TOWARDS THE CLASSIFICATION OF FINITE SIMPLE GROUPS WITH
EXACTLY THREE OR FOUR SUPERCHARACTER THEORIES

arXiv:1605.08971v1 [] 29 May 2016

A. R. ASHRAFI AND F. KOOREPAZAN-MOFTAKHAR

Abstract. A supercharacter theory for a finite group G is a set of superclasses each of which is a
union of conjugacy classes together with a set of sums of irreducible characters called supercharacters
that together satisfy certain compatibility conditions. The aim of this paper is to give a description of
some finite simple groups with exactly three or four supercharacter theories.
Keywords: Sporadic group, alternating group, Suzuki group.
2010 Mathematics Subject Classification: Primary: 20C15; Secondary: 20D15.

1. Introduction
Throughout this paper G is a finite group, d(n) denotes the number of divisors of positive integer
n, Irr(G) stands for the set of all ordinary irreducible character of G and Con(G) is the set of all
conjugacy classes of G. For other notations and terminology concerning character theory, we refer
to the famous book of Isaacs [11]. Following Diaconis and Isaacs [7], a pair (X , K) together with
the choices of characters χX is called a supercharacter theory of G if the following conditions are
satisfied:
(1)
(2)
(3)
(4)

X is a partition of Irr(G) and K is a partition of Con(G);
{1} ∈ K;
the characters χX , X ∈ X , are constant on the members of K;
|X | = |K|.

The elements of X and K are called supercharacters and superclasses of G, respectively. It is
easy to see that m(G) = (Irr(G), Con(G)) and M (G) = (X , K), where X = {{1}, Irr(G) \ {1}} and
K = {{1}, Con(G) \ {1}} are supercharacter theories of G which are called the trivial supercharacter
theories of G. The set of all supercharacter theories of a finite group G is denoted by Sup(G) and set
s(G) = |Sup(G)|. These notions were first introduced by Andre for finite unitriangular groups using
polynomial equations defining certain algebraic varieties [1, 2, 3].
Following Hendrickson [9], we assume that P art(S) denotes the set of all partitions of a set S. If
X and Y are two elements of P art(S) then we say that X is a refinement of Y or Y is coarser
than X and we write “X  Y”, if [a]X ⊆ [a]Y for all a ∈ S. For two supercharacter theories (X , K)
and (Y, L), we define (X , K) ∨ (Y, L) = (X ∨ Y, K ∨ L). It is well-know that (P art(S), ) is a lattice
which is called the partition lattice of S. By [10, Proposition 2.16], the join of two supercharacter
theories of a group G is again a supercharacter theory for G, but it is possible to find a pair of
supercharacter theories such that their meet is not a supercharacter theory. This shows that the set
1

of all supercharacter theories of a finite group with usual join and meet don’t constitute a lattice in
general.
Burkett et al. [5] gave a classification of finite groups with exactly two supercharacter theories. They
proved that a finite group G has exactly two supercharacter theories if and only if G is isomorphic
to the cyclic group Z3 , the symmetric group S3 or the simple group Sp(6, 2). The aim of this paper
is to continue this work towards a classification of finite simple groups with exactly three or four
supercharacter theories.
If p and 2p + 1 are primes then p is called a Sophie Germain prime and 2p + 1 is said to be a safe
prime. The safe primes are recorded in on-line encyclopedia of integer sequences as A005385, see [16]
for details. The first few members of this sequence is 5, 7, 11, 23, 47, 59, 83, 107, 167, 179, 227, 263,
347, 359, 383, 467, 479, 503, 563, 587, 719, 839, 863, 887, 983, 1019, 1187, 1283, 1307, 1319, 1367,
1439, 1487, 1523, 1619, 1823, 1907. These two sequences of prime numbers have several applications
in public key cryptography and primality testing and it has been conjectured that there are infinitely
many Sophie Germain primes, but this remains unproven [6].
2. Supercharacter Theory Construction for Sporadic Groups
The aim of this section is to compute some supercharacter theories for sporadic simple groups. We
use the following simple lemma in our calculations:
Lemma 2.1. Let G be a finite group, χ be a non-real valued irreducible character of G and x ∈ G
such that χ(x) is a non-real number. We also assume that each row and column of G has at most two
non-real numbers. Define:
X

= {{χ1 }, {χ, χ}, Irr(G) − {χ1 , χ, χ}}

K = {{e}, {xG , (x−1 )G }, Con(G) − {e, xG , (x−1 )G }}.

Then (X , K) is a supercharacter theory for G.
Proof. To prove, it is enough to investigate the main condition of supercharacter theory for the conjugacy classes xG , (x−1 )G and irreducible characters χ, χ. By definition of σX on X = {xG , (x−1 )G },
σX (x−1 ) = χ(1)χ(x−1 ) + χ(1)χ(x−1 )
= χ(1)χ(x) + χ(1)χ(x)
= χ(1) χ(x) + χ(1)χ(x)
= σX (x).
Therefore, σX is constant on the part {xG , (x−1 )G } of K, as desired.



The following lemma is important for constructing supercharacter theories on simple groups.
Lemma 2.2. Suppose G is a finite group, A = {χ(x) | χ ∈ Irr(G)} and Q(A) denotes the filed
generated by Q and A. Then the following holds:
(1) If X (G) = {{χ, χ} | χ ∈ Irr(G)} and K(G) = {{xG , (x−1 )G } | x ∈ G} then (X , K) is a
supercharacter theory of G.
(2) If Γ = Gal( Q(A)
Q ), X (G) is the set of all orbits of Γ on Irr(G) and K(G) is the set of all orbits
of Γ on Con(G) then (X , K) is a supercharacter theory of G.
2

Proof. The proof follows from [7, p. 2360] and [15].



To calculate the supercharacter of a finite group G, we first sort the character table of G by the
following GAP commands [19]:
u:=CharacterTable(G);
t:=CharacterTableWithSortedCharacters(u);
Then we prepare a GAP program to check whether or not a given pair (X , K) of a partition K
for conjugacy classes and another partition X for irreducible characters constitutes a supercharacter
theory. To find this pair of partitions, we usually apply Lemma 2.1.
Theorem 2.3. The Mathieu groups M11 , M12 , M22 , M23 and M24 have at least five supercharacter
theories.
Proof. We will present five supercharacter theories for each Mathieu group as follows:
• The Mathieu group M11 . Suppose the irreducible characters and conjugacy classes of the
11
11
11
Mathieu group M11 are Irr(M11 ) = {χ1 , χ2 , . . . , χ10 } and Con(M11 ) = {xM
, xM
, . . . , xM
1
2
10 },
respectively. We now define:
K1

=

n
o
11
11
11
}(2 ≤ i ≤ 8), {xM
{1}, {xM
, xM
9
10 } ,
i

X1

=

K2

=

{{1}, {χi }(2 ≤ i ≤ 5), {χ6 , χ7 }, {χi }(8 ≤ i ≤ 10)} ,
n
o
M11
M11
M11
11
},
{x
{1}, {xM
,
x
}(9
≤
i
≤
10)
,
}(2
≤
i
≤
6),
{x
8
7
i
i

X2

=

{{1}, {χ2}, {χ3 , χ4 }, {χi }(5 ≤ i ≤ 10)} .

