Perfect groups whose character degrees square divide its order

Question

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.

Answer 1

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 $60$.

Answer 2

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.

So the answer to your bonus question is no.

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.

Another related result is in "Gagola, Stephen M., Jr.; Lewis, Mark L. A character-theoretic condition characterizing nilpotent groups. Comm. Algebra 27 (1999), no. 3, 1053-1056." (This also relies on CFSG.)

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 $G$.