Perfect groups whose character degrees square divide its order

This post is an extension of that one from the non-abelian finite simple groups to the finite perfect groups. According to the mentioned post and its second comment, for all non-abelian finite simple group $$G$$, there is an irreducible complex character $$\chi$$ such that $$\chi(1)^2$$ does not divide $$|G|$$. The proof uses the classification of finite simple groups (CFSG), but we are interested in a CFSG-free proof, so if you find one, please post an answer to the mentioned post. Anyway, here we are interested in the finite perfect groups:

Question: Is there a finite perfect group $$G$$ such that for all irreducible complex character $$χ$$ then $$\chi(1)^2$$ divides |G|?

Bonus question 1: Is there such a group which is a direct product of non-abelian finite simple groups?
Bonus question 2: If so, what is the minimum number of components for such a direct product? Two?

Above bonus questions can be seen as a game with the prime factorization of the order of the non-abelian finite simple groups and their character degrees. Let us call a finite group $$G$$ thin if for all prime $$p$$ then there is an irreducible complex character $$\chi$$ such that $$\nu_p(\chi(1)) = \nu_p(|G|)$$, where $$\nu_p$$ is the p-adic valuation. The seven first non-abelian finite simple groups ($$\mathrm{PSL}(2,q)$$ with $$q=5,7,9,8,11,13,17$$) are thin. We wonder whether every $$\mathrm{PSL}(2,q)$$ is thin. Anyway, for our purpose here, we must consider non-thin groups only. A finite group $$G$$ will be called $$p$$-fat if for all irreducible complex character $$\chi$$ then $$\nu_p(\chi(1)) < \nu_p(|G|)$$. Note that a finite group is non-thin if and only if it is $$p$$-fat for some prime $$p$$. The first non-thin non-abelian finite simple group is $$A_7$$, which is $$2$$-fat. Now the notion of $$p$$-fat group is not enough for our need. A finite group $$G$$ will be called $$p$$-superfat if for all irreducible complex character $$\chi$$ then $$2\nu_p(\chi(1)) < \nu_p(|G|)$$. The group $$A_7$$ is $$2$$-superfat. The next $$p$$-superfat non-abelian finite simple group is $$M_{22}$$, which is also $$2$$-superfat. We wonder whether for all prime $$p$$ there is a $$p$$-superfat non-abelian finite simple group. Anyway, the goal here is to find relevant direct product of such superfat groups.

• google "character of $p$-defect 0" Nov 12, 2022 at 12:48
• Feb 4 at 5:35

If you extend to perfect groups, this is pretty easy. Let $$G$$ be your favourite simple group, and let $$A$$ be a very large abelian group. A semidirect product $$X=A\rtimes G$$ has character degrees dividing $$|G|$$. To make this group perfect, $$A$$ must be a product of irreducible $$\mathbb{F}_pG$$-modules for various primes dividing $$|G|$$. By making $$|A|$$ a multiple of $$|G|$$ we make $$|X|$$ a multiple of $$|G|^2$$, while all character degrees divide $$|G|$$.

To do this concretely, let $$G=A_5$$. We choose simple $$\mathbb{F}_pG$$-modules of dimension $$4$$, $$4$$ and $$3$$ for $$p=2,3,5$$ respectively (these are minimal). Then $$|X|=|G|\cdot 2^43^45^3.$$ To form the semidirect product, form the semidirect products $$H_p=M_p\rtimes G$$ as usual, where $$M_p$$ is the $$\mathbb{F}_pG$$-module, and then take their direct product. This is still too big, so now take a diagonal subgroup: keep the $$M_p$$ and take a diagonal subgroup $$G$$ acting on each of them. This will still be perfect.

