Group Theory – Perfect Groups Whose Character Degrees Square Divide Its Order

charactersfinite-groupsgapgroup-theoryrepresentation-theory

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.

Best Answer

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 $60$.

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