Group and function Hopf algebras of the symmetric group on three letters

Applying the stated composition convention gives $ab=(123)$, since $1\mapsto 2$, $2\mapsto 3$, and $3\mapsto 1$; it gives $ba=(132)$, since $1\mapsto 3$, $3\mapsto 2$, and $2\mapsto 1$. These are distinct permutations.

The group elements form a basis of $kG$, so the products of the basis vectors $a,b$ are the distinct basis vectors $ab$ and $ba$ from step 1.1; hence $kG$ is not commutative. For every basis vector $g$, $\tau \mathrm{\Delta }(g)=\tau (g\otimes g)=g\otimes g=\mathrm{\Delta }(g)$, and linearity then gives cocommutativity on all of $kG$.

Multiplication in $k^{S_3}$ is pointwise, so $fg=gf$ at every group element and this algebra is commutative. Let $f={\delta }_{ab}$, the function equal to $1$ at $ab$ and $0$ elsewhere. Then $\mathrm{\Delta }(f)(a,b)=f(ab)=1$, whereas $(\tau \mathrm{\Delta }(f))(a,b)=\mathrm{\Delta }(f)(b,a)=f(ba)=0$ by step 1.1. Therefore $\mathrm{\Delta }(f)\mathrm{\ne }\tau \mathrm{\Delta }(f)$, so $k^{S_3}$ is not cocommutative.

The structures in [F1] make both algebras Hopf algebras, while steps 2.1–2.2 show that commutativity and cocommutativity distinguish the two examples exactly as stated.