Is a bialgebra with all group-like elements invertible a Hopf algebra? We know that in a Hopf algebra all group-like elements are invertible. Is the converse also true? Here is the precise formulation of my question :
Let $B$ be a bialgebra and $GLE$ = { $g \in B ~|~ g \neq 0, \Delta (g) = g \otimes g$ } the set of group-like elements. We know that this set is a monoid and that if B has an antipode, namely if $B$ is a Hopf algebra, then $GLE$ is a group.
Now, suppose that in the bialgebra $B$, every group-like element is invertible, does $B$ then have an antipode? (it would be easy to define an antipode $S: B \rightarrow B$ on $GLE$ by $S(g) = g^{-1}$, but what about the other elements?)
If there is a known counterexample, then what would be the extra-condition required to assure the existence of the antipode?
 A: Sufficient conditions are known for the existence of an antipode on a bialgebra $A$ over a Commutative ring $R$ that do not assume that $A$ is either commutative or cocommutative. Assume that $A$ 
is graded, $A_i = 0$ for $i<0$, and $A_0$ is $R$-free.  Let $\epsilon\colon A\longrightarrow R$ be the counit (or augmentation).  Assume that $\epsilon^{-1}(1)$ is a group under the multiplication of  $A$.  For each grouplike element $g$, let $A_g = Rg \oplus \bar{A_g}$, where the set $\bar{A_g}$ of positive
degree elements of $A_g$ is the set of all elements $x\in A$ such that 
$$\psi(x) = x\otimes g + \sum x'\otimes x'' + g\otimes x$$
where the $x'$ and $x''$ are both of positive degree.  Assume that $A$ is the direct sum over $g\in GLE$ of the $A_g$.  Then $A$ has an antipode $\chi$.   If $A$ is commutative or cocommutative, 
then $\chi^2$ is the identity, but not in general otherwise.  The conditions are motivated by thinking 
about the homology of an $H$-space $X$ with coefficients in $R$.
A: The answer is "no". A counterexample is given by Radford in Example 2 (p. 567) in the paper 
$\quad\quad$On Bialgebras which are simple Hopf Modules, Amer. Math. Soc. 80(1980),563-568
He takes the coalgera $C = \mathbb{C}^\ast = Hom_{\mathbb{R}}(\mathbb{C},\mathbb{R})$ over $\mathbb{R}$. Then the tensor algebra $T(C) = \mathbb{R} \oplus C \oplus (C \otimes C) \oplus ...$ is a cocommutative bialgebra that is no Hopf algebra and has $1$ as the only group-like element (thus forming a group). 
Additional conditions that imply the existence of an antipode are (over fields) 


*

*Commutative bialgebras: That's a result of Takeuchi (can be found in this paper), also reproved by Radford in the cited paper. 

*Cocummutative pointed bialgebras (see Sweedler's book "Hopf Algebras", Prop. 9.2.5). 
