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?

  • $\begingroup$ ${}{}{}{}(1,0)$? $\endgroup$ – Angina Seng Jun 4 at 17:40
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    $\begingroup$ What you call “factorize it” is usually called “mod out by” (taking a quotient, not writing it as a product). $\endgroup$ – Arturo Magidin Jun 4 at 17:48

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 $n$.

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  • $\begingroup$ Oh ye, it was obvious, thx.. $\endgroup$ – Just do it Jun 4 at 17:44
  • $\begingroup$ And I can express any element of group from $(n+1,n)$? Only $(k,0)$, no? $\endgroup$ – Just do it Jun 4 at 17:59
  • $\begingroup$ @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 $-1$. $\endgroup$ – Alex Kruckman Jun 4 at 19:02
  • $\begingroup$ 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.. $\endgroup$ – Just do it Jun 4 at 19:22
  • $\begingroup$ @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 $n$. $\endgroup$ – Alex Kruckman Jun 4 at 19:33

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