a combinatoric proof for p-regular Hook partitions

1. THE MULLINEUX MAP AND
p-REGULARIZATION FOR HOOK
PARTITIONS.
FANGYA TAN
Submitted in partial fulfillment for the degree of
Bachelor of Science in Mathematics, University at
Buffalo.
Advisor: David J. Hemmer
Date: May 2009.
2000 Mathematics Subject Classification. Primary 20C30.
Thanks to Dr. Hemmer.

2. MULLINEUX MAP 2
1. Introduction
Problems in the representation theory of the symmetric groups Sn
are often closely related to combinatorics of objects such as Young
tableaux and tabloids, and symmetric functions, and many theorems
have combinatorial proofs. In this thesis we discuss a purely combina-
torial fact (Theorem 1.1 below) about partitions which has no known
combinatorial proof. At present the only known proof uses the repre-
sentation theory of the symmetric group. We will present a combina-
torial proof for a special case of this theorem. An excellent reference
for the representation theory of the symmetric group is [4].
In characteristic zero, irreducible modules for the symmetric group
Sn are indexed by partitions λ of n and are called Specht modules Sλ
.
It is known that
Sλ
⊗ sgn ∼= S(λT )∗
where λT
is the transpose partition.
In characteristic p, the irreducible modules are indexed by p-regular
partitions, and are denoted Dλ
. It is not always the case that λT
is
p-regular when λ is. Thus the problem of determining which µ labels
the irreducible module Dλ
⊗ sgn is more complicated. The famous
Mullineux conjecture [5] gave a conjectural description of M(λ), defined
by:
Dλ
⊗ sgn ∼= DM(λ)
where Dλ ∼= Sλ
/ rad Sλ
.
James showed [3] that when λ is not p-regular, there is a p-regular
partition λR
, such that Sλ
contains a single copy of DλR
and the other
composition factors are all of the form Dµ
where λR
> µ.
Using representation theory, Bessenrodt, Olsson and Xu proved the
following:
Theorem 1.1. [2, Prop. 4.2]
(1.1) M(λR
) ≥ (λT
)R
.
Bessenrodt, Olsson and Xu remarked [2, Rmk. p.453] that there
is no known combinatorial proof of (1.1). In this thesis we present a
combinatorial proof of (1.1) in the case where λ is a hook partition.
2. Notation and Definitions
Definition 2.1. A partition λ of a positive integer n is a nonincreasing
sequence of positive integers λ = (λ1, λ2, . . . , λs) whose sum is n, i.e.
λi ≥ λi+1 ≥ 0 and λi = n.

3. MULLINEUX MAP 3
We can use a Young diagram, denoted [λ], to represent a partition
λ as illustrated in Figure 1.1, which gives the Young diagram for the
partition λ = (8, 3, 2).
Figure 1. Young diagram of (8,3,2)
For a partition λ, the conjugate, or transpose, partition λT
is defined
so that [λT
] is [λ] with the rows and columns reversed.
If λ = (8, 3, 2) as above, then the Young diagram for λT
= (3, 3, 2, 15
)
is as illustrated in diagram.
Figure 2. Young diagram of λT
We define a partition λ to be a p-regular partition if it is a partition
with no part repeating p or more times, so λ = (λ1 ≥ λ2 ≥ ...λm) and
λi = λi+p for all i.
Example: λ = (8, 23
, 1) is p-regular if and only if p > 3.
Furthermore,we define a partial order on the set of partitions of n,
called the dominance order. Say a partition λ dominates µ, and write
λ µ, if
λ1 + λ1 + ... + λi ≥ µ1 + µ2 + ... + µi
for all i.
Example: (7, 5, 3) ¤ (6, 6, 3).

