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$\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\Aut{Aut}$Consider the class of finite groups $G$ having a zero entry in each column of its character table (except the first one), i.e. for all $g \neq e$ there is an irreducible character $\chi$ such that $\chi(g) = 0$.

I have been led to consider such character table, here are the examples found (see the GAP codes in Appendix):

  • at order less than $384$, this class reduces to $A_5$, $S_5$, $\PSL(2,7)$, $\Aut(\PSL(2,7))$ and $A_6$,
  • every simple group of order less than $3000000$ is in this class except $A_7$, $M_{22}$,
  • every perfect group of order less than $3600$ in this class is simple,
  • if $5 \le n \le 19$, then $A_n$ is not in this class iff $n \in \{ 7, 11, 13, 15, 16, 18, 19\}$. Idem for $S_n$.

Question: What are the finite groups in this class? Which simple groups are not in?

The group $\PSL(2,q)$ is in this class iff it is simple, iff $q \ge 4$ (the generic character table is known).

Observe that all the examples found above are almost simple, but:

Proposition: This class is stable by direct product.
proof: Immediate by Theorem 4.21 in Isaacs' Character Theory of Finite Groups stating that the irreducible characters of a direct product are exactly the product of the irreducible characters of the components. $\square$

Note that $A_5 \times A_5$ is a non almost-simple finite group in this class, maybe the smallest one.


Appendix

Small groups:

gap> for o in [2..383] do n:=NrSmallGroups(o);; for d in [1..n] do G:=SmallGroup(o,d);; L:=Irr(CharacterTable(G));; l:=Length(L);; a:=0;; for j in [2..l] do LL:= List([1..l], i-> L[i][j]);; if not 0 in LL then a:=1;; break; fi; od; if a=0 then Print([G,[o,d]]); fi; od; od;
[ AlternatingGroup( [ 1 .. 5 ] ), [ 60, 5 ] ][ SymmetricGroup( [ 1 .. 5 ] ), [ 120, 34 ] ][ Group( [ (3,4)(5,6), (1,2,3)(4,5,7) ] ), [ 168, 42 ] ][ Group( [ (1,4,6,8,5,2,7,3), (1,3,8,6,5,4,7) ] ), [ 336, 208 ] ][ AlternatingGroup( [ 1 .. 6 ] ), [ 360, 118 ] ]

Simple Groups:

gap> it:=SimpleGroupsIterator(10,3000000);; for G in it do L:=Irr(CharacterTable(G));; l:=Length(L);; a:=0;; for j in [2..l] do LL:= List([1..l], i-> L[i][j]);; if not 0 in LL then a:=1;; break; fi; od; if a=1 then Print([G]); fi; od;
[ A7 ][ M22 ]

Perfect Groups:

gap> for o in [60..3599] do n:=NumberPerfectGroups(o);; for d in [1..n] do G:=PerfectGroup(o,d);; L:=Irr(CharacterTable(G));; l:=Length(L);; a:=0;; for j in [2..l] do LL:= List([1..l], i-> L[i][j]);; if not 0 in LL then a:=1;; break; fi; od; if a=0 then Print([G,[o,d]]); fi; od; od;
[ A5, [ 60, 1 ] ][ L3(2), [ 168, 1 ] ][ A6, [ 360, 1 ] ][ L2(8), [ 504, 1 ] ][ L2(11), [ 660, 1 ] ][ L2(13), [ 1092, 1 ] ][ L2(17), [ 2448, 1 ] ][ L2(19), [ 3420, 1 ] ]

Alternating groups:

gap> for n in [5..19] do s:=Concatenation("A",String(n));;  L:=Irr(CharacterTable(s));; l:=Length(L);; a:=0;; for j in [2..l] do LL:= List([1..l], i-> L[i][j]);; if not 0 in LL then a:=1;; break; fi; od; if a=1 then Print([s]); fi; od;
[ "A7" ][ "A11" ][ "A13" ][ "A15" ][ "A16" ][ "A18" ][ "A19" ]
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  • $\begingroup$ Is there an application that makes these groups particularly natural or interesting, or is it just curiosity? $\endgroup$
    – LSpice
    Aug 17 '21 at 13:18
  • 4
    $\begingroup$ @LSpice: there is an application in tensor category: a pivotal categorification (over a field possibly of positive characteristic) of the character ring of such a finite group must be pseudo-unitary. $\endgroup$ Aug 17 '21 at 13:46
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    $\begingroup$ This doesn’t answer your question, but a lovely old theorem of Burnside says every row has a zero except the ones coming from 1-dimensional reps. $\endgroup$ Aug 17 '21 at 17:12
  • $\begingroup$ Sebastien, What happens when the "character" table does not come from a group --- do you still get something pseudo-unitary ? $\endgroup$ Aug 17 '21 at 19:29
  • 2
    $\begingroup$ @InesInstitoris: yes (if I am not mistaken) a pivotal categorification of a commutative fusion ring having such a character table must be pseudo-unitary. I don't how to prove that without the pivotal assumption. $\endgroup$ Aug 17 '21 at 20:02
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Partial answer: the finite group $G$ is clearly in this class if it has a $p$-block of defect zero for every prime $p$ which divides $|G|$. This is a sufficient condition which may not be necessary. No non-trivial finite solvable group satisfies the sufficient condition. Most (but not all) finite simple groups satisfy the sufficient condition (for example, $M_{22}$ and $M_{24}$ do not, and many alternating groups do not).

