Assume that $G$ is a $p$-group, $N$ a normal subroup and $k$ a field (infinite or finite) of $char(k)=p \gneq 0$. What can we say about the representations of $G$ if we already known about the representations of $N$? For instance if the group algebra $kN$ is representation-infinite (or otherwise of infinite representation type) can we say something about the representation type of $kG$? For the moment i don't care if we need to give some further assumptions about the groups or the base field in order to answer the above question.


I'm assuming that the field is infinite.

Then, there is the characterisation of representation type by Bondarenko and Drozd:

Let $G$ be a finite group and $k$ an infinite field of characteristic $p$ (dividing the order of the group). Then

  • $kG$ has finite representation type if and only if $G$ has cyclic Sylow $p$-subgroups
  • $kG$ has tame representation type if and only if $p=2$ and the Sylow $2$-subgroups are dihedral, semidihedral or generalised quaternion.
  • Otherwise, $kG$ has wild representation type.

So, if $G$ is a $p$-group, then $kG$ has finite representation type if and only if $G$ is cyclic. So assume that a normal subgroup $N$ has infinite representation type, then $N$ - being a $p$-group as well - is not cyclic. Therefore $G$ can't be cyclic as subgroups of cyclic groups are cyclic.

For tame representation type, a similar approach works. For generalised quaternion groups, it is known that every subgroup of such group is either generalised quaternion or cyclic (see e.g. Keith Conrad: Generalized Quaternions, Corollary 4.8). For dihedral groups, subgroups are either dihedral or cyclic (see e.g. this math.stackexchange question. I would guess, that a similar property holds for semidihedral groups, but these seem to be more difficult as they can have dihedral and generalised quaternion subgroups (see e.g. enter link description here)

  • $\begingroup$ Thank you very much for your response, i do know the above "amazing" theorem, and also the first dot is the Higman's Griterion. So can we say something like, in each dimension $d \geqslant 2$ we can construct infinite numbers of indecomposable representations? $\endgroup$ – user321268 May 9 '16 at 11:49
  • $\begingroup$ @mayer_vietoris I doubt that for each dimension $d\geq 2$ we can construct an infinite number of representations. But combining with Brauer-Thrall 2 (assuming the underlying field is algebraically closed) one can deduce that there are infinitely many $d$ such that there are infinitely many indecomposable modules of length $d$. $\endgroup$ – Julian Kuelshammer May 9 '16 at 11:54
  • $\begingroup$ Can you suggest materials for amazing theorems? I cannot find them, especially for tame and wild representations of group algebras. Thank you so much. $\endgroup$ – Nguyen Dang Son Aug 14 '20 at 4:34
  • 1
    $\begingroup$ The original reference to Bondarenko-Drozd. A possible textbook on the matter is Benson's "Representations and cohomology" $\endgroup$ – Julian Kuelshammer Aug 14 '20 at 6:33
  • $\begingroup$ Thank you for your information. $\endgroup$ – Nguyen Dang Son Aug 14 '20 at 19:59

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