional context, you are asking for a proof of the fact that $$\mathbb Z$$ is a so-called Principal Ideal Domain (a ring in which any ideal is principal). This is actually true for any euclidean domain.