# generating set of $\mathbb{Z}$

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 $$$$. What is playing the role of this $$a$$ in the group from factorset? That is not $$(1,1)$$ because of $$(k,k) = (0,0)$$ --- identity element.

• ${}{}{}{}(1,0)$? – Angina Seng Jun 4 at 17:40
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 $$n$$.
• And I can express any element of group from $(n+1,n)$? Only $(k,0)$, no? – Just do it Jun 4 at 17:59
• @Justdoit 1 generates the integers as a group, which means you're allowed to take inverses... to generate the integers as a monoid, you also need $-1$. – Alex Kruckman Jun 4 at 19:02
• So, we have two generating elements: $(1,0)$ and $(0,1)$ in the group structure on the factor of $\mathbb{N}^2$? Not one, like in $\mathbb{Z}$? Or I do not really understand.. – Just do it Jun 4 at 19:22
• @Justdoit No: As a group, $\mathbb{N}^2/\sim$ is generated by a single element, $(1,0)$, just like how the (isomorphic) group $\mathbb{Z}$ is generated by $1$. This is because the subgroup of $\mathbb{N}^2/\sim$ generated by $(1,0)$ contains the inverse of $(1,0)$, which is $(0,1)$, and hence it contains $(n,0) = (1,0) + \dots + (1,0)$ and $(0,n) = (0,1)+\dots+(0,1)$, and every element of $\mathbb{N}^2$ is equivalent to $(n,0)$ or $(0,n)$ for some natural number $n$. – Alex Kruckman Jun 4 at 19:33