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Let $G$ be a finite group with size $n$. Two basic results from finite dimensional complex linear representation theory reads:

  1. the dimension $d$ of a irreducible representation divides $n$.

  2. The sum of squars of dimensions over all nonisomorphic irreducible representation is $n$.

Question

(1) What I'm interested in is the inverse. Given a natural number $n$, for any tuple of natural numbers $(d_1,\cdots,d_l)$ such that

  1. $d_i | n$ for each $i$.

  2. $n = \sum_{i=1}^l d_i^2 $

  3. $d_1 = 1$

is there a finite group $G$ whose simple modules have the prescribed dimensions $(d_1,\cdots,d_l)$?

(2) Is it hopeful to have a list of $n$ such that some admissible tuples don't correspond to some finite group?

(3) Are there programs that give us a list of all finite groups with prescribed order, and also give us the character table of a specific group? The closest I can find about finite groups are sage and atlas, but they do not seem to support what I wish.

Example

For example, when $n = 8$, the only admissible tuples are $(1,\cdots,1)$, $(1,1,1,2)$. In both cases, there are groups with prescribed dimensions of simple modules.

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  • 1
    $\begingroup$ Atlas is not a program - it's a database, which contains information in various formats, including GAP. GAP also has packages AtlasRep which provides an Interface to the Atlas of Group Representations, and CTbliLib - the CharacterTable library. For the list of finite groups of a given order, see the GAP Small Groups Library. See this question for more GAP resources. $\endgroup$ – Alexander Konovalov Dec 23 '19 at 19:37
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All groups of order $15$ are cyclic, so a counterexample to (1) is $15 = 6 \times 1^2 + 3^2$. Of course there are extra conditions that you could impose and which fail to hold in this example, such as the number of linear characters dividing the group order, but I expect there would still be counterexamples.

I would guess that the answer to (2) is no.

For (3), try GAP (open source) or Magma. You cannot just ask for the groups of an arbitrary large order, but ordered up to $2000$ are all known, and there are functions for computing character tables.

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  • $\begingroup$ Number of irreps divides the group order?! I'm bad at memorizing facts.. is this something that I can find in basic texts? My motivation to this question is actually to find a complete set of rules of prescription. I believe there should be one, and I want to see the relation between both sides (group / representation) clearly. $\endgroup$ – Student Dec 24 '19 at 3:18
  • $\begingroup$ I said the number of linear representations (i.e. representations of degree $1$) divides the group order. It is equal to $|G/[G,G]|$. $\endgroup$ – Derek Holt Dec 24 '19 at 11:08
  • $\begingroup$ I wasn't aware of that fact either, thank you! Would you mind pointing out a reference? $\endgroup$ – Student Dec 24 '19 at 13:39

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