By Lemma 2.1 and [10, Proposition 2.16], C1 = (X1 , K1 ), C2 = (X2 , K2 ) and C3 = C1 ∨ C2 are
supercharacter theories of M11 . Since C1 , C2 , C3 , m(M11 ) and M (M11 ) are distinct, s(M11 ) ≥ 5.
• The Mathieu group M12 . We assume that the irreducible characters and conjugacy classes of
12
12
the Mathieu group M12 are Irr(M12 ) = {χ1 , χ2 , . . . , χ15 } and Con(M12 ) = {xM
, . . . , xM
1
15 },
respectively. Define:
K1

=

o
n
M12
12
12
}(2 ≤ i ≤ 13), {xM
{1}, {xM
14 , x15 } ,
i

X1

=

K2

=

{{1}, {χi }(2 ≤ i ≤ 3), {χ4 , χ5 }, {χi }(6 ≤ i ≤ 15)} ,
o
n
M12
M12
12
12
12
12
12
}(13 ≤ i ≤ 15) ,
{1}, {xM
}(2 ≤ i ≤ 5), {xM
, xM
}, {xM
}(8 ≤ i ≤ 10), {xM
11 , x12 }, {xi
7
6
i
i

X2

=

{{1}, {χ2 , χ3 }, {χi }(4 ≤ i ≤ 8), {χ9 , χ10 }, {χi }(11 ≤ i ≤ 15)} .

Then by Lemma 2.1 and [10, Proposition 2.16], the pairs C1 = (X1 , K1 ), C2 = (X2 , K2 ) and
C3 = C1 ∨ C2 are three supercharacter theories of M12 different from m(M12 ) and M (M12 ).
This proves that s(M12 ) ≥ 5, as desired.
22
22
• The Mathieu group M22 . Suppose Irr(M22 ) = {χ1 , . . . , χ12 } and Con(M22 ) = {xM
, . . . , xM
1
12 }.
We define:
K1

=

n
o
M22
M22
22
}(2
≤
i
≤
10),
{x
{1}, {xM
,
x
}
,
11
12
i

X2

=

K2

=

{{1}, {χi}(2 ≤ i ≤ 9), {χ10 , χ11 }, {χ12 }} ,
o
n
M22
M22
22
}(10
≤
i
≤
12)
,
},
{x
,
x
{1}, {xiM22 }(2 ≤ i ≤ 7), {xM
9
8
i

X2

=

{{1}, {χ2}, {χ3 , χ4 }, {χi }(5 ≤ i ≤ 12)} .
3

Then by Lemma 2.1 the pairs C1 = (X1 , K1 ) and C2 = (X2 , K2 ) are supercharacter theories of
M22 . We now apply [10, Proposition 2.16] to prove that C3 = C1 ∨ C2 is another supercharacter
theory for M22 which shows that s(M22 ) ≥ 5.
23
23
• The Mathieu group M23 . Suppose Irr(M23 ) = {χ1 , . . . , χ17 } and Con(M23 ) = {xM
, . . . , xM
1
17 }
are the irreducible characters and conjugacy classes of the Mathieu group M23 , respectively.
If we can present three supercharacter theories for M23 different from m(M23 ) and M (M23 )
then it can be easily proved that s(M23 ) ≥ 5, as desired. Define:
K1

=

n
o
M23
23
23
}(2 ≤ i ≤ 15), {xM
{1}, {xM
16 , x17 } ,
i

X1

=

K2

=

{{1}, {χi}(2 ≤ i ≤ 9), {χ10 , χ11 }, {χi }(12 ≤ i ≤ 17)} ,
n
o
M23
M23
M23
23
}(2
≤
i
≤
9),
{x
{1}, {xM
,
x
}(12
≤
i
≤
17)
,
},
{x
10
11
i
i

X2

=

{{1}, {χi}(2 ≤ i ≤ 11), {χ12 , χ13 }, {χi }(14 ≤ i ≤ 17)} .

To complete the proof, it is enough to apply Lemma 2.1 and [10, Proposition 2.16] for proving
that C1 = (X1 , K1 ), C2 = (X2 , K2 ) and C3 = C1 ∨ C2 are supercharacter theories of M23 .
• The Mathieu group M24 . We now assume that Irr(M24 ) = {χ1 , χ2 , . . . , χ26 } and Con(M24 ) =
24
24
24
{xM
, xM
, . . . , xM
1
2
26 }. If we define:
n

o
M24
M24
M24
24
}(2
≤
i
≤
20),
{x
{1}, {xM
,
x
}(23
≤
i
≤
26)
,
},
{x
21
22
i
i

K1

=

X1
K2

= {{1}, {χi}(2 ≤ i ≤ 4), {χ5 , χ6 }, {χi }(7 ≤ i ≤ 26)} ,
n
o
M24
M24
24
}(2
≤
i
≤
24),
{x
=
{1}, {xM
,
x
}
,
25
26
i

X2

= {{1}, {χi}(2 ≤ i ≤ 9), {χ10 , χ11 }, {χi }(12 ≤ i ≤ 26)} .

then by a similar calculation as other cases, we can prove that C1 = (X1 , K1 ), C2 = (X2 , K2 )
and C3 = C1 ∨C2 are three supercharacter theories different from m(M24 ) and M (M24 ), proving
this case.
This completes the proof.



Theorem 2.4. The Leech lattice groups have at least five supercharacter theories.
Proof. There are seven Leech lattice simple groups. These are HS, J2 , Co1 , Co2 , Co3 , M cL and Suz.
Our main proof will consider seven cases as follows:
• The Higman-Sims group HS. To establish five supercharacter theories for HS, we assume that
Irr(HS) = {χi }1≤i≤24 and Con(HS) = {xHS
i }1≤i≤24 . Define:
K1

=


HS
HS
{1}, {xHS
i }(2 ≤ i ≤ 22), {x23 , x24 } ,

X1

=

K2

=

{{1}, {χi }(2 ≤ i ≤ 10), {χ11 , χ12 }, {χi }(13 ≤ i ≤ 24)} ,

HS
HS
HS
{1}, {xHS
i }(2 ≤ i ≤ 18), {x19 , x20 }, {xi }(21 ≤ i ≤ 24) ,

X2

=

{{1}, {χi }(2 ≤ i ≤ 13), {χ14 , χ15 }, {χi }(16 ≤ i ≤ 24)} .