4. MULLINEUX MAP 4
Definition 2.2. Let F be an arbitrary field, and let Mλ
be the F-
vector space spanned by the various λ-tabloids (see [4, p.41] for defini-
tion).
If we let Sn act on the set of λ-tabloids by
πt := πt(π ∈ Sn),
then this action is well defined. Extending this action to be linear on
Mλ
turns Mλ
an FSn-module.
Mλ
is the permutation module of Sn action on the cosets of the
Young subgroup Sλ1 ⊗ · · · × Sλs of Sn. Mλ
is a cyclic FSn-module,
generated by any one tabloid. If t is a tableau, then define kt by,
kt =
π∈Ct
(sgnπ)π
The polytabloid, et, associated with the tableau t is given by
et = tkt.
Definition 2.3. The Specht module Sλ
for the partition λ is the sub-
module of Mλ
spanned by polytabloids.
A polytabloid not only depends on tableau t but also tabloid t. All
the tabloids involved in et have coefficient ±1.
Example. If
t = 125(2.1)
43
then Kt = (1 − (14))(1 − (23)), and
et =
125
43
−
425
13
−
135
42
−
435
12
Since the Specht module Sλ
is a submodule of Mλ
. We have
Sλ
≤ Mλ
.
Definition 2.4. For λ a p-regular partition of n, we denote Dλ
to be
corresponding p-modular irreducible representation of Sn. The collec-
tion of Dλ
for λ ap-regular partition of n, where p is a prime, gives a
complete set of irreducible Sn modules..
Definition 2.5. If we tensor an irreducible Dλ
of Sn with the one
dimensional sign representation we obtain a new modular irreducible
representation. In characteristic zero, the new irreducible modular is
described by taking the transpose of the corresponding partition. In
most of other cases, it is complicate to determine the dimensions of the

5. MULLINEUX MAP 5
simple modules. The Mullineux conjecture [5] defined the Mullineux
map to be a bijective λ → λM
on the set of p-regular partitions and
conjectured this map is the same as the map λ → λP
, where Dλ
⊗sgn =
DλP
.
The Mullineux conjecture was proved by Kleshchev, so we have:
(2.2) Dλ
⊗ sgn ∼= DM(λ)
Next we describe an algorithm for computing the Mullineux map.
Definition 2.6. Mullineux symbol called the residue symbol for p-
regular partition.
Let us suppose λ is an arbitrary p-regular partition with total of
n nodes. The partition is composed of p-segments. Each p-segment
contains p nodes, except the last one, it might have p nodes or less.
We can cross out the first p-segments,which has p nodes in the longest
row, in our case we will start from head of λ, then we cross out another
p-segment by starting in the row below the original p-segment, in our
case, it would be the first row of the foot. We will carry on the process
until we left nothing to cross out. We assume a1 is the total number of
nodes in the p-rim of λ = λ(1)
and l1 stands for the number of rows in
λ. We obtain a new p-regular partition λ(2)
= n−a1 after removing the
p rim of λ(1)
. We let a2, l2 be the length of the p-rim and the number
of parts of λ(2)
. Continuing this procedure we acquire a sequence of
partitions λ = λ(1)
, λ(2)
, ..., λ(m)
,where λ(m)
= 0, and λ(m+1)
= 0. We
denote the corresponding Mullineux symbol of λ as,
(2.3) Gp(λ) =
a1 a2 · · · am
l1 l2 · · · lm
the Mullineux map Gp(λM
) is a p-regular partition satisfying
(2.4) Gp(λM
) =
a1 a2 · · · am
s1 s2 · · · sm
wheresi = ai − li + εi.
and εi is defined by:
εi =
1 if si ai
0 if p ai.
For more details on this algorithm, see [2].
Example: Let p = 3, λ = (8, 3, 2) then following the procedure above
we obtain Figure 3,
From Figure 3, we obtain the Mullineux symbols of λ and λM
:

6. MULLINEUX MAP 6
× × × × × × × ×
×
×
× ×
×
−→
3 3 2 2 2 1 1 1
2
2
1 1
1
Figure 3. Example of computing Mullineux symbol
G3(λ) =
6 5 2
3 3 1
→ G3(λM
) =
6 5 2
3 3 2
From the Mullineux symbol for λM
, we can reconstruct the partition:
• • • • • •
•
•
•
•
• •
• •
−→
3
3
2
2
2
2
2
2 1 1 1
1 1
1
Figure 4. M(832) = (643)
Definition 2.7. The residue of row i column j is j −i mod p. A ladder
is a line through the Young diagram which has slope p − 1. One easily
sees that a ladder passes through nodes of same residue.
Given an arbitrary λ, if λ is not p-regular,we can realize the p-
regularization λR
by pushing all nodes up along ladders. In other
words, we can p-regularize λ by sliding nodes up a ladder. Let us
suppose there is node λi, where i ≥ p. We start by moving d to the
right on column but it staying in the same row then sliding λi up to
p − 1 node, we call it λi. If λi is still greater the P, then we continue
the procedure again until λ is P-regular.
Example: The procedure is illustrated as follows, we have a partition
λ = (9, 17
), with prime p = 5
*
After applying the p-regularization algorithm, we obtain λR
= (9, 23
, 1)
as,
Definition 2.8. A partition or Young Diagram of the form (l − a, 1a
)
is called a hook.

