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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

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    $\begingroup$ You need at least the Hopf algebra to be co-commutative for it to come from a group algebra. $\endgroup$ – Lord Shark the Unknown May 5 '18 at 15:09
  • $\begingroup$ Thank you I edited adding co-commutativity $\endgroup$ – Dac0 May 5 '18 at 15:46
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    $\begingroup$ Stoll not true. The symmetric algebra over a vector space is not a group algebra. $\endgroup$ – darij grinberg May 5 '18 at 15:46
  • $\begingroup$ @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. $\endgroup$ – Jules Lamers May 6 '18 at 0:01
  • $\begingroup$ Can someone create an answer that fully clarify the thing? Now I'm getting confused... $\endgroup$ – Dac0 May 6 '18 at 3:24
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Yes it is true. But not for the commutative hopf algebras in general. You need some more assumptions: first we need the field to be algebraically closed and of characteristic zero. You also need cocommutativity and finite dimensionality of the hopf algebra to have a full "correspondence".
To be more precise:

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 $\mathcal{H}$, $\mathcal{Ab}_{fin}$.

In my understanding, it is actually this equivalence of categories, which inspired the introduction of the term quantum groups (implying that the hopf algebra theory may be considered as a kind of "quantum" generalization of the group theory).

Maybe you can also find some interest in this -somewhat related- post: https://math.stackexchange.com/a/2756755/195021

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