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Recall that a Nakayama algebra (also called serial algebras in the literature) is an algebra such that every indecomposable module has a unique composition series. In the book "Classical artinian rings and related topics" from 2009 by Yoshitomo Baba und Kiyoichi Oshiro one can find the following two questions at the end of the book:

  1. Let G be a finite group and K be an algebraic closure of a field k. If kG is a Nakayama algebra, is KG a Nakayama algebra? And how is the converse?

  2. For which algebraically closed fields $K$ and for which $n$ are the group algebras $KS_n$ and $KA_n$ Nakayama algebras, where $S_n$ is the symmetric and $A_n$ the alternating group?

(the original formulation for 2. is: "Let K be an algebraically closed field. Are there infinitely many $KS_n$ or $KA_n$ which are non-semisimple Nakayama algebras?")

Is an answer to those problems known now?

edit: It would be interesting to know whether there are quick computer programs available to test whether $KS_n$ or $KA_n$ are Nakayama algebras for $K$ being the algebraic closure of a prime field and given $n$.

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    $\begingroup$ I think for perfect fields 1. should be possible to settle using the classification of Nakayama algebras, which is for example given as Thm. 10.3.2 in [Drozd, Kirichenko: Finite dimensional algebras]. $\endgroup$ Oct 24 '18 at 20:15
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For question 2 and $S_n$, suppose $\text{char}(K)=p$ and $KS_n$ is a Nakayama algebra. Then $KS_n$ must have finite representation type, and so $S_n$ must have cyclic Sylow $p$-subgroups, which means $n<2p$ (and for $KS_n$ to be non-semisimple we must have $n\geq p$).

Then every non-simple block has cyclic defect, and it is well-known that cyclic defect blocks of symmetric groups have Brauer trees that are lines with $p-1$ edges and multiplicity one. But such a Brauer tree algebra is only a Nakayama algebra if the number of edges is at most two.

So $KS_n$ is a non-semisimple Nakayama algebra if and only if either

  • $p=2$ and $2\leq n<4$, or
  • $p=3$ and $3\leq n<6$.

(Probably there’s a similar answer for $KA_n$.)

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  • $\begingroup$ Thanks. Is it more general also known when an indecomposable projective module over the symmetric group is uniserial? $\endgroup$
    – Mare
    Dec 24 '18 at 16:03
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    $\begingroup$ @Mare I don’t know, sorry. But it wouldn’t surprise me if there were no (or very few) uniserial projectives except in blocks with cyclic defect group. $\endgroup$ Dec 24 '18 at 16:45

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