Let $X$={$a_i| i\in I$} be a set. Then the free abelian group $F$ on $X$ is isomorphic to the group $G$ defined by generators $X$ and the relations {$a_ia_ja_i^{-1}a_j^{-1}=e| i,j\in I$}.

There is already a solution on this website free abelian group, and Exercise on characterization of free abelian groups, but I did not find them so routine as my brief solution:

$G$ is abelian since its generators commute. If $a_1^{p_1}...a_s^{p_s}$ as a member of $G$ (where $a_i$'s are distinct as members of $X$) is in $ker \phi$ where $\phi$ is the Van Dyck’s epimorphism from $G$ to $F$ then $p_1a_1+...p_sa_s$=0 in $F$ so that each $p_i$ is 0 since $F$ is free abelian, so that $\phi$ is a monomorphism as well.

Is my solution correct?


It depends on what you mean by "classification".

Your proof that $G$ is free abelian is correct, but you might want to reconsider the exposition a bit, since it sounds vaguely circular (even though it isn't).

Of course, there are a lot of ways of describing the free abelian groups. They are also just the free $\Bbb{Z}$-modules, which are just direct sums of copies of $\Bbb{Z}$.

Since $\Bbb{Z}$ is commutative, direct sums of different cardinalities give non-isomorphic free modules, since commutative rings have invariant basis number. To me, you need this additional result (that all the items on the list are distinct) to say that you have classified the free abelian groups.

I should add that the only proofs I know of the last paragraph require ZFC. There are probably models of ZF that break it.

  • $\begingroup$ My mere intention was to show that $G$ and $F$ are isomorphic, so the title of my question is badly worded $\endgroup$ – user555729 Jun 1 '18 at 6:36
  • $\begingroup$ Do you mean for the classification I should add that $G$ is determined up to isomorphism by the cardinality of $X$? $\endgroup$ – user555729 Jun 1 '18 at 6:55
  • $\begingroup$ Well, you already have that. I mean that you also want that the cardinality of $X$ is determined by the isomorphism class of $G$. $\endgroup$ – C Monsour Jun 1 '18 at 7:02

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