Then by Lemma 2.1, one can see that C1 = (X1 , K1 ) and C2 = (X2 , K2 ) are supercharacter
theories of HS. We now apply [10, Proposition 2.16], to prove that C3 = C1 ∨ C2 is another
supercharacter theory of HS. These supercharacter theories are different from m(HS) and
M (HS) which concludes that s(HS) ≥ 5.
4

1
• The Conway group Co1 . Suppose Irr(Co1 ) = {χi }1≤i≤101 and Con(Co1 ) = {xCo
}1≤i≤101 .
i
Define:

K1

=

n
o
Co1
Co1
1
1
}(2 ≤ i ≤ 96), {xCo
{1}, {xCo
}(99 ≤ i ≤ 101) ,
97 , x98 }, {xi
i

X1

=

K2

=

{{1}, {χi }(2 ≤ i ≤ 26), {χ27 , χ28 }, {χi }(29 ≤ i ≤ 101)} ,
n
o
Co1
Co1
1
1
}(2 ≤ i ≤ 77), {xCo
{1}, {xCo
}(80 ≤ i ≤ 101) ,
78 , x79 }, {xi
i

X2

=

{{1}, {χi }(2 ≤ i ≤ 16), {χ17 , χ18 }, {χi }(19 ≤ i ≤ 101)} .

By Lemma 2.1, C1 = (X1 , K1 ) and C2 = (X2 , K2 ) are supercharacter theories of Co1 and by
[10, Proposition 2.16], C3 = C1 ∨ C2 . Since these supercharacter theories are different from
m(Co1 ) and M (Co1 ), s(Co1 ) ≥ 5. Hence the result follows.
2
• The second Conway group Co2 . Let Irr(Co2 ) = {χi }1≤i≤60 and Con(Co2 ) = {xCo
}1≤i≤60 .
i
Define:
K1

=

n
o
Co2
Co2
Co2
Co2
Co2
2
}(2
≤
i
≤
45),
{x
{1}, {xCo
,
x
}(48
≤
i
≤
58),
{x
},
{x
,
x
}
,
46
47
59
60
i
i

X1

=

K2

=

{{1}, {χi }(2 ≤ i ≤ 11), {χ12 , χ13 }, {χi }(14 ≤ i ≤ 30), {χ31 , χ32 }, {χi }(33 ≤ i ≤ 60)} ,
n
o
Co2
Co2
Co2
2
}(2
≤
i
≤
52),
{x
{1}, {xCo
,
x
}(55
≤
i
≤
60)
,
},
{x
53
54
i
i

X2

=

{{1}, {χi }(2 ≤ i ≤ 9), {χ10 , χ11 }, {χi }(12 ≤ i ≤ 60)} .

By Lemma 2.1 and [10, Proposition 2.16], C1 = (X1 , K1 ), C2 = (X2 , K2 ) and C3 = C1 ∨ C2
are supercharacter theories of the Conway group Co2 . Since these supercharacter theories are
different from m(Co2 ) and M (Co2 ), s(Co2 ) ≥ 5.
• The third Conway group Co3 . We assume that Irr(Co3 ) = {χi }1≤i≤42 and Con(Co3 ) =
3
{xCo
}1≤i≤42 . Define:
i
K1

=

o
n
Co3
Co3
Co3
Co3
3
3
3
}(38 ≤ i ≤ 42) ,
}(26 ≤ i ≤ 35), {xCo
{1}, {xCo
}(2 ≤ i ≤ 23), {xCo
36 , x37 }, {xi
24 , x25 }, {xi
i

X1

=

K2

=

{{1}, {χi }(2 ≤ i ≤ 5), {χ6 , χ7 }, {χi }(8 ≤ i ≤ 17), {χ18 , χ19 }, {χi }(20 ≤ i ≤ 42)} ,
o
n
Co3
Co3
3
3
}(40 ≤ i ≤ 42) ,
}(2 ≤ i ≤ 37), {xCo
{1}, {xCo
38 , x39 }, {xi
i

X2

=

{{1}, {χi }(2 ≤ i ≤ 15), {χ16 , χ17 }, {χi }(18 ≤ i ≤ 42)} .

Then m(Co3 ), M (Co3 ), C1 = (X1 , K1 ), C2 = (X2 , K2 ) and C3 = C1 ∨ C2 are five supercharacter
theories for Co3 . Thus s(Co3 ) ≥ 5, as required.
cL }
• The McLaughlin group M cL. Suppose Irr(M cL) = {χi }1≤i≤24 and Con(M cL) = {xM
1≤i≤24 .
i
We now define:
K1

=


McL
McL
{1}, {xMcL
}(2 ≤ i ≤ 15), {xMcL
}(18 ≤ i ≤ 24) ,
i
16 , x17 }, {xi

X1

=

K2

=

{{1}, {χi }(2 ≤ i ≤ 6), {χ7 , χ8 }, {χi }(9 ≤ i ≤ 24)} ,

McL
McL
{1}, {xMcL
}(2 ≤ i ≤ 12), {xMcL
}(15 ≤ i ≤ 24) ,
i
13 , x14 }, {xi

X2

=

{{1}, {χi }(2 ≤ i ≤ 20), {χ21 , χ22 }, {χi }(23 ≤ i ≤ 24)} .

Since m(M cL), M (M cL), C1 = (X1 , K1 ), C2 = (X2 , K2 ) and C3 = C1 ∨C2 are five supercharacter
theories of M cL, s(M cL) ≥ 5.
5

• The Suzuki group Suz. The irreducible characters and conjugacy classes for the Suzuki group
Suz are Irr(Suz) = {χi }1≤i≤43 and Con(Suz) = {xSuz
}1≤i≤43 , respectively. Define:
i
K1

=


Suz
Suz
{1}, {xSuz
}(2 ≤ i ≤ 40), {xSuz
i
41 , x42 }, {x43 } ,

X1

=

K2

=

{{1}, {χi }(2 ≤ i ≤ 24), {χ25 , χ26 }, {χi }(27 ≤ i ≤ 43)} ,

Suz
Suz
}(37 ≤ i ≤ 43) ,
{1}, {xSuz
}(2 ≤ i ≤ 34), {xSuz
i
35 , x36 }, {xi

X2

=

{{1}, {χi }(2 ≤ i ≤ 12), {χ13 , χ14 }, {χi }(15 ≤ i ≤ 43)} .

Since this group has five supercharacter theories m(Suz), M (Suz), C1 = (X1 , K1 ), C2 =
(X2 , K2 ) and C3 = C1 ∨ C2 , we conclude that s(Suz) ≥ 5.
This completes the proof.



