Suppose we are given a $n$-dimensional Hopf algebra $H$ over field $\mathbb K$. I want to prove that

$H$ contains n distinct group-like elements if and only if there exists a group $G$ such that $H\simeq \mathbb K[G]$ as Hopf algebras.

We know that if $H$ is a cocommutative Hopf algebra then there exists such group $G$ such that $H\simeq \mathbb K[G]$ as Hopf algebras. But I don't know if this is related to the above statement?

Can someone let me know that how can I prove it?



If $H \cong \mathbb{K}[G]$ for some group $G$, then one can just take the elements of $G$ as the distinct group-like elements.

For the other implication recall than in any coalgebra the set of group-like elements is linearly independent, and that in any Hopf algebra the set of group-like elements forms a group with respect to the multiplication of $H$.

If an $n$-dimensional Hopf algebra $H$ contains $n$ distinct group-like elements, then it follows that $G$, the set of group-like elements in $G$, is linearly independent and contains at least $n$ elements. Thus $G$ is a a basis of $H$. Thus we may identify $H$ with $\mathbb{K}[G]$ as $\mathbb{K}$-vector spaces.

The multiplication of $G$ is the restriction of the multiplication of $H$, and so the multiplication of $H$ turns out to be the unique $\mathbb{K}$-bilinear extension of the multiplication of $G$. This shows that the multiplication of $H$ coincides with the one of $\mathbb{K}[G]$. (It also follows their units coincide.)

The comultiplication and counit of $H$ are given on the basis $G$ by $\Delta(g) = g \otimes g$ and $\varepsilon(g) = 1$ (because the elements $g \in G$ are group-like). This hows that the comultiplication and counit of $H$ also coincide with the one of $\mathbb{K}[G]$.

Alltogether we have found that $H \cong \mathbb{K}[G]$.

  • $\begingroup$ Many thanks for your complete and knowledgeable answer! $\endgroup$ – Nikita Jun 10 '18 at 14:16

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