I did this for $$A_5$$ and $$p=2,3$$ to get started, and $$X$$ can be generated by the permutations $$(1, 6, 16)(2, 8, 13)(3, 15, 11)(4, 9, 10)(5, 14, 12)(17, 62, 54, 48, 93, 82, 76, 40, 32)(18, 65, 81, 46, 96, 28, 77, 34, 59)(19, 95, 25, 50, 36, 56, 72, 64, 87)(20, 89, 52, 51, 39, 83, 70, 67, 33)(21, 92, 79, 49, 42, 29, 71, 61, 60)(22, 41, 26, 44, 63, 57, 75, 91, 85)(23, 35, 53, 45, 66, 84, 73, 94, 31)(24, 38, 80, 43, 69, 30, 74, 88, 58)(27, 47, 90, 55, 78, 37, 86, 97, 68)$$ and $$(1, 2, 10, 9)(3, 6, 8, 13)(4, 12, 15, 7)(5, 16, 14, 11)(17, 36, 59, 51, 92, 84)(18, 86, 60, 38, 93, 44)(19, 49, 52, 82, 94, 34)(20, 39, 53, 45, 95, 87)(21, 80, 54, 41, 96, 47)(22, 43, 55, 85, 88, 37)(23, 42, 56, 48, 89, 81)(24, 83, 57, 35, 90, 50)(25, 73, 67)(26, 63, 68, 78, 74, 30)(27, 32, 69, 65, 75, 71)(28, 76, 61)(29, 66, 62, 72, 77, 33)(31, 70, 64)(40, 97, 46, 58, 79, 91).$$

This isn't quite good enough as we have not added the abelian $$5$$-subgroup, but it's enough to understand the construction.

Edit: I just tacked on the $$5$$-subgroup as well, and an explicit example is the group generated by

(1,2,3,4,6)(5,7,10,11,14)(8,12,9,13,15)(16,17,19,20,24)(18,21,26,29,22)(23,27,
25,28,30)(32,42,47,38,55)(33,43,51,40,52)(34,44,48,41,53)(35,45,49,37,54)(36,
46,50,39,56)(57,60,58,59,61)


and

(1,2,4)(3,5,8)(6,9,10)(7,11,14)(12,13,15)(16,18,22)(17,20,25)(19,23,21)(24,28,
31)(26,27,30)(32,57,46,36,60,44,34,58,45,35,59,43,33,61,42)(37,56,50,38,52,51,
41,53,48,40,55,47,39,54,49)


It has order $$2^63^55^4$$ and character degrees divide $$60$$.

• Character degrees (with multiplicities) are: [ [ 1, 1 ], [ 3, 2 ], [ 4, 1 ], [ 5, 7 ], [ 10, 4 ], [ 12, 10 ], [ 15, 14 ], [ 20, 59 ], [ 30, 164 ], [ 60, 2651 ] ] Nov 12, 2022 at 19:16
• Where do you find the irreducible $\mathbb{F}_p G$-modules? Nov 13, 2022 at 2:47
• Do you think that the example you found is the smallest possible one? Nov 13, 2022 at 5:08
• @SebastienPalcoux I know the dimensions of simple modules for $A_5$, and also you can construct them on Magma with IrreducibleModules(G,F). I don't know if it's the smallest example, but I think Alexander Hulpke is checking this. Nov 13, 2022 at 11:55
• @SebastienPalcoux. I have checked up to order 1350000 that there was no such example (and then I needed the computer for something else). So this example is at worst too large by a factor $<10$. Nov 20, 2022 at 15:26

I cannot give a complete answer, but here are some thoughts.

In "Gagola, Stephen M., Jr.; A character theoretic condition for $$F(G)>1$$. Comm. Algebra 33 (2005), no. 5, 1369-1382." the following result is proven (relying on CFSG):

Theorem: Suppose that $$G$$ is a nontrivial finite group such that $$\chi(1)^2 \mid |G|$$ for all complex irreducible characters $$\chi$$ of $$G$$. Then $$G$$ has a nontrivial abelian normal subgroup.

Now for an example as in your main question, the group $$G$$ must have a nontrivial abelian normal subgroup $$N$$. If you can prove that the "squares of character degrees divide order" property holds for $$G/N$$ as well, it would follow by induction that no such $$G$$ exists.
Theorem: Suppose that $$G$$ is a finite group. Then $$G$$ is nilpotent if and only if $$\chi(1)^2 \mid [G:\ker \chi]$$ for all complex irreducible characters $$\chi$$ of $$G$$.