Theorem 2.5. The Monster sections have at least five supercharacter theories.
Proof. The Monster sections are eight simple groups He, HN , T h, F i22 , F i23 , F i′24 , B and M . We
will present five supercharacter theories in each case as follows:
• The Held group He. Suppose Irr(He) = {χi }1≤i≤33 and Con(He) = {xHe
i }1≤i≤33 are irreducible characters and conjugacy classes of the group He, respectively. Define:


He
He
He
{1}, {xHe
i }(2 ≤ i ≤ 27), {x28 , x29 }, {xi }(30 ≤ i ≤ 33) ,

K1

=

X1
K2

= {{1}, {χi }(2 ≤ i ≤ 29), {χ30 , χ31 }, {χi }(32 ≤ i ≤ 33)} ,

He
He
He
=
{1}, {xHe
i }(2 ≤ i ≤ 25), {x26 , x27 }, {xi }(28 ≤ i ≤ 33) ,

X2

= {{1}, {χi }(2 ≤ i ≤ 6), {χ7 , χ8 }, {χi }(9 ≤ i ≤ 33)} .

Since m(He), M (He), C1 = (X1 , K1 ), C2 = (X2 , K2 ) and C3 = C1 ∨ C2 are five supercharacter
theories for He, s(He) ≥ 5.
• The Harada-Norton group HN . This group has exactly 54 conjugacy classes and irreducible
characters. Suppose Irr(HN ) = {χi }1≤i≤54 and Con(HN ) = {xHN
}1≤i≤54 . We also define:
i
K1

=


HN
HN
{1}, {xHN
}(2 ≤ i ≤ 38), {xHN
}(41 ≤ i ≤ 54) ,
i
39 , x40 }, {xi

X1

=

K2

=

{{1}, {χi }(2 ≤ i ≤ 50), {χ51 , χ52 }, {χi }(53 ≤ i ≤ 54)} ,

HN
{1}, {xHN
}(2 ≤ i ≤ 52), {xHN
i
53 , x54 } ,

X2

=

{{1}, {χi }(2 ≤ i ≤ 34), {χ35 , χ36 }, {χi }(37 ≤ i ≤ 54)} .

By Lemma 2.1 and [10, Proposition 2.16], C1 = (X1 , K1 ), C2 = (X2 , K2 ) and C3 = C1 ∨ C2 are
supercharacter theories of HN and apart from supercharacter theories m(HN ) and M (HN ),
it concludes that m(HN ) ≥ 5.
• The Thompson group T h. The Thompson group T h has exactly 48 conjugacy classes and
irreducible characters. We assume that Irr(T h) = {χi }1≤i≤48 and Con(T h) = {xTi h }1≤i≤48
and define:


{1}, {xTi h}(2 ≤ i ≤ 24), {xT25h , xT26h }, {xTi h }(27 ≤ i ≤ 39), {xT40h , xT41h }, {xTi h }(42 ≤ i ≤ 48) ,

K1

=

X1
K2

= {{1}, {χi }(2 ≤ i ≤ 8), {χ9 , χ10 }, {χi }(11 ≤ i ≤ 34), {χ35 , χ36 }, {χi }(37 ≤ i ≤ 48)} ,

=
{1}, {xTi h}(2 ≤ i ≤ 36), {xT37h , xT38h }, {xTi h }(39 ≤ i ≤ 48) ,

X2

= {{1}, {χi }(2 ≤ i ≤ 21), {χ22 , χ23 }, {χi }(24 ≤ i ≤ 48)} .
6

Again apply Lemma 2.1 and [10, Proposition 2.16] to deduce that C1 = (X1 , K1 ), C2 = (X2 , K2 )
and finally C3 = C1 ∨ C2 are supercharacter theories for T h. Thus s(T h) ≥ 5.
• The Fischer group F i22 . The Fischer group F i22 has exactly 65 conjugacy classes and irreducible characters. Set Irr(F i22 ) = {χi }1≤i≤65 and Con(F i22 ) = {xFi i22 }1≤i≤65 . We also
define:
K1

=

n
o
i22
i22
F i22
F i22
i22
F i22
F i22
{1}, {xF
}(2 ≤ i ≤ 35), {xF
}(38 ≤ i ≤ 60), {xF
}(63 ≤ i ≤ 65) ,
36 , x37 }, {xi
61 , x62 }, {xi
i

X1

=

K2

=

{{1}, {χi }(2 ≤ i ≤ 39), {χ40 , χ41 }, {χi }(42 ≤ i ≤ 50), {χ51 , χ52 }, {χi }(53 ≤ i ≤ 65)} ,
n
o
i22
i22
F i22
F i22
{1}, {xF
}(2 ≤ i ≤ 54), {xF
}(57 ≤ i ≤ 65) ,
i
55 , x56 }, {xi

X2

=

{{1}, {χi }(2 ≤ i ≤ 42), {χ43 , χ44 }, {χi }(45 ≤ i ≤ 65)} .

Since C1 = (X1 , K1 ), C2 = (X2 , K2 ) and finally C3 = C1 ∨ C2 , are three supercharacter theories
of F i22 different from m(F i22 ) and M (F i22 ), s(F i22 ) ≥ 5.
• The Fischer group F i23 . This group has exactly 98 conjugacy classes and irreducible characters.
Set Irr(F i23 ) = {χi }1≤i≤98 and Con(F i23 ) = {xFi i23 }1≤i≤98 . Define:
K1

=

n
o
i23
i23
F i23
F i23
{1}, {xF
}(2 ≤ i ≤ 79), {xF
}(82 ≤ i ≤ 98) ,
i
80 , x81 }, {xi

X1

=

K2

=

{{1}, {χi }(2 ≤ i ≤ 16), {χ17 , χ18 }, {χi }(19 ≤ i ≤ 98)} ,
o
n
i23
F i23
F i23
i23
F i23
F i23
i23
}(82 ≤ i ≤ 98) ,
}(65 ≤ i ≤ 79), {xF
}(2 ≤ i ≤ 62), {xF
{1}, {xF
80 , x81 }, {xi
63 , x64 }, {xi
i

X2

=

{{1}, {χi }(2 ≤ i ≤ 14), {χ15 , χ16 }, {χ17 , χ18 }, {χi }(19 ≤ i ≤ 98)} ,

C1

=

(X1 , K1 ), C2 = (X2 , K2 ), C3 = C1 ∨ C2 .

