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Kindly ask this question:

Can we say that $\mathbb{Z}$ × $\mathbb{Z}$ is an extention of $\mathbb{Z}$ by $\mathbb{Z}$?

Samir
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1 Answers1

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Yes. By the definition of a group extension, because $G=\mathbb{Z}\times\mathbb{Z}$ has $N=\mathbb{Z}\times\{0\}\cong\mathbb{Z}$ as a normal subgroup, and $Q=G/N\cong \mathbb{Z}$ as the corresponding quotient group, there is a short exact sequence $$0\rightarrow \mathbb{Z}\,\xrightarrow{f}\,\mathbb{Z}\times\mathbb{Z}\,\xrightarrow{g}\,\mathbb{Z}\rightarrow 0,$$ i.e. the group $\mathbb{Z}\times\mathbb{Z}$ is an extension of $\mathbb{Z}$ by $\mathbb{Z}$. The maps $f$ and $g$ are given by $f(n)=(n,0)$ and $g(n,m)=m$.

Zev Chonoles
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  • I'm probably being picky, but 0's seem more appropriate than 1's here. – wckronholm Jun 01 '11 at 16:28
  • @Zev: Is this the only form of extention of $Z$ by itself? – Samir Jun 01 '11 at 16:29
  • @Samir: No, we can also do $$0\rightarrow \mathbb{Z},\xrightarrow{f},\mathbb{Z}\times\mathbb{Z},\xrightarrow{g}, \mathbb{Z}\rightarrow 0$$ where $$f(n)=(n,n)\hskip0.2in g(n,m)=n-m,$$ or $$f(n)=(n,-n)\hskip0.2in g(n,m)=n+m,$$ or $$f(n)=(0,n)\hskip0.2in g(n,m)=n.$$ – Zev Chonoles Jun 01 '11 at 16:34
  • @Zev: So in any form of these extentions, we can conclude that they are torsion-free? – Samir Jun 01 '11 at 16:37
  • @Samir: I'm sorry, I don't understand; conclude that what is torsion free? The groups $\mathbb{Z}$ and $\mathbb{Z}\times\mathbb{Z}$ are torsion-free regardless of whether we are thinking about them in an extension or not. – Zev Chonoles Jun 01 '11 at 16:40
  • @Zev: Thanks. I meant Z×Z . :) – Samir Jun 01 '11 at 16:42
  • @Samir: Since you are a new user, I would just like to bring up that an answer can be accepted. I'm not pressuring you to accept my answer, and as the advice there indicates, waiting a while to accept can increase the chance that someone with a better answer or explanation might come along and post, so don't accept mine unless or until you want to. – Zev Chonoles Jun 01 '11 at 17:37