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There is a theorem that if a Hopf algebra $H$ is commutative or cocommutative, then $S^2=id_H$, where $S$ denotes the antipode.

May I know if the converse is true?

(i.e. if $S^2=id_H$, does it follow that $H$ is commutative or cocommutative?)

If no, what would be the simplest counter-example?

Sincere thanks for any help!

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    $\begingroup$ The simplest counterexample is a tensor product of a commutative non-cocommutative Hopf algebra with a cocommutative non-commutative Hopf algebra. $\endgroup$ – darij grinberg Aug 12 '15 at 20:17
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The Eight Dimensional Kac-Paljutkin Hopf Algebra has an involutive antipode but is neither commutative nor cocommutative.

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The converse is not true, Sweedler's 4-dimensional Hopf algebra (first example of a non-comm and non-cocomm Hopf algebra) is a counter example. This example is completely taken from section 4.3.6 of "Hopf Algebras" by Dascalescu, Nastasescu and Raianu.

Let $k$ be a field of characteristic $\neq 2$. As a vector space, define $H = \langle 1, c, x, cx \rangle$ to be 4-dimensional over $k$. Define the multiplication by $c^2 = 1, x^2=0, xc = -cx$ and the unit to be $1$. Define the comultiplication $\Delta(c) = c \otimes c$ and $\Delta(x) = c \otimes x + x \otimes 1$, and the counit $\epsilon(c) = 1$ and $\epsilon(x) = 0$. Finally, let the antipode be $S(c) = c^{-1} = c$ and $S(x) = -cx$.

By using that $S$ is an antimorphism we have $S(cx) = S(x)S(c) = -cxc = xcc = x$ and thus $S^2(cx) = S(x) = -cx \neq cx$ because $k$ is of characteristic $\neq 2$.

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    $\begingroup$ This is a nice example (please don't delete it!), but not an answer to the question... $\endgroup$ – darij grinberg Aug 12 '15 at 20:16

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