Since {m(F i23 ), M (F i23 ), C1 , C2 , C3 } ⊆ Sup(F i23 ), s(F i23 ) ≥ 5.
• The Fischer group F i′24 . The largest Fischer group has exactly 108 conjugacy classes and
F i′
irreducible characters. Set Irr(F i′24 ) = {χi }1≤i≤108 and Con(F i′24 ) = {xi 24 }1≤i≤108 . Define:
K1

=

o
n
F i′
F i′
F i′
F i′
{1}, {xi 24 }(2 ≤ i ≤ 105), {x10624 , x10724 }, {x10824 } ,

X1

=

K2

=

{{1}, {χi}(2 ≤ i ≤ 98), {χ99 , χ100 }, {χi }(101 ≤ i ≤ 108)} ,
o
n
F i′
F i′
F i′
F i′
{1}, {xi 24 }(2 ≤ i ≤ 80), {x81 24 , x82 24 }, {xi 24 }(83 ≤ i ≤ 108) ,

X2

=

{{1}, {χi}(2 ≤ i ≤ 100), {χ101 , χ102 }, {χi }(103 ≤ i ≤ 108)} .

Since m(F i′24 ), M (F i′24 ), C1 = (X1 , K1 ), C2 = (X2 , K2 ) and C3 = C1 ∨ C2 are supercharacter
theories of F i24 , s(F i24 ) ≥ 5.
• The Baby Monster group B. This group has exactly 184 conjugacy classes and irreducible
characters. Suppose Irr(B) = {χi }1≤i≤184 , Con(B) = {xB
i }1≤i≤184 and define:
K1

=


B
B
B
{1}, {xB
i }(2 ≤ i ≤ 177), {x178 , x179 }, {xi }(180 ≤ i ≤ 184) ,

X1

=

K2

=

{{1}, {χi }(2 ≤ i ≤ 177), {χ178 , χ179 }, {χi }(180 ≤ i ≤ 184)} ,

B
B
B
{1}, {xB
i }(2 ≤ i ≤ 171), {x172 , x173 }, {xi }(174 ≤ i ≤ 184) ,

X2

=

{{1}, {χi }(2 ≤ i ≤ 20), {χ21 , χ22 }, {χi }(23 ≤ i ≤ 184)} .

Since m(B), M (B), C1 = (X1 , K1 ), C2 = (X2 , K2 ) and C3 = C1 ∨ C2 are supercharacter theories
of B, s(B) ≥ 5, as desired.
7

• The Monster group M . The largest sporadic group M has exactly 198 conjugacy classes and
irreducible characters. Set Irr(M ) = {χi }1≤i≤194 and Con(M ) = {xM
i }1≤i≤194 .
K1

=


M
M
{1}, {xM
i }(2 ≤ i ≤ 192), {x193 , x194 } ,

X1

=

K2

=

{{1}, {χi}(2 ≤ i ≤ 46), {χ47 , χ48 }, {χi }(49 ≤ i ≤ 194)} ,

M
M
M
{1}, {xM
i }(2 ≤ i ≤ 188), {x189 , x190 }, {xi }(191 ≤ i ≤ 194) ,

X2

=

{{1}, {χi}(2 ≤ i ≤ 123), {χ124 , χ125 }, {χi }(126 ≤ i ≤ 194)} .

We can see that the group M has at least 5 supercharacter theories as m(M ), M (M ), C1 =
(X1 , K1 ), C2 = (X2 , K2 ) and C3 = C1 ∨ C2 . This shows that s(M ) ≥ 5.
This completes the proof.



Theorem 2.6. The Pariahs have at least five supercharacter theories.
Proof. The Pariahs are six sporadic groups J1 , O′ N , J3 , Ru, J4 and Ly. Our main proof will consider
six separate cases as follows:
• Janko group J1 . The first Janko group J1 has exactly 15 conjugacy classes and irreducible
characters. Set Irr(J1 ) = {χi }1≤i≤15 and Con(J1 ) = {xJi 1 }1≤i≤15 .
K1

=

o
n
{1}, {xJi 1 }(2 ≤ i ≤ 3), {xJ4 1 , xJ5 1 }, {xJi 1 }(6 ≤ i ≤ 7), {xJ8 1 , xJ9 1 }, {xJ101 }, {xJ111 , xJ121 }, {xJi 1 }(13 ≤ i ≤ 15) ,

X1

=

K2

=

{{1}, {χ2 , χ3 }, {χi }(4 ≤ i ≤ 6), {χ7 , χ8 }, {χi }(9 ≤ i ≤ 12), {χ13 , χ14 }, {χ15 }} ,
o
n
{1}, {xJi 1 }(2 ≤ i ≤ 12), {xJ131 , xJ141 , xJ151 } ,

X2

=

{{1}, {χi }(2 ≤ i ≤ 8), {χ9 , χ10 , χ11 }, {χi }(12 ≤ i ≤ 15)} .

Since m(J1 ), M (J1 ), C1 = (X1 , K1 ), C2 = (X2 , K2 ) and C3 = C1 ∨C2 are supercharacter theories
of J1 , s(J1 ) ≥ 5.
• The O′ Nan group O ′ N . The O′ Nan group O ′ N has exactly 30 conjugacy classes and irreducible
′
characters. Set Irr(O ′ N ) = {χi }1≤i≤30 and Con(O ′ N ) = {xiO N }1≤i≤30 .
K1

=

n
o
′
′
O′ N
N
O′ N
O′ N
{1}, {xiO N }(2 ≤ i ≤ 21), {x22
, xO
,
x
},
{x
}(25
≤
i
≤
30)
,
23
24
i

X1

=

K2

=

{{1}, {χi}(2 ≤ i ≤ 25), {χ26 , χ27 , χ28 }, {χi }(29 ≤ i ≤ 30)} ,
n
o
′
′
O′ N
N
O′ N
{1}, {xiO N }(2 ≤ i ≤ 26), {x27
, xO
}(29 ≤ i ≤ 30) ,
28 }, {xi

X2

=

{{1}, {χi}(2 ≤ i ≤ 28), {χ29 , χ30 }} .

Apply Lemma 2.1 and [10, Proposition 2.16] to deduce that C1 = (X1 , K1 ), C2 = (X2 , K2 ) and
C3 = C1 ∨ C2 are supercharacter theories of O′ N . Hence s(O′ N ) ≥ 5, as desired.
• The Janko group J3 . We assume that Irr(J3 ) = {χi }1≤i≤21 and Con(J3 ) = {xJi 3 }1≤i≤21 .
Define:
K1

=

n
o
{1}, {xJi 3 }(2 ≤ i ≤ 9), {xJ103 , xJ113 , xJ123 }, {xJi 3 }(13 ≤ i ≤ 21) ,

X1

=

K2

=

{{1}, {χi }(2 ≤ i ≤ 13), {χ14 , χ15 , χ16 }, {χi }(17 ≤ i ≤ 21)} ,
n
o
{1}, {xJi 3 }(2 ≤ i ≤ 19), {xJ203 , xJ213 } ,

X2

=

{{1}, {χ2 , χ3 }, {χi }(4 ≤ i ≤ 21)} .