7. MULLINEUX MAP 7
0 1 2 3 4 0 1 2 3
4
3
2
1
0
4
3
¢
¢
¢
¢
¢
¢
¢
¢
¢¢
¢
¢
¢
¢
¢
¢
¢
¢
¢¢
¢
¢
¢
¢
¢
¢
¢
¢
¢¢
0 1 2 3 4 0 1 2 3
4
3
2
1
0
4
3
Figure 5. Ladder algorithm
Choose a partition λ. For a node (s, t) ∈ λ, we denote by Hst the
(s, t) hook of λ, which consists of the node (s,t) together with all nodes
to the right of it in row s and all nodes below it in column t. The node
(s, λs) is called the hand of the hook while the node (λt, t) is called the
foot. The (s, t) hook length satisfies, Hst = λs + λt + 1 − s − t.
A rim p-hook is a rim hook of length p.
Bessenrodt-Olsson used representation theory to prove Theorem 1.1,
which relates the Mullineux map and the map λ → λR
. The starting
point is the following theorem of James.
Theorem 2.9 (5, Theorem 1.2.1). If λ is p-regular then Sλ
has a
unique top composition factor Dλ ∼= Sλ
/radSλ
. More generally, Sλ
has
a single composition factor isomorphic to DλR
. Moreover, if Dµ
is a
composition factor of radSλ
, then µ £ λR
.

8. MULLINEUX MAP 8
It is a well known fact that in characteristic p,the irreducible Sn
modules is Dλ
, where λ is p-regular. in characteristic zero, if we tensor
the simple modules with signature representation, which is taking the
transpose of the corresponding partition, then we will have a bijection.
We conclude that Sλ
is simple and self-dual.
(2.5) Sλ
⊗ sgn ∼= (Sλ
)∗
We also have
(2.6) DλR
⊗ sgn ∼= DM(λR)
It is obviously that DλR
∈ Sλ
, if we tensor it with the signature
representation, then DλR
⊗ sgn ∈ Sλ
⊗ sgn will holds. Finally, the
following inequality suffices,
DλR
⊗ sgn ≤ Sλ
)∗
,
we can also state it as
DM(λR)
≤ (Sλ
)∗
.
This inequality yields (λT
)R
≤ M(λR
), the conjecture was made by
James.
3. Hooks
Our goal is to give a combinatorial proof of Theorem 1.1 for the case
when λ is a hook partition. In this section we consider some preliminary
lemmas about hooks. Assume in this section that λ is a hook with an
arm length of a nodes, a leg length b nodes and p represents a prime
greater than 2. We formulated λ as λ = (a, 1b
).
For a hook λ = (a, 1b
), we define ¯λ by adding p nodes to the first
part of λ, so ¯λ = (a + p, 1b
).
We define ˜λ by adding p nodes to the foot of λ. In formula, ˜λ =
(a, 1b+p
) Now we attempt to explore the relations between (λT
)R
and(¯λT
)R
,
M(¯λT
)R
and M(λT
)R
, (λT
)R
and (˜λT
)R
, also M(λT
)R
and M(˜λT
)R
.
Lemma 3.1. Take (λT
)R
and add a rim p-hook with head in the first
or second row, we obtain (¯λT
)R
.
Proof. Let us assume we have a p-hook λ = (a, 1b
).
Case 1: Let a ≤ p, b p, with prime p.
By assumption, we know the partition is p-regular. It is clear that
λT
= (b + 1, 1a−1
) and ¯λ = (a + p, 1b
). Let us take the transpose of ¯λ,
then we get a new partition ¯λT
= (b+1, 1a+p−1
. Since a+p−1 ≥ p, we

