Infinite-Representation Type Group Algebras 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. 
 A: 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)
