$\frac{\mathbb{Z}*\mathbb{Z}}{N} \cong \mathbb{Z}\oplus\mathbb{Z}$ If $N$ is the normal subgroup of $\mathbb{Z}*\mathbb{Z}$ generated by $aba^{-1}b^{-1}$. How to show that $\frac{\mathbb{Z}*\mathbb{Z}}{N} \cong \mathbb{Z}\oplus\mathbb{Z}$?
 A: This is best seen as an application of the following general principle from category theory (which is trivial to prove in one line, as for operators on a Hilbert space): Let $C \xrightarrow{G} D \xrightarrow{G'} E$ be functors. If $F$ is left adjoint to $G$ and $F'$ is left adjoint to $G'$, then $FF'$ is left adjoint to $G'G$.
Here, we have the forgetful functors $G$ from abelian groups to groups, and $G'$ from groups to sets. Hence, $G'G$ is the forgetful functor from abelian groups to sets. The left adjoint of $G$ is the abelianization, which mods out the commutator subgroup. The left adjoint of $G'$ is the free group construction. It follows that the abelianization of the free group on $S$ is canonically isomorphic to the free abelian group on $S$. In your question $S=\{a,b\}$.
Try to prove this with elements directly, it will get incredibly cumbersome! Using universal properties you can often forget about elements completely, only the morphisms matter.
This principle has lots of other applications. For example, the abelianization of the tensor algebra of some module $M$ is the symmetric algebra on $M$, i.e. the free commutative algebra over $M$. The abelianization of the free Lie algebra on a set $X$ is just the free vector space on $X$. The monoid algebra over the free commutative monoid on some set $X$ is the free commutative algebra on $X$, i.e. the polynomial ring with variables $X$. etc.
A: Hint: Note that $(\mathbb{Z}\ast\mathbb{Z})/N$ is an abelian group generated by two elements $\{a,b\}$ for which the following universal property holds: given any abelian group $A$ and any set map $s:\{a,b\}\to A$, there exists a unique group homomorphism $(\mathbb{Z}\ast\mathbb{Z})/N\to A$ which extends $s$.
More intuitively, but not quite as rigorous (without further justification) is that taking the abelinization of a presented group $\langle S\mid R\rangle$ produces the group $\langle S\mid R^\ast\rangle$ where $R^\ast=R\cup\{xyx^{-1}y^{-1}\}$. So, $F_2=\mathbb{Z}\ast\mathbb{Z}$ is the presented group $\langle a,b\mid\rangle$ and so $F_2^\text{ab}=\langle a,b\mid aba^{-1}b^{-1}\rangle=\mathbb{Z}^2$. 
