monoid bialgebra is a Hopf algebra if and only if monoid is a group

Assume that $H$ is a Hopf algebra with comultiplication $\mathrm{\Delta }$, counit $\epsilon$ and antipode $S$. It is well known, that if $c\in H$ and $\mathrm{\Delta }(c)=\sum _{i=1}^n{a}_i\otimes b_i$, then $\sum _{i=1}^{n}S(a_i)b_i=\epsilon (c)1=\sum _{i=1}^n{a}_{i}S(b_i)$ (actualy, this condition defines the antipode), where on the left and right side we have multiplication in $H$.

Now let $G$ be a monoid and $k$ a field. It is well known that $kG$ is a bialgebra (please, see parent object for details), but one may ask, when $kG$ is a Hopf algebra? We will try to answer this question.

Proposition. A monoid bialgebra $kG$ is a Hopf algebra if and only if $G$ is a group.

Proof. ,,$\Leftarrow$” If $G$ is a group, then define $S:kG\to kG$ by $S(g)=g^{-1}$. It is easy to check, that $S$ is the antipode, thus $kG$ is a Hopf algebra.

,,$\Rightarrow$” Assume that $kG$ is a Hopf algebra, i.e. we have the antipode $S:kG\to kG$. Then, for any $g\in G$ we have $S(g)g=gS(g)=1$ (because $\mathrm{\Delta }(g)=g\otimes g$ and $\epsilon (g)=1$). Here $1$ is the identity in both $G$ and $kG$. Of course $S(g)\in kG$, so

$$S(g)=\sum _{h\in G}{\lambda }_{h}h.$$

Thus we have

$$1=\left(\sum _{h\in G}{\lambda }_{h}h\right)g=\sum _{h\in G}{\lambda }_{h}hg.$$

Of course $G$ is a basis, so this decomposition is unique. Therefore, there exists $g^{\prime }\in G$ such that ${\lambda }_{g^{\prime }}=1$ and ${\lambda }_{h^{\prime }}=0$ for $h^{\prime }\ne g^{\prime }$. We obtain, that $1=g^{\prime }g$, thus $g$ is left invertible. Since $g$ was arbitrary it implies that $g$ is invertible. Thus, we’ve shown that $G$ is a group. $\mathrm{\square }$