Let $m,n \in \Bbb Z$. Show that $\gcd(m,n)$ is a generator for the subgroup $\langle m,n\rangle$.

Question

Let $m,n \in \Bbb Z$. Show that $\gcd(m,n)$ is a generator for the subgroup $\langle m,n\rangle$.

The subgroup $\langle m,n\rangle$ must contain at least the elements $m,m^2,n,n^2, mn,nm,m^{-1}$ and $n^{-1}$. Let $\gcd(m,n)=d$. We know that $d \mid m$ and that $d \mid n$. These two imply that $d \mid m^2$ and that $d \mid n^2$. Also $d$ divides both $mn$ and $nm$.

What can I do to show that $\langle d \rangle =\{d^k \mid k \in \Bbb Z\} = \langle m,n\rangle$? It seems that I could use the fact that $\langle m,n\rangle$ is the $smallest$ subgroup that contains $\{m,n\}$ and since $d$ divides both elements they are some multiples of $d$?

Answer 1

Let $m,n \in \mathbb{Z}$ and $d=\gcd(m,n)$.

$\bullet$ Let $a \in \langle m,n\rangle$ : then there exists $u,v \in \mathbb{Z}$ such that $$a=um+vn$$

Because $d\mid m$ and $d\mid n$, then $d\mid um+vn$, so $d\mid a$, i.e. $a \in d\mathbb{Z}$. This shows that $$\boxed{\langle m,n\rangle \subset d\mathbb{Z}}$$

$\bullet$ Conversely, let $a \in d\mathbb{Z}$ : there exists $k \in \mathbb{Z}$ such that $a=dk$. By Bezout's identity, there exists $u,v \in \mathbb{Z}$ such that $d=um+nv$. Then $a=ukm+vkn$, so $a \in \langle m,n\rangle$. This proves that $$\boxed{d\mathbb{Z} \subset \langle m,n\rangle}$$

These two inclusions put together prove that $d\mathbb{Z}= \langle m,n\rangle$, so $d$ is a generator of $\langle m,n\rangle$.