The dual Hopf algebra of a finite group algebra

The group basis makes ${\dim }_k(kG)=\mathrm{\mid }G\mathrm{\mid }<\mathrm{\infty }$, so the zero ideal has finite codimension and [F3] gives $k{G}^{\circ }=(kG{)}^{\ast }$; [F2] gives its dual Hopf structure. Since $G$ is finite, the functions ${\delta }_g$ form a basis of $k^G$: every function $f$ is ${\sum }_{g\in G}f(g){\delta }_g$, and evaluation at each group element proves linear independence.

For $a,b,u\in G$, the dual multiplication gives $({\delta }_a{\delta }_b)(u)=({\delta }_a\otimes {\delta }_b)(u\otimes u)={\delta }_a(u){\delta }_b(u)$, so ${\delta }_a{\delta }_b=0$ if $a\mathrm{\ne }b$ and ${\delta }_a{\delta }_a={\delta }_a$. Thus multiplication is pointwise, and the constant-one function is ${\sum }_{g\in G}{\delta }_g$.

The coproduct is determined by $\mathrm{\Delta }({\delta }_g)(u\otimes v)={\delta }_g(uv)$. The right-hand side is $1$ exactly for pairs with $uv=g$, hence $\mathrm{\Delta }({\delta }_g)={\sum }_{ab=g}{\delta }_a\otimes {\delta }_b$; the sum is finite because $G$ is finite.

The dual counit is $\epsilon ({\delta }_g)={\delta }_g(1_{kG})={\delta }_g(e)$, and the dual antipode is $S^{\ast }({\delta }_g)(h)={\delta }_g(S(h))={\delta }_g(h^{-1})={\delta }_{g^{-1}}(h)$. These formulas give the stated counit and antipode, so [F2] verifies that the listed maps form a Hopf algebra.