The question (exercise 7.10.4) is as follows:
The group $G = \langle x, y; xyx^{-1}y^{-1}\rangle$ is called a free abelian group. Prove that if $u, v$ are elements of an abelian group $A$, then there is a unique homomorphism $\phi:G\to A$ such that $\phi(x)=u, \phi(y)=v$.
So at first I thought that this was pretty straightforward. If $u,v$ are elements of an abelian group $A$ then they clearly satisfy the relation $uvu^{-1}v^{-1}=1$ and so if we let $S=\{u,v\}, R=\{uvu^{-1}v^{-1}\}$, $\mathcal{F}$ the free group on $S$, and $\mathcal{R}$ to be the normal subgroup of $\mathcal{F}$ generated by $R$, and finally $\mathcal{A} = \mathcal{F}/\mathcal{R}$, then we have a homomorphism $\phi:\mathcal{A}\to A$ that sends $u,v$ to $u,v$ (I believe this is just a direct application of corollary 7.10.4 from Artin). And $\mathcal{A}$ and $G$ are the same group, so we're done.
But, the solution here seems to do a lot of extra work that I don't understand the need for.
Thanks!
Best Answer
Consider the mapping $\varphi : \{x,y\} \to A$ such that $x \mapsto u$ and $y \mapsto v$. Then by the universal property of the free group there is a group homomorphism $\widetilde\varphi : F(\{x,y\}) \to A$ that extends $\varphi$. Now, as $A$ is abelian, $$\widetilde\varphi(xyx^{-1}y^{-1}) = uvu^{-1}v^{-1} = 1$$ meaning that $xyx^{-1}y^{-1} \in \ker \widetilde\varphi$, and then $\ker \widetilde\varphi$ contains the normal subgroup $N$ generated by $xyx^{-1}y^{-1}$. In fact, $N = \ker \widetilde\varphi$ (this is the lot of extra work), and then $\phi : G \to A$ is exactly ($G=F(\{x,y\})/N$) the induced map $F(\{x,y\})/N \to A$ of $\widetilde\varphi$, that is, the one that satisfies that $\phi \circ \pi = \widetilde\varphi$, where $\pi : F(\{x,y\}) \to F(\{x,y\})/N$ is the canonical projection.