So I learned about both free groups and free abelian groups, their constructions, universal properties etc. It seems that they are completely different concepts, even though they are both free objects in their respective categories. I am still a bit confused about the following: if we start with a free group and adjoin commutative relations between their generators, do we get a free abelian group? In other words, is $$ \langle F(a_1, \dots, a_n) \mid a_ia_j=a_ja_i \text{ for } 1 \leq i,j, \leq n \rangle = \mathbb{Z}^{(a_1, \dots, a_n)}$$ true? Are there some examples that will help clear up my confusion? Thanks.

  • $\begingroup$ I think for the relations you mean: $a_ia_ja_i^{-1}a_j^{-1}$. $\endgroup$ – Nick Dec 14 '16 at 15:35
  • $\begingroup$ @Nick You are right, I just edited it. $\endgroup$ – user228960 Dec 14 '16 at 15:37
  • $\begingroup$ To answer your question: yes. See the wikipedia page for "free abelian group" under the heading "presentation". It addresses this. $\endgroup$ – Nick Dec 14 '16 at 15:40
  • $\begingroup$ Yes, the abelianization of a free group is a free abelian group. $\endgroup$ – Dustan Levenstein Dec 14 '16 at 15:40
  • $\begingroup$ I guess that should have been obvious. Thanks for your comments. $\endgroup$ – user228960 Dec 14 '16 at 15:53

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