Later edit: In the other direction, if $G$ is a finite group such that $F^{\ast}(G)$ is a $p$-group for some prime $p$, then whenever $S$ is a Sylow $p$-subgroup of $G$, and $z$ is any non-identity element of $Z(S)$, then no irreducible character of $G$ vanishes at $z$, so $G$ does not have the desired property. This is because $G$ has only one $p$-block, the principal block, and no irreducible character in the principal $p$-block vanishes at a central element of a Sylow $p$-subgroup of $G$ by the fundamental congruence of central characters due to R. Brauer.

Third edit: A well-known paper of Granville and Ono, "Defect zero $p$-blocks for finite simple groups", Trans AMS, 348,1, January 1996, gives a complete determination of finite non-Abelian simple groups $G$ which do not have a $p$-block of defect zero for some prime $p$. All other non-Abelian finite simple groups $G$ have the property of the question, that is, for each non-identity element $x \in G$, there is an irreducible complex character $\chi$ of $G$ with $\chi(x) = 0$.

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  • 1
    $\begingroup$ I confirm that your sufficient condition is not necessary, but there are just $7$ non-abelian and non-alternating finite simple groups which are both in this class and do not satisfy your sufficient condition: $M_{12}$,$J_2$,$HS$,$Suz$,$Ru$,$C_1$, $BM$ (see my answer showing that every non-abelian and non-alternating finite simple group is in this class except $M_{22}$, $M_{24}$ and $C_3$). We still do not know whether there exists a non almost-simple group in this class. We also do not know whether an almost-simple group is in this class if and only if the corresponding simple group is in. $\endgroup$ Aug 19 '21 at 5:50
  • $\begingroup$ I think it is easy to construct examples in the class which are not almost simple. I will edit my answer to give an example. An example $G$ with $F^{\ast}(G)$ not a direct product of non-Abelian simple groups would be interesting. $\endgroup$ Aug 19 '21 at 8:09
  • $\begingroup$ You are right, it is easy to construct such examples: $A_5 \times A_5$ is in this class but is not almost simple. $\endgroup$ Aug 19 '21 at 8:17
  • $\begingroup$ Yes indeed, so I will not re-edit. $\endgroup$ Aug 19 '21 at 8:20
  • $\begingroup$ I wonder whether $A_5 \times A_5$ is the group of smallest order among the non almost-simple groups in this class. $\endgroup$ Aug 19 '21 at 9:06
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By using the answer of Geoff Robinson, we can prove that every non-abelian and non-alternating finite simple group is in this class except $M_{22}$, $M_{24}$ and $C_3$. The result follows from the following computation together with Corollary 2 in Granville and Ono - Defect zero $p$-blocks for finite simple groups.

gap> S:=["M12","M22","M24","J2","HS","Suz","Ru","C1","C3","BM"];;
gap> for s in S do L:=Irr(CharacterTable(s));; l:=Length(L);; a:=0;; for j in [2..l] do LL:= List([1..l], i-> L[i][j]);; if not 0 in LL then a:=1;; break; fi; od; if a=1 then Print([s]); fi; od;
[ "M22" ][ "M24" ][ "C3" ]

Corollary 2. Every finite simple group $G$ has a $p$-block of defect $0$, for every prime $p$, except in the following cases:

  • $G$ has no $2$-block of defect $0$ if it is isomorphic to $M_{12}$, $M_{22}$, $M_{24}$, $J_2$, $\mathit{HS}$, $\mathit{Suz}$, $\mathit{Ru}$, $C1$, $C3$, $\mathit{BM}$, or $A_n$ where $n \ne 2m^2 + m$ nor $2m^2 + m + 2$ for any integer $m$.

  • $G$ has no $3$-block of defect $0$ if it is isomorphic to $\mathit{Suz}$, $C3$, or $A_n$ with $3n + 1 = m^2 r$ where $r$ is squarefree and divisible by some prime $q \equiv 2 \bmod 3$.

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