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 \(ba\).

picture with the triangle