9. MULLINEUX MAP 9
b
a
× × × . . . ×
×
×
...
×
Figure 6. λ
need to regularize ¯λT
. By applying the ladder algorithm, all the nodes
greater than p are placed in the second column but never in the third
since a ≤ p. Finally we obtain a new diagram for ¯λT
.
¯λT
satisfies,
× × × . . .
×
×
×
×
×
×
...
×
¯λT
b
‚
−→
× × × . . .
•
•
×
×
×
×
...
...
•×
(¯λT
)R
Figure 7. P regularize of ¯λT
the sum of foot “×” adds up to a + p − 1. If we already have a − 1
“×”, then we are forced to have another p “×”, which is the same as
adding a rim p-hook to λT
. We let m be the differences between a − 1
and p − 1, then if we add a rim p-hook to ¯λT
from the second row, we
have p−1−(a−1)+(a−1)+a = p+a−1. Lemma 1 is true for the case 1.
Case 2: let a p, b p with prime p.
Let’s suppose λ is (a, 1b
), then λT
is (b + 1, 1a−1
).
We illustrate the relation between λT
and λT
)R
as the diagrams in
Figure 8.
We add p nodes to the first row of λ to get ¯λ = (a + p, 1b
), then
we take the transpose, (¯λ)T
= (b + 1, 1a+p−1
). We can get (¯λT
)R
by

10. MULLINEUX MAP 10
× × × . . .
×
×
×
×
×
×
...
×
×
λT
−→
×× × . . .
•
•
•
×
×
×
...
×
(λT
)R
Figure 8. p-regularization of λT
implementing Ladder algorithm p-times. This implies that if we were
to add a rim p-hook to (λT
)R
then we would obtain (¯λT
)R
. Lemma 1
is true for case 2.
If ¯λ = (a + p, 1b
), we add a rim p-hook to ¯λT
from the first row to
obtain (¯λT
)R
if and only if (a + p) ≥ (1 + b)2
. Otherwise, we add a rim
p-hook to ¯λT from the second row.
We let m be the residue of (a−1)/p. the foot of (λT
)R
is n(p−1)+m.
the foot of (¯λT
)R
is n(p−1)+(p−1−m)+m+m+1 = n(p−1)+p+m,
which is the same as adding a rim p-hook to (λT
)R
.
Case 3: let a ≤ p, b ≥ p, with prime p.
If λ = (a, 1b
), then λT
= (b + 1, 1a−1
) is already p-regular. It yields
(¯λ)T
= (b + 1, 1a+p−1
). We can draw our conclusion from Case 1, that
if we add a rim p-hook from (λT
)R
with head in the first or second row,
we are able to get (¯λT
)R
.
Case 4: let a p, b p, with prime p.
If λ = (a, 1b
), then λT
= (b+1, 1a−1
). We need to p-regularize λT
since
a p. By applying the ladder theorem, we obtain (λT
)R
. The rest of
the proof is similar to Case 2.
Lemma 3.2. Take M(λR
), where b (p − 1)a. Adding a rim p-hook
from the p-th row, we obtain M(¯λR
).

11. MULLINEUX MAP 11
Proof. Case 1: Let a p, b p, with prime p. If λ = (a, 1b
), we
denote the Mullineux symbol of λ as
M(λ) =
(1) if (p (a + b))
(2) if (p | (a + b)).
(1) :
a + b
b + 1
−→
a + b
a
¯λ = (a + p, 1b
). Our Mullineux symbol of ¯λ is
p + b a
b + 1 1
−→
p + b a
p a
(2) :
a + b
b + 1
−→
a + b
a − 1
Our Mullineux symbol for ¯λ = (a + p, 1b
) is
p + b a
b + 1 1
−→
p + b a
p − 1 a − 1
× × × . . . ×
×
×
...
×
M(λR
)
× × × . . .
•
×
×
×
...
×
•
•
•
...
•
M(¯λR
)
Figure 9. M(λ)
We start drawing from the p-th row and up to make the first column
has total p nodes, and then tracing it along the rim. We successfully
get M(¯λR
) by adding a rim p-hook to M(λR
).
Case 2, let a = p, b p. For λ = (a, 1b
), the Mullineux map satisfies
a + b
b + 1
−→
a + b
a
.

