Let $~$n \geq 3$~$. Then the Dihedral group on $~$n$~$ vertices, $~$D_{2n}$~$, is not abelian.

Proof

The most natural dihedral group presentation is $~$\langle a, b \mid a^n, b^2, bab^{-1} = a^{-1} \rangle$~$. In particular, $~$ba = a^{-1} b = a^{-2} a b$~$, so $~$ab = ba$~$ if and only if $~$a^2$~$ is the identity. But $~$a$~$ is the rotation which has order $~$n > 2$~$, so $~$ab$~$ cannot be equal to $~$ba$~$.

[todo: picture with the triangle]