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Let $A$ be a finite dimensional algebra over a field $K$. $A$ is called periodic if $A$ as an $A$-bimodule is a periodic module, that is $\Omega^n(A) \cong A$ for some $n \geq 1$. Being periodic implies that the algebra is selfinjective.

Question: Is a classification of periodic Hopf algebras known?

For group algebras the answer is very nice: When $KG$ is periodic for all algebraically closed fields $K$ if and only if $G$ has the property that every abelian subgroup is cyclic.

Other examples of periodic Hopf algebras are Taft algebras.

I do not know any other non-trvial examples (trivial examples are semisimple Hopf algebras over perfect fields).

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For finite groups, you can do better than that. Firstly, the field doesn't have to be algebraically closed. Secondly, you can do it a prime at a time. Given a field $k$ of characteristic $p$, the group algebra $kG$ is periodic if and only if every abelian $p$-subgroup of $G$ is cyclic. This is only dependent on the characteristic, not the field.

For a finite dimensional cocommutative Hopf algebra (finite group scheme) $G$ over $k$, periodicity is equivalent to the statement that over an arbitrary extension field $K$ the group algebra $KG_K$ does not contain a flatly embedded subalgebra isomorphic to $K[t_1,t_2]/(t_1^p,t_2^p)$. A good example of a connected periodic finite group scheme is the first Frobenius kernel of $SL_2$.

Without either commutativity or cocommutativity it's a mess, and I very much doubt that there's a known sensible answer, but I could be wrong.

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    $\begingroup$ You might be interested in the following paper: "Complexity, periodicity and one-parameter subgroups" by Rolf Farnsteiner, Trans AMS 365 (3) (2013), 1487-1531. $\endgroup$
    – Dave Benson
    Apr 24 at 7:04
  • $\begingroup$ Thanks, it would be interesting to see which representation-finite selfinjective algebras can appear as blocks of Hopf algebras as they are always periodic. But I found no answer so far although this seems to be a natural question. $\endgroup$
    – Mare
    Apr 24 at 18:14
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    $\begingroup$ Even for blocks of group algebras of finite groups, this is a tough question. Walter Feit wrote a paper in which he showed that the Brauer tree of such a block is either a star or has at most 248 vertices. In particular, there are many Brauer tree algebras (all of finite representation type and symmetric) that are not Morita equivalent to blocks of finite group algebras. $\endgroup$
    – Dave Benson
    Apr 24 at 18:37
  • $\begingroup$ Is it known whehter every Brauer tree algebras appears as a block of a Hopf algebra? $\endgroup$
    – Mare
    Apr 24 at 18:50
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    $\begingroup$ I'm pretty sure this isn't known, but if I were a betting man, I'd bet there are Brauer tree algebras that are not Morita equivalent to a block of any finite dimensional Hopf algebra. The basic problem is that we don't know much about finite dimensional Hopf algebras that are neither commutative nor cocommutative. We don't even know whether they have finitely generated cohomology. $\endgroup$
    – Dave Benson
    Apr 24 at 18:55

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