Since m(J3 ), M (J3 ), C1 = (X1 , K1 ), C2 = (X2 , K2 ) and C3 = C1 ∨C2 are supercharacter theories
of J3 , s(J3 ) ≥ 5.
8

• The Rudvalis group Ru. Set Irr(Ru) = {χi }1≤i≤36 and Con(Ru) = {xRu
i }1≤i≤36 and define:
K1

=


Ru
Ru
{1}, {xRu
i }(2 ≤ i ≤ 34), {x35 , x36 } ,

X1

=

K2

=

{{1}, {χi }(2 ≤ i ≤ 33), {χ34 , χ35 }, {χ36 }} ,

Ru
Ru
Ru
Ru
{1}, {xRu
i }(2 ≤ i ≤ 31), {x32 , x33 , x34 }, {xi }(35 ≤ i ≤ 36) ,

X2

=

{{1}, {χi }(2 ≤ i ≤ 16), {χ17 , χ18 , χ19 }, {χi }(20 ≤ i ≤ 36)} .

Apply again Lemma 2.1 and [10, Proposition 2.16] to deduce that the Rudvalis group Ru
has at least five supercharacter theories as m(Ru), M (Ru), C1 = (X1 , K1 ), C2 = (X2 , K2 ) and
C3 = C1 ∨ C2 . Hence s(Ru) ≥ 5, as required.
• The Janko group J4 . This group has exactly 62 conjugacy classes and irreducible characters.
Define Irr(J4 ) = {χi }1≤i≤62 , Con(J4 ) = {xJi 4 }1≤i≤62 and define:
K1

=

n
o
{1}, {xJi 4 }(2 ≤ i ≤ 49), {xJ504 , xJ514 , xJ524 }, {xJi 4 }(53 ≤ i ≤ 62) ,

X1

=

K2

=

{{1}, {χi }(2 ≤ i ≤ 52), {χ53 , χ54 , χ55 }, {χi }(56 ≤ i ≤ 62)} ,
n
o
{1}, {xJi 4 }(2 ≤ i ≤ 42), {xJ434 , xJ444 , xJ454 }, {xJi 4 }(46 ≤ i ≤ 49), {xJ504 , xJ514 , xJ524 }, {xJi 4 }(53 ≤ i ≤ 62) ,

X2

=

{{1}, {χi }(2 ≤ i ≤ 52), {χ53 , χ54 , χ55 }, {χ56 , χ57 , χ58 }, {χi }(59 ≤ i ≤ 62)} .

The Janko group J4 has at least five supercharacter theories as m(J4 ), M (J4 ), C1 = (X1 , K1 ),
C2 = (X2 , K2 ) and C3 = C1 ∨ C2 . Therefore, s(J4 ) ≥ 5.
• The Lyons group Ly. The Lyons group Ly has exactly 53 irreducible characters and conjugacy
classes. Set Irr(Ly) = {χi }1≤i≤53 and Con(Ly) = {xLy
i }1≤i≤53 and define:
K1

=

n
o
Ly
Ly
Ly
Ly
Ly
Ly
{1}, {xLy
i }(2 ≤ i ≤ 37), {x38 , x39 , x40 , x41 , x42 }, {xi }(43 ≤ i ≤ 53) ,

X1

=

K2

=

{{1}, {χi }(2 ≤ i ≤ 38), {χ39 , χ40 , χ41 , χ42 , χ43 }, {χi }(44 ≤ i ≤ 53)} ,
n
o
Ly
Ly
Ly
Ly
Ly
Ly
Ly
Ly
Ly
{1}, {xLy
i }(2 ≤ i ≤ 37), {x38 , x39 , x40 , x41 , x42 }, {xi }(43 ≤ i ≤ 50), {x51 , x52 , x53 } ,

X2

=

{{1}, {χi }(2 ≤ i ≤ 25), {χ26 , χ27 , χ28 }, {χi }(29 ≤ i ≤ 38), {χ39 , χ40 , χ41 , χ42 , χ43 }, {χi }(44 ≤ i ≤ 53)} .

Since the group Ly has at least five supercharacter theories as m(Ly), M (Ly), C1 = (X1 , K1 ),
C2 = (X2 , K2 ) and C3 = C1 ∨ C2 , s(Ly) ≥ 5.
This completes the proof.



Lemma 2.7. The Janko group J2 has at least three supercharacter theories.
Proof. A simple investigation shows that the supercharacter theories of the Janko group J2 are m(J2 ),
M (J2 ), C = (X , K) such that
K

=

n
o
{1}, {xJi 2 }(2 ≤ i ≤ 6), {xJ7 2 , xJ8 2 }, {xJ9 2 , xJ102 }, {xJi 2 }(11 ≤ i ≤ 14), {xJ152 , xJ162 }, {xJ172 , xJ182 }, {xJ192 }, {xJ202 , xJ212 } ,

X

=

{{1}, {χ2 , χ3 }, {χ4 , χ5 }, {χi }(6 ≤ i ≤ 7), {χ8 , χ9 }, {χi }(10 ≤ i ≤ 13), {χ14 , χ15 }, {χ16 , χ17 }, {χi }(18 ≤ i ≤ 20)} ,

proving the lemma.



We run a GAP program to be sure that s(J2 ) = 3, but our program after some days stopped,
because it needs a huge amount of RAM.
Conjecture 2.8. s(J2 ) = 3.
9

3. Supercharacter Theory Construction for Alternating and Suzuki Groups
The intention of this section is to move a step towards a classification of the finite simple groups
with exactly three and four supercharacter theories. We start with the cyclic group of order p.
Lemma 3.1. Suppose p is prime. Then,
(1) s(Zp ) = 3 if and only if p = 5,
(2) s(Zp ) = 4 if and only if p is a Sophie Germain prime.
Proof. It is clear that p is an odd prime. By [9, Theorem 6.32 and Table 1], s(Zp ) = d(p − 1).

(1) s(Zp ) = 3. In this case, s(Zp ) = d(p − 1) and so p ≥ 5. Since p − 1 is an even integer, the case
of p > 5 cannot be occurred and so p = 5, as desired.
(2) s(Zp ) = 4. Since d(p − 1) = 4, p − 1 = q 3 , q is prime, or p − 1 = 2r, where r is prime. If
p − 1 = q 3 then q = 2 and p = 9, a contradiction. So, p − 1 = 2r, where r is prime. This shows
that p is a Sophie Germain prime.

This completes the proof.



