Hopf-* structure of the finite-group double

For basis elements $d_{h,g},d_{k,l},d_{m,n}$, the product $(d_{h,g}{d}_{k,l})d_{m,n}$ is nonzero exactly when $h=gk{g}^{-1}$ and $h=(gl)m(gl{)}^{-1}$, while $d_{h,g}(d_{k,l}{d}_{m,n})$ is nonzero exactly when $k=lm{l}^{-1}$ and $h=gk{g}^{-1}$. These conditions are equivalent since $(gl)m(gl{)}^{-1}=g(lm{l}^{-1})g^{-1}$; when they hold both products equal $d_{h,gln}$. Bilinearity gives associativity on $D(G)$.

For each $d_{h,g}$, $1_D{d}_{h,g}=d_{h,g}$ because only the summand $d_{h,e}$ survives, and $d_{h,g}{1}_D=d_{h,g}$ because exactly the summand indexed by $g^{-1}hg$ survives. Thus $1_D$ is a two-sided unit.

Both iterated coproducts of $d_{h,g}$ are ${\sum }_{abc=h}{d}_{a,g}\otimes d_{b,g}\otimes d_{c,g}$, and applying either counit to $\mathrm{\Delta }(d_{h,g})$ leaves $d_{h,g}$, by selecting respectively the factor indexed by $e$. Hence $\mathrm{\Delta }$ is coassociative and $\epsilon$ is a counit. Also $\mathrm{\Delta }(1_D)={\sum }_{a,b}{d}_{a,e}\otimes d_{b,e}=1_D\otimes 1_D$ and $\epsilon (1_D)=1$.

For basis elements $d_{h,g},d_{k,l}$, the product $\mathrm{\Delta }(d_{h,g})\mathrm{\Delta }(d_{k,l})$ is zero unless $h=gk{g}^{-1}$. In that case its terms correspond bijectively to factorizations $rs=k$, through $(r,s)\mapsto (gr{g}^{-1},gs{g}^{-1})$, and their sum is $\mathrm{\Delta }(d_{h,gl})=\mathrm{\Delta }(d_{h,g}{d}_{k,l})$. Thus $\mathrm{\Delta }$ preserves multiplication. Moreover $\epsilon (d_{h,g}{d}_{k,l})=\epsilon (d_{h,g})\epsilon (d_{k,l})$: both sides equal $1$ exactly when $h=k=e$.

In $m(S\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(d_{h,g})$, a summand indexed by $ab=h$ is nonzero exactly when $b=a^{-1}$. This is impossible unless $h=e$; if $h=e$, the resulting sum is ${\sum }_a{d}_{g^{-1}{a}^{-1}g,e}=1_D$. Therefore $m(S\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(d_{h,g})=\epsilon (d_{h,g})1_D$.

In $m(\mathrm{i}\mathrm{d}\otimes S)\mathrm{\Delta }(d_{h,g})$, the summand indexed by $ab=h$ is nonzero exactly when $a=b^{-1}$. Again it vanishes for $h\mathrm{\ne }e$, and for $h=e$ it is ${\sum }_a{d}_{a,e}=1_D$. Therefore $m(\mathrm{i}\mathrm{d}\otimes S)\mathrm{\Delta }(d_{h,g})=\epsilon (d_{h,g})1_D$, proving the Hopf algebra assertion by [F2].

The formula in [F1] gives $(d_{h,g}^{\ast }{)}^{\ast }=d_{h,g}$ and $1_D^{\ast }=1_D$. On basis products, $(d_{h,g}{d}_{k,l}{)}^{\ast }=d_{k,l}^{\ast }{d}_{h,g}^{\ast }$: both sides vanish unless $h=gk{g}^{-1}$, and in that case both equal $d_{l^{-1}kl,l^{-1}{g}^{-1}}$. Conjugate-linear extension makes $\ast$ an involution reversing multiplication.

The two expressions $\mathrm{\Delta }(d_{h,g}^{\ast })$ and $\mathrm{\Delta }(d_{h,g}{)}^{\ast }$ agree after the bijection $(a,b)\mapsto (g^{-1}ag,g^{-1}bg)$ between factorizations of $h$ and of $g^{-1}hg$. Also $\epsilon (d_{h,g}^{\ast })=\overline{\epsilon (d_{h,g})}$. Hence the stated involution is compatible with the coproduct and counit.