# Reference on correspondence between commutative Hopf Algebras and Groups

Is it true that every commutative Hopf algebra is related to a Group in such a way that the co-multiplication is originated from the multiplication of the group, the antipode from the inverse?

Making it more explicitly, can all commutative and co-comutative Hopf algebra $H$ be written in this form: $H=\mathbb{C}[G]$, with the usual group algebra structure $$\eta: 1 \to e, \quad m: g \otimes h\mapsto gh$$ the coalgebra structure $$\varepsilon: g \mapsto 1, \quad \Delta: g \mapsto g \otimes g$$ where all of the maps above are defined on the basis of group elements?

If that's the case can anybody give me a reference for that fact? Thank you in advance

• You need at least the Hopf algebra to be co-commutative for it to come from a group algebra. – Lord Shark the Unknown May 5 '18 at 15:09
• Thank you I edited adding co-commutativity – Dac0 May 5 '18 at 15:46
• Stoll not true. The symmetric algebra over a vector space is not a group algebra. – darij grinberg May 5 '18 at 15:46
• @darijgrinberg More generally I guess any Hopf algebra with 'algebra-like' elements, ie $\Delta X = X\otimes1+1\otimes X$, is a counter example. Another such is the universal enveloping algebra of a Lie algebra, for which the coproduct is algebra-like on all elements of the Lie algebra. – Jules Lamers May 6 '18 at 0:01
• Can someone create an answer that fully clarify the thing? Now I'm getting confused... – Dac0 May 6 '18 at 3:24

it can be shown that there is an equivalence of Categories, between the Category of commutative, cocommutative, finite dimensional Hopf algebras $\mathcal{H}$ (over an algebraically closed field, of characteristic zero) and the category $\mathcal{Ab}_{fin}$ of the finite, abelian groups. It is possible to construct fully faithful and essentially faithful functors between $\mathcal{H}$ and $\mathcal{Ab}_{fin}$.
Start from an object of $\mathcal{H}$ i.e. a commutative, cocommutative, finite dimensional Hopf algebra $\mathcal{H}$, over an algebraically closed field, of characteristic zero. The set $G(H)$ of its grouplike elements forms a finite abelian group i.e. an element of $\mathcal{Ab}_{fin}$. It is relatively easy to see that a hopf algebra morphism induces an abelian group homomorphism.
So you get a functor $$\mathcal{G} : \mathcal{H} \Rrightarrow \mathcal{Ab}_{fin}$$ On the other hand, start from a finite abelian group $G$ and take its group hopf algebra $kG$. It is clearly commutative, cocommutative and finite dimensional, i.e. an object of $\mathcal{H}$. On the other hand, an abelian group homomorphism induces -by linear extension, due to the universal property of the group algebra- a morphism of hopf algebras between the corresponding group hopf algebras.
So you get a functor $$\mathcal{F} : \mathcal{Ab}_{fin} \Rrightarrow \mathcal{H}$$ Now, it can be shown that: $$\begin{array}{cccc} \mathcal{G} \mathcal{F} = Id_{\mathcal{A}b_{fin}} & & & \mathcal{F} \mathcal{G} \cong Id_{\mathcal{H}}\\ \end{array}$$ Consequently, the functors $\mathcal{G}$, $\mathcal{F}$ constitute an equivalence of the categories $\mathcal{H}$, $\mathcal{Ab}_{fin}$.