$a\mathbb{Z} +b\mathbb{Z} = \text{hcf}(a,b) \mathbb{Z}$

Question

So I have shown that $a\mathbb{Z}+b\mathbb{Z}=k\mathbb{Z} $ for some k since it’s a subgroup of $\mathbb{Z} $. I have shown that $h\mathbb{Z} \subset k\mathbb{Z} $. Here’s $h=\text{hcf}(a,b) \mathbb{Z} $. I just need to show the opposite opposite inclusion using elementary arguments.

Answer 1

Since $k\in a\mathbb Z+b\mathbb Z,$ there are integers $X,Y$ such that $k=aX+bY=\gcd(a,b)\cdot (\frac{a}{\gcd(a,b)}X+\frac{b}{\gcd(a,b)}Y)$.

Hence, we have that $k\in\gcd(a,b)\mathbb Z,$ and thus $k\mathbb Z\subseteq\gcd(a,b)\mathbb Z.$