Exercise on characterization of free abelian groups

Question

I was wondering if someone can please check my work on a homework problem. This is from the graduate Hungerford text. Chapter 2.1, number 3.

Let $X=\{a_i\ |\ i\in I\}$ be a set. Then the free abelian group on $X$ is isomorphic to the group defined by the generators $X$ and the relations $$\{a_ia_ja_i^{-1}a_j^{-1}=e\ |\ i,j\in I\}.$$ Proof: Let $F$ be the free abelian group on $X$ and let $G$ be the group defined by the generators $X$ and the relations $\{a_ia_ja_i^{-1}a_j^{-1}=e\ |\ i,j\in I\}$. Since any two generators in $X$ commute, it follows that $G$ is abelian. Now since $F$ is free abelian on $X$, and since $G$ is an abelian group, the inclusion $\iota:X\rightarrow G$ induces a homomorphism of abelian groups $f:F\rightarrow G$ such that $fh=\iota$ where $h:X\rightarrow F$. We have $f(h(X))=\iota(X)=X$. So $X$ is in the image of $f$. But since $X$ generates $G$, the image of $f$ is all of $G$ so that $f$ is epic. I am having a lot of trouble proving that $f$ is also monic. Can someone read this over and tell me if I am on the right track. Thank you very much.

Answer 1

What you’ve written so far seems good to me, (modulo the now-fixed mix-up).

To go on: Try to find a group arrow $G → F$. Use the universal property of the free group on $X$ (of which $G$ is a factor group). Then use the first isomorphism theorem (or the universal property of factor groups, whatever you call it).

Then don’t show that $f$ is monic, but that it has a left inverse (this makes the argument more solid since you then don’t need to show “monic + epic = iso in $\mathrm{Grp}$”, but only “left invertible + epic = iso” which is category-theoretic and straightforward).