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Let $G$ be a finite group. It is well-known that for all irreducible complex character $\chi$ then $\deg(\chi)$ divides $\lvert G\rvert$. Motivated by some problems with modular tensor categories, we are interested in considering the finite groups $G$ such that for all $\chi$ (as above) then $\deg(\chi)^2$ divides $\lvert G\rvert$. Every abelian group satisfies this property, and the smallest non-abelian ones are $D_4$, $Q_8$, $D_6$, $\mathit{Dic}_3$. Now, we are specifically interested in the finite simple groups.

Question: Is there a non-abelian finite simple group $G$ such that for all irreducible complex character $\chi$ then $\deg(\chi)^2$ divides $\lvert G\rvert$?

We expect that the answer is no, and we need a proof without using the classification of finite simple groups (CFSG). A proof using CFSG starts as follows:

  • if $G$ is of Lie type over a finite field of characteristic $p$, then its Steinberg character has degree $d$ equal to the largest power of $p$ dividing $\lvert G\rvert$, in particular $d^2$ does not divide $\lvert G\rvert$.

  • if $G$ is sporadic, the proof reduces to the GAP computation in Appendix.

  • if $G$ is the alternating group $A_n$, then we can prove it for $n \le 23$ using the highest degree of an irreducible character, listed in A060955, see Appendix. Note that simple $A_n$ always has an irreducible character of degree $n-1$, so if $n-1$ is a prime $p$ then $p^2$ cannot divide $\lvert A_n\rvert = n!/2$. That permits to prove the case $n=24$.

We are interested in a general argument for the alternating groups (the only remaining case using CFSG).

Now, as written above, we need a proof without using CFSG.


Appendix

Sporadic case:

gap> L:=[ "M11", "M12", "M22", "M23", "M24", "J1", "J2", "J3", "J4", "Co1", "Co2", "Co3", "Fi22", "Fi23", "Fi24'", "Suz", "HS", "McL", "He", "HN", "Th", "B", "M", "ON", "Ly", "Ru", "Tits" ];;  
gap> for N in L do
   >    c:=0; 
   >    CD:=CharacterDegrees(CharacterTable(N)); 
   >    o:=Sum(List(CD,t->t[2]*t[1]^2)); 
   >    for d in CD do 
   >       if RemInt(o,d[1]^2)<>0 then 
   >           c:=1;
   >           break; 
   >       fi; 
   >    od; 
   >    if c=0 then 
   >       Print(N); 
   >       Print(L); 
   >    fi; 
   > od;  
gap> 

Alternating case

sage: L=[5, 10, 35, 70, 216, 567, 2310, 5775, 21450, 69498, 243243, 1153152, 3620864, 16336320, 64664600, 249420600, 997682400, 5462865408, 21422145536]  
sage: for i in range(19):
....:     n=i+5
....:     if factorial(n)%(2*L[i]^2)==0:
....:         print(i)
....:  
sage:  
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    $\begingroup$ There is a theorem that if $G$ is finite simple and $p$ is a prime dividing $|G|$, then $G$ has a complex irreducible character $\chi$ with $p \mid \chi(1)$. So for $G = A_n$ apply this for $n/2 < p < n$ (Bertrand's postulate) to get $\chi(1)^2 \nmid |G|$. See Theorem 5.4 in "Michler, Gerhard O. A finite simple group of Lie type has p-blocks with different defects, p≠2. J. Algebra 104 (1986), no. 2, 220-230." $\endgroup$
    – spin
    Nov 8, 2022 at 5:28
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    $\begingroup$ @SebastienPalcoux: I am curious why do you need to avoid CFSG? $\endgroup$
    – spin
    Nov 8, 2022 at 7:11
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    $\begingroup$ @spin I want a proof without CFSG to generalize it to tensor categories (we are very very far from a generalization of CFSG for tensor categories). $\endgroup$ Nov 8, 2022 at 9:17
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    $\begingroup$ I suspect it would be difficult to prove this without CFSG, or at least no such proof is known. $\endgroup$
    – spin
    Nov 8, 2022 at 12:58
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    $\begingroup$ @TheoJohnson-Freyd: Yes I mean nonabelian simple $G$. It does settle many cases, but the proof of the result I cited also relies on CFSG. If $G$ is alternating or sporadic, there is a prime $p$ such that $p \mid |G|$ and $p^2 \nmid |G|$. But for groups of Lie type it is not necessarily true. There are even examples where $|G|$ is a square, see for example here. $\endgroup$
    – spin
    Nov 9, 2022 at 4:55

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