The following well-known results are crucial in the classification of alternating simple groups with
exactly three or four supercharacter theories.
Theorem 3.2. The following are hold:
(1) (Berggren [4]) Every irreducible characters of the alternating group An are real valued if and
only if n ∈ {1, 2, 5, 6, 10, 14}.
(2) (Grove [8, Proposition 8.2.1]) If K is a conjugacy class in Sn , K ⊂ An , and σ ∈ K, then K
is a conjugacy class in An if and only if some odd elements of Sn commutes with σ; if that is
not the case, then the conjugacy class K splits as the union of two An −classes, each of size
|K|/2. If λ is the (partition) type of σ then K splits if and only if the parts of λ are all odd
and all different from each other.
(3) Suppose x ∈ An is a product of r pair-wise disjoint cycles including all fixed points as singleton
P
rj −1
is odd.
cycles. Then xAn is non-real if and only if m
j=1 2
Theorem 3.3. The simple alternating group An has exactly three supercharacter theories if and only
if n = 5 or 7. There is no simple alternating groups with exactly four supercharacter theories.
Proof. Suppose n ≥ 5. It is easy to see that the alternating groups A5 and A7 have exactly three
supercharacter theories. Our main proof will consider three separate cases as follows:
(1) All character values of An are real. Since n ≥ 5, Theorem 3.2(1) implies that n = 5, 6, 10 or
14. In what follows three non-trivial supercharacter theories for the alternating groups A6 , A10
and A14 are presented.
(a) By a GAP program, one can see that the group A5 has exactly 3 supercharacter theories
m(A5 ), M (A5 ) and C = (X , K) such that
K

=

n
o
A5
A5
A5
5
{1}, {xA
2 }, {x3 }, {x4 , x5 } ,

X

=

{{1}, {χ2, χ3 }, {χi }(4 ≤ i ≤ 5)} .

A5
A5
A5
A5
5
Here, Irr(A5 ) = {χ1 , χ2 , χ3 , χ4 , χ5 } and Con(A5 ) = {xA
1 , x2 , x3 , x4 , x5 }.
10

A6
A6
6
(b) Suppose Irr(A6 ) = {χ1 , χ2 , . . . , χ7 } and Con(A6 ) = {xA
1 , x2 , . . . , x7 }. Define C1 =
(X1 , K1 ) and C2 = (X2 , K2 ) as follows:

K1

=

n
o
A6
A6
A6
A6
A6
6
{1}, {xA
2 }, {x3 , x4 }, {x5 }, {x6 , x7 } ,

X1

=

K2

=

{{1}, {χ2, χ3 }, {χ4 , χ5 }, {χ6 }, {χ7 }} ,
n
o
A6
A6
A6
A6
A6
6
{1}, {xA
,
x
},
{x
,
x
},
{x
},
{x
}
,
2
5
3
4
6
7

X2

=

{{1}, {χ2, χ3 , χ7 }, {χ4 }, {χ5 }, {χ6 }} ,

If C3 = C1 ∨ C2 then C3 is a supercharacter theory for A6 different from C1 , C2 , m(A6 )
and M (A6 ). Therefore, A6 has at least five supercharacter theories.
A10
A10
10
(c) Suppose Irr(A10 ) = {χ1 , χ2 , . . . , χ24 }, Con(A10 ) = {xA
1 , x2 , . . . , x24 }, C1 = (X1 , K1 ),
C2 = (X2 , K2 ) and C3 = C1 ∨ C2 , where
K1

=

n
o
A10
A10
A10
10
},
{x
{1}, {xA
,
x
}(22
≤
i
≤
24)
,
}(2
≤
i
≤
19),
{x
21
20
i
i

X1

=

K2

=

{{1}, {χi}(2 ≤ i ≤ 19), {χ20 , χ21 }, {χi }(22 ≤ i ≤ 24)} ,
n
o
A10
10
10
}(2 ≤ i ≤ 22), {xA
{1}, {xA
23 , x24 } ,
i

X2

=

{{1}, {χi}(2 ≤ i ≤ 11), {χ12 , χ13 }, {χi }(14 ≤ i ≤ 24)} .

Then Ci = (Xi , Ki ), 1 ≤ i ≤ 3, are three supercharacter theories of A10 different from
m(A10 ) and M (A10 ).
(d) We claim that the alternating group A14 has at least 5 supercharacter theories. These are
m(A14 ), M (A14 ), C1 = (X1 , K1 ), C2 = (X2 , K2 ) and C3 = C1 ∨ C2 such that
K1

=

o
n
A14
14
14
}(2 ≤ i ≤ 70), {xA
{1}, {xA
71 , x72 } ,
i

X1

=

K2

=

{{1}, {χi}(2 ≤ i ≤ 19), {χ20 , χ21 }, {χi }(22 ≤ i ≤ 72)} ,
n
o
A14
A14
14
14
}(2 ≤ i ≤ 67), {xA
{1}, {xA
}(70 ≤ i ≤ 72) ,
68 , x69 }, {xi
i

X2

=

{{1}, {χi}(2 ≤ i ≤ 56), {χ57 , χ58 }, {χi }(59 ≤ i ≤ 72)} .

(2) The alternating group An has exactly one pair of non-real valued irreducible character. In
this case, we will prove n ∈ {7, 9, 11, 15, 18, 19, 23}. Since the number of non-real irreducible
characters is equal to the number of conjugacy classes that are not preserved by the inversion
mapping, we need those n for which An has exactly two such conjugacy classes. Suppose
x ∈ An . By [12, Proposition 12.17(2)], xSn = xAn ∪(12)x(12)An if and only if CSn (x) = CAn (x).
n−r
is odd, where r is the number
By Theorem 3.2(2), the last one is satisfied if and only if
2
of cycles in decomposition of x. Our aim is to find all natural numbers n such that An has
exactly one non-real class. Our main proof will consider the following separate cases:
(a) n ≡ 0 (mod 4). If n ≥ 8 then [1, n − 1] and [3, n − 3] are two partitions with given
properties. So, n = 4 which contradicts by our main assumption that n ≥ 5.
(b) n ≡ 1 (mod 4). If n ≥ 13 then [1, 3, n − 4] and [1, 5, n − 6] are two partitions with this
property that all parts are odd and n−3
2 (r = 3) is odd. Thus n = 5 or 9. By Theorem
3.2(1), the alternating group A5 does not have non-real class and the alternating group
A9 has exactly a unique pair of non-real class.
(c) n ≡ 2 (mod 4). If n ≥ 22 then [1, 3, 5, n − 9] and [1, 3, 7, n − 11] are two partitions with
given properties and so n = 6, 10, 14 or 18. By Theorem 3.2(1), the alternating groups
11

A6 , A10 and A14 don’t have non-real conjugacy class, but the alternating group A18 has
a unique pair of conjugate non-real characters.
(d) n ≡ 3 (mod 4). If n ≥ 31 then [1, 3, 5, 7, n − 16] and [1, 3, 5, 9, n − 18] are two partitions
with this property that all parts are odd and n−5
2 (r = 5) is odd. Thus n = 7, 11, 15, 19, 23
or 27. Since [27] and [1, 5, 7, 9, 11] are two partitions with mentioned properties, the case
of n = 27 cannot be happened. Other cases are solution of our problem.
Hence, the alternating group An , n ≥ 5, has a unique non-real conjugacy class if and only
if n ∈ {7, 9, 11, 15, 18, 19, 23}. A simple calculations by GAP shows that the alternating group
A7 has exactly three supercharacter theories m(A7 ), M (A7 ) and C = (X , K), where
K

=

n
o
A7
A7
7
}(2
≤
i
≤
7),
{x
{1}, {xA
,
x
}
,
8
9
i

X

=

{{1}, χ2 , {χ3 , χ4 }, {χi }(5 ≤ i ≤ 9)} .

