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I am interested in determining the the number of indecomposable modules of finite groups over finite fields of a fixed dimension. Specifically, I have the following conjecture:

Conjecture. Suppose we have a finite group $G$ of order $n$, and $F_p$ the field of size $p$, $p$ a prime. Let $s(k)$ be the number of non-isomorphic indecomposable modules of dimension $k$ and over $F_p$, for the group algebra $F_pG$. Then $s(k)$ is bounded by a polynomial in $n$ and $p^k$.

I have no idea whether this conjecture is true or not -- I suspect it not correct in general, but could not prove it. The second Brauer-Thrall conjecture (now theorem) is somehow related but does not address the case of finite fields. I would be very grateful if anyone could give some hints of references for this. Thank you!

Best, Jimmy

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    $\begingroup$ It is not completely clear what you are asking. Are you looking for methods or algorithms to calculate the dimensions of the indecomposable modules, or are you just asking for references? $\endgroup$
    – Derek Holt
    Jan 24, 2015 at 15:52
  • $\begingroup$ As Derek points out, the formulation is not quite clear. Usually a conjecture as general as this is based on some computational evidence for typical small groups, so it's relevant to ask what examples you've studied. (Also, more tags such as 'finite-groups' and 'rt.representation-theory' are needed.) $\endgroup$ Jan 24, 2015 at 16:24
  • $\begingroup$ @DerekHolt I am looking for a counter example to the conjecture (now provided by Jeremy Rickard). Indeed my formulation is not clear; sorry for that. $\endgroup$
    – Jimmy
    Jan 24, 2015 at 23:34
  • $\begingroup$ @JimHumphreys Thank you for your suggestion. I only checked the Klein four group case, for which there is only one parameter so the conjecture seems hold. Indeed I should have checked more and in particular the wild cases, but I got lost in reading the paper proving the Brauer-Thrall conjecture due to my lack of background in modular representation theory. $\endgroup$
    – Jimmy
    Jan 24, 2015 at 23:36

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Are you sure you mean $p^k$, and not something like $p^{k^2}$?

If you look at indecomposable $d$-dimensional representations of the free algebra $\mathbb{F}_p\langle x,y\rangle$, the number grows faster than $p^d$. For a lower bound consider those where $x$ acts as a fixed single Jordan block $X$ (just to ensure indecomposability) and $y$ as a general matrix $Y$. There are $p^{d^2}$ of these, and two will be isomorphic iff the two $Y$s are conjugate by an invertible matrix centralizing $X$, which gives around $p^{d^2/2}$ non-isomorphic representations.

For most groups $G$ (in particular this is not hard to do explicitly for $G$ the semidirect product of $C_p^3$ by $C_2$ for odd $p$, with the generator of $C_2$ acting by inverting elements of $C_p^3$, but probably also any group $G$ such that $\overline{\mathbb{F}}_pG$ has wild representation type) there's a map from $\mathbb{F}_p\langle x,y\rangle$-modules to $\mathbb{F}_pG$-modules that preserves indecomposability and non-isomorphism and multiplies dimension by a constant. The construction I have in mind for the particular $G$ I mentioned multiplies dimensions by $2$.

So this shows that even for fixed $G$ the number of $k$-dimensional non-isomorphic indecomposable $\mathbb{F}_pG$-modules grows faster than any polynomial in $p^k$.

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  • $\begingroup$ Thank you Prof. Jeremy Rickard. This is what I am looking for. The conjecture comes from an algorithm I have related to test isomorphism of finite groups. If it holds then that algorithm will run in time efficient for my purpose. Now that it is not true so I need to look for other means. Thank you again. $\endgroup$
    – Jimmy
    Jan 24, 2015 at 23:39
  • $\begingroup$ May I ask what might happen for the tame case? Thank you. $\endgroup$
    – Jimmy
    Jan 24, 2015 at 23:49

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