 Solution

Take the group algebra $H=\mathbb{C}[S_3]$ withon its group basis, extended linearly. Multiplication of group basis elements makes $\Delta ,\epsilon$ algebra homomorphisms, and $g^{-1}g=g{g}^{-1}=1$ verifies the antipode identities. The coproduct is fixed by the tensor flip on every basis element, so this is a cocommutative Hopf algebra. But $(12)(23)$ and $(23)(12)$ are distinct basis elements, proving noncommutativity. Thus $\boxed{\mathbb{C}[S_3]\text{ is noncommutative but cocommutative}}$.