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?



Is a bialgebra with all group-like elements invertible a Hopf algebra?
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](https://www.ams.org/journals/proc/1980-080-04/S0002-9939-1980-0587928-4/S0002-9939-1980-0587928-4.pdf)


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)


1. Commutative bialgebras: That's a result of Takeuchi (can be found in this [paper](https://projecteuclid.org/journals/journal-of-the-mathematical-society-of-japan/volume-23/issue-4/Free-Hopf-algebras-generated-by-coalgebras/10.2969/jmsj/02340561.full)), also reproved by Radford in the cited paper.
2. Cocummutative pointed bialgebras (see Sweedler's book "Hopf Algebras", Prop. 9.2.5).


