I have some troubles with identification of the generating set in the next group:
If I want to create a group $\mathbb{Z}$ from commutative monoid $\mathbb{N}$ I should take $\mathbb{N}^2$ and factorize it by $(n_1,m_1) = (n_2,m_2)$ if $n_1+m_2 = n_2+m_1$. After that, the operation $-$ is obvious. I try to figure out what is the generating element in this new group. I know, that a $\mathbb{Z}$ isomorphic to a free one-element group $<a>$. What is playing the role of this $a$ in the group from factorset? That is not $(1,1)$ because of $(k,k) = (0,0)$ --- identity element.
Can somebody help me, please?
The pair $(a,b)$ represents the integer $a-b$. So the integer $1$ is represented by the pair $(n+1,n)$ for any natural number $n$.