The free group given by $\langle a,b:a^2=b^3=e\rangle$ is not abelian. Let $G=<a,b:a^2=b^3=e>$. I'm trying to show that $G$ is not abelian. $G$ is by definition given by $F(\{a,b\})/N$ where $F(\{a,b\})$ is the free group on 2 letters and $N$ is the smallest normal group containing $\{a^2,b^3\}$. I'm pretty sure that $aba^{-1}b^{-1}\notin N$ (that would show that $G$ is not abelian). I think $N=<A\cup B>$ where $A=\{xa^2x^{-1}:x\in F(\{a,b\})$ and $B=\{xb^3x^{-1}:x\in F(\{a,b\})$. I'm not sure how to prove $aba^{-1}b^{-1}\notin N$, It looks like any element in $N$ will never have a string like $aba^{-1}$ in it's expansion.
 A: The simplest way to do this is by quotienting $G$ by another relation; see if you can find another relation $r$ to apply such that $H=G/\langle r\rangle$ (to abuse notation slightly) is non-abelian.  Note that if $G$ were abelian, then any quotient group formed by adding another relation would also have to be abelian (why?), so conversely once you find a non-abelian $H$ then you've shown that $G$ must be non-abelian.
A: Another solution is to find an explicit group $H$ with two elements $p,q$ of order $2$ and $3$ respectively, but such that they do not commute. Then you can define a morphism $G \to H$ sending $a$ to $p$ and $b$ to $q$: if $a$ and $b$ commuted then $p$ and $q$ would commute too.
To find such an explicit example, consider in $\mathbb{R}^3$ a rotation $\rho$ of angle $2 \pi /3$ and a reflection $r$.
Complement: The group $\langle a,b \mid a^2=b^3=1 \rangle$ is in fact the free product $\mathbb{Z}_3 \ast \mathbb{Z}_2$. Now, it is possible to make it act on its Bass-Serre tree $T$ - here, the first barycentric subdivision of a regular ternary tree - and it is clear that $a$ and $b$ do not commute looking at their actions on $T$.
