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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.

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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)

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  • $\begingroup$ 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, 2016 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, 2016 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, 2020 at 4:34
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    $\begingroup$ The original reference to Bondarenko-Drozd. A possible textbook on the matter is Benson's "Representations and cohomology" $\endgroup$
    – Julian Kuelshammer
    Aug 14, 2020 at 6:33
  • $\begingroup$ Thank you for your information. $\endgroup$
    – Nguyen Dang Son
    Aug 14, 2020 at 19:59

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