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I am supposed to find a generator for the group $m\mathbb{Z}+n\mathbb{Z}=\{a+b|a\in m\mathbb{Z}, b \in n \mathbb{Z}\}$, with addition.

My intuition: Since elements of $m\mathbb{Z}+n\mathbb{Z}$ will be numbers of the form $mc+nd$ for $c,d\in\mathbb{Z}$, they are linear combinations of $m$ and $n$. Therefore, the smallest linear combination is $\gcd(m,n)$ which should be the generator.

However, how would I go about proving this? Intuition is not nearly enough, I think.

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    Are you also asking about how to prove that the gcd is the least such element, or do you already know that proof? – 2017-01-30
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    Yes, I have proven that. – 2017-01-30
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    Then you need only prove containment in both directions - see my answer. – 2017-01-30

3 Answers 3

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Hint:

You have to prove first that $\gcd(m,n)\in m\mathbf Z+n\mathbf Z$, second that any element of $m\mathbf Z+n\mathbf Z$. is a multiple of $d=\gcd(m,n)$.

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You only have to prove that $\gcd(m,n)$ is an element of $n\mathbb Z+ m\mathbb Z$ and that every element in $n\mathbb Z + m\mathbb Z$ is a multiple of $\gcd(m,n)$.

The first observation is known as Bezout's identity. The second observation is clear, because the sum of two multiples of a number $d$ is always a multiple of $d$.

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    Is it necessary to prove that it is also the smallest nonzero element? – 2017-01-30
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$$\begin{align} d\Bbb Z = m\Bbb Z + n\Bbb Z\ \, \\ \iff d\Bbb Z \supseteq m\Bbb Z + n\Bbb Z \ \,&{\rm and}\,\ d\Bbb Z \subseteq m\Bbb Z + n\Bbb Z \\ \iff d\Bbb Z \supseteq m\Bbb Z,\ \ n\Bbb Z\ \ \ &{\rm and}\,\ d\Bbb Z\subseteq m\Bbb Z + n\Bbb Z\\ \iff\qquad\ \ d\ \ \mid\ \ m,\ \ n\quad\ \ &{\rm and}\ \ d\ =\ m\,j + n\,k\,\ \text{for some $\,j,k\in\Bbb Z$} \end{align}$$