The alternating group A9 has at least 5 supercharacter theories m(A9 ), M (A9 ), C1 = (X1 , K1 ),
C2 = (X2 , K2 ) and C3 = C1 ∨ C2 . The partitions of conjugacy classes and irreducible characters
of C1 and C2 are defined as follows:
K1

=

n
o
A9
A9
A9
9
}(2
≤
i
≤
12),
{x
{1}, {xA
,
x
}(15
≤
i
≤
18)
,
},
{x
13
14
i
i

X1

=

K2

=

{{1}, {χ2}, {χ3 , χ4 }, {χi }(5 ≤ i ≤ 18)} ,
n
o
A9
A9
9
{1}, {xA
i }(2 ≤ i ≤ 16), {x17 , x18 } ,

X2

=

{{1}, {χi }(2 ≤ i ≤ 6), {χ7 , χ8 }, {χi }(9 ≤ i ≤ 18)} .

The alternating groups A11 , A15 , A18 , A19 and A23 has at least one non-real irreducible character and one non-rational irreducible real character. So, by Lemma 2.1 it has at least five
supercharacter theories m, M , C1 , C2 and C3 = C1 ∨ C2 , proving this case.
(3) The alternating group An has at least two pairs of non-real valued irreducible characters. We
first assume that n > 24. If A = {χ(x) | χ ∈ Irr(G)} and Q(A) denotes the filed generated
by Q and A then by [14, Theorem], the character table of An has both irrational and non-real
character values. On the other hand, by [13, Theorem 2.5.13], each row or column of the
character table of An contains at most one pair of irrational numbers. Now by Lemma 2.1, we
have at least five supercharacter theories, as required.
Next we assume that n ≤ 24. Set Γ1 = {5, 6, 7, 9, 10, 11, 14, 15, 18, 19, 23} and Γ2 =
{8, 12, 13, 16, 17, 20, 21, 22, 24}. If n ∈ Γ1 then the number of supercharacter theories of An
are investigated in Cases (1) and (2). So, we have to prove that s(An ) ≥ 5, when n ∈ Γ2 . By
an easy calculation with GAP, one can see that if n ∈ Γ2 then An has at least two pairs of
non-real valued irreducible characters and by Lemma 2.1, s(An ) ≥ 5.
Hence the result.



In the end of this paper we prove that the simple Suzuki group Sz(q), q = 22n+1 has at least six
super character theories.
12

Theorem 3.4. The Suzuki group Sz(q) has at least 6 supercharacter theories as: m(Sz(q)), M (Sz(q)),
C1 = (X1 , K1 ), C2 = (X2 , K2 ), C3 = (X3 , K3 ) and C4 = (X4 , K4 ), where
K1

=

{{1}, {ρ, ρ−1}, Con(G(q)) − {ρ, ρ−1 }},

X1

=

{{1}, {W1, W2 }, Irr(G(q)) − {W1 , W2 }},

K2

=

{{1}, {π0}, Con(G(q)) − {π0 }},

X2

=

{{1}, {Xi}, Irr(G(q)) − {Xi }},

K3

=

{{1}, {π1}, Con(G(q)) − {π1 }},

X3

=

{{1}, {Yj }, Irr(G(q)) − {Yj }},

K4

=

{{1}, {π2}, Con(G(q)) − {π2 }},

X4

=

{{1}, {Zk }, Irr(G(q)) − {Zk }}.

Proof. The schematic form of the character table of Sz(q), q = 22n+1 , is shown in Table 1, see [17, 18]
for details.
Table 1. The Schematic Form of the Character Table of Sz(q).
Irreducible Characters
Degrees
#Irreducible Characters
2
X
q
1
2
Xi
q +1
q/2 − 1
Yj
(q − r + 1)(q − 1)
(q + r)/4
Zk
(q + r + 1)(q − 1)
(q − r)/4
Wl
r(q − 1)/2
2
In this table, the first column designates the characters, the second column indicates the degrees,
and the last one is the number of characters of each degree.
Suppose 2q = r 2 . The Suzuki group Sz(q) contains cyclic groups of order q − 1, q + r + 1 and
q − r + 1. These subgroups are denoted by A0 , A1 and A2 , respectively. We also assume that πi is a
typical non-identity element of Ai , i = 0, 1, 2, σ = (0, 1) and ρ = (1, 0).
Let ε0 , ε1 and ε2 be a (q − 1)th , a primitive (q + r + 1)th and a (q − r + 1)th root of unity. We
also assume that ξ0 , ξ1 and ξ2 are generators of A0 , A1 and A2 , respectively. Define εi0 , εi1 and εi2 as
follows:
−ij
εi0 (ξ0j ) = εij
0 + ε0 ; ( i = 1, . . . , q/2 − 1),

ikq
−ik
εi1 (ξ1k ) = εik
+ ε−ikq
; ( i = 1, . . . , q + r),
1 + ε1 + ε1
1
ikq
−ik
+ ε−ikq
.
εi2 (ξ2k ) = εik
2 + ε2 + ε2
2

The functions εi0 , εi1 and εi2 are characters of A0 , A1 and A2 , respectively. Following Suzuki [17], the
character table of Sz(q), is computed in Table 2.
We now apply Lemma 2.1 to construct four supercharacter theories given the statement of this
theorem. By considering m(Sz(q)) and M (Sz(q)), it can be proved that s(Sz(q)) ≥ 6.

13

Table 2. The character table of Sz(q).

X
Xi
Yj
Zk
Wl

1
σ
ρ, ρ−1
π0
π1
π2
2
q
0
0
1
−1
−1
q2 + 1
1
1
εi0 (π0 )
0
0
j
(q − r + 1)(q − 1) r − 1
−1
0
−ε1 (π0 )
0
k
(q + r + 1)(q − 1) −r − 1
−1
0
0
−ε2 (π0 )
√
0
1
−1
r(q − 1)/2
−r/2 ±r −1/2

Acknowledgement. The authors are indebted to professors Marston Conder, Geoffrey R. Robinson
and Jeremy Rickard for some critical discussion through Group Pub Forum and MathOverFlow on
Theorem 3.3. The research of the authors are partially supported by the University of Kashan under
grant no 572760/1.

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Department of Pure Mathematics, Faculty of Mathematical Sciences, University of Kashan, Kashan
87317−53153, I. R. Iran

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