# Existence of groups with prescribed dimensions of irreps

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.

• 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. – Alexander Konovalov Dec 23 '19 at 19:37

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.
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.
• I said the number of linear representations (i.e. representations of degree $1$) divides the group order. It is equal to $|G/[G,G]|$. – Derek Holt Dec 24 '19 at 11:08