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]