Group-like elements in a finite set coalgebra

The assignments extend linearly by [F4]. For each basis vector, $\alpha (\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(e_g)=e_g\otimes (e_g\otimes e_g)=(\mathrm{i}\mathrm{d}\otimes \mathrm{\Delta })\mathrm{\Delta }(e_g)$, and both counit maps send $\mathrm{\Delta }(e_g)=e_g\otimes e_g$ to $e_g$. Since the maps are linear and the $e_g$ span $C_G$, [F1] holds; if $G$ is empty, all maps are the unique zero maps and the coalgebra identities hold on the zero space.

Write any $x\in C_G$ uniquely as $x={\sum }_{g\in G}{a}_g{e}_g$. If $x$ is group-like, then ${\sum }_g{a}_g{e}_g\otimes e_g=\mathrm{\Delta }(x)=x\otimes x={\sum }_{g,h}{a}_g{a}_h{e}_g\otimes e_h$. By [F3], coefficient comparison gives $a_g=a_g^2$ for each $g$ and $a_g{a}_h=0$ for distinct $g,h$. From $a_g(a_g-1)=0$, either $a_g=0$, or $a_g\mathrm{\ne }0$ and multiplying by its inverse from [F5] gives $a_g=1$. The counit equation ${\sum }_g{a}_g=\epsilon (x)=1$ forces at least one coefficient to be $1$, while the off-diagonal equations force at most one. Thus $x=e_g$ for one $g\in G$; conversely every $e_g$ satisfies [F2]. For $G=\mathrm{\varnothing }$, $x=0$ and $\epsilon (x)=0$, so none is group-like.