# If $\Delta$ is cocommutative in $k(G)$, why does that imply $G$ is abelian?

Suppose $G$ is a finite group and $k(G)$ is it's group function Hopf algebra. I read that for $k(G)$ is quasi-triangulated requires that $G$ be abelian.

If $R$ is the distinguished element of $k(G)\otimes k(G)=k(G\times G)$ required by quasi-triangulation, and since $k(G)$ is commutative, then the requirement $\tau\circ\Delta f=R(\Delta f)R^{-1}$ in fact becomes $$\tau\circ\Delta f=R(\Delta f)R^{-1}=RR^{-1}\Delta f=\Delta f$$ meaning the coproduct $\Delta$ is cocommutative.

How does this relate back to imply that $G$ is abelian? The coproduct is $\Delta f(x,y)=f(xy)$.

$k(G)$ cocommutative means $\tau \circ \Delta = \Delta$ with $\tau: k^{G \times G} \to k^{G\times G},\,\tau(f)(x,y)=f(y,x)$.
Let $f: G \to k$. Then $(\tau \circ \Delta)(f)=\Delta f$ means $f(xy)=f(yx)$ for all $x,y \in G$.
For fixed x,y choose $f=\delta_{xy}$, i.e. $\delta_{xy}(xy)=1$ and $\delta_{xy}(g)=0$ for $g \ne xy$. Hence $\delta_{xy}(yx)=\delta_{xy}(xy)=1$ which implies $yx=xy$ by the definition of $\delta_{xy}$. This shows the commuativity of $G$.