12. MULLINEUX MAP 12
‚
× × . . . ×
×
×
...
×
M(λ)
−→
× × . . .
•
×
×
×
...
×
•
•
•
•
•
...
M(¯λ)
Since p = a, we start tracing along the rim from the pth row of
M(λR
) and up to the second row in order to get M(¯λR
). This is the
same as adding a rim p-hook to M(λR
) as we desired.
Case 3, let a ≥ p, b ≤ p, with prime p.
We already have λR
by assumption, we denote the Mullineux Symbol
as,




p + b
n times
p...p a − np
b + 1
n times
1...1 1



 −→




p + b p
n − 1 times
p...p a − np
p p − 1
n − 1 times
p − 1...p − 1 a − np



 .
The Mullineux symbol of ¯λR
= (a + p, 1b
) satisfies,




p + b
n times
p...p a − (n + 1)p
b + 1
n times
1...1 1



 −→




p + b p
n − 1 times
p...p a − (n + 1)p
p p − 1
n times
p − 1...p − 1 a − (n + 1)p



 .
For simplification reason, we replace the Mullineux map of λR
and
¯λR
by,
M(λR
) :
p + b a − p
b + 1 1
−→
p + b a − p
p a − p
M(¯λR
)
p + b p a − 2p
b + 1 1 1
−→
p + b p a − 2p
p p − 1 a − 2p
.
Comparing M(¯λR
), we observe that there is an extra p-nodes placed
in first and second row of M(λR
). We can conclude from this fact that
if we add a rim p-hook from the first row of M(λR
), we will get M(¯λR
).

13. MULLINEUX MAP 13
× × . . . ×
×
×
...
×
×
−→
M(¯λ)
× • . . .
•
• × . . . ×
×
×
...
×
×
M(¯λR
)
Figure 10. p-regularize M(λ)
Lemma 3.3. Take (λT
)R
and add p to the first row we obtain (˜λT
)R
Proof. Case 1, let a ≤ p, b p, with primep.
We begin with λ = (a, 1b
) → λT
= (b + 1, 1a−1
).
˜λ = (a, 1b+p
) → ˜λT
= (b + p + 1, 1a−1
).
× × × . . . ×
×
×
...
×
M(λT
)R
−→
× × × . . . • •. . .×
×
×
...
×
M(˜λT
)R
Figure 11. 1
From observing Figure 14, we can get ˜λT
by adding p nodes to the
first row of λT
. Note that the extra nodes b + p has no effect on the
p-regularization after we did the transpose.
Case 2, let a p, b p, with prime p.
We have λ = (a, 1b
). Since a p, we need to p-regularize λ. By
applying the ladder algorithm, we acquire (λT
)R
.
Now, we add an extra foot of length p to λ, we have ˜λ = (a, 1b+p
).
Next, we take the transpose of ˜λ, which is ˜λT
= (b + 1 + p, 1p−1
). It
is similar to what we saw previously, we get (˜λT
)R
by adding p to the
first row of (λT
)R
.
Case 3, let b p, a p.

14. MULLINEUX MAP 14
We begin with λ = (a, 1b
).
We acquire λT
= (b + 1, 1a−1
) by taking the transpose. We have
(λT
)R
= (b + 1, 1a−1
), because a p, λ is p-regular already.
we acquire ˜λ by adding p-nodes to the foot of λ, which is, (a, 1b+p
).
(˜λ)T
= (b + p + 1, 1a−1
). since a p, so (˜λ)T
is still p-regular. (˜λT
)R
=
(b + p + 1, 1a−1
).
Comparing the formulas with (λT
)R
and (˜λT
)R
, we observe if we add
p nodes to the first row to (λT
)R
, we automatically have (˜λT
)R
.
Lemma 3.4. Take M(λR
),add p nodes to first row, we get M(˜λR
).
Proof. Case 1: let a ≤ p, b ≤ p. We denote the Mullineux symbol of
λ as,
M(λ) =
a + b
b + 1
We obtain˜λR
by adding p nodes to the leg of λ and regularize it.
Finally M(˜λR
)satisfies,




p
n times
p...p a − (np)
b + 1
n times
1...1 1



 −→




b + p
n times
p...p a − (np)
p
n times
p − 1...p − 1 a − (np) − 1



 .
.
× × . . . ×
×
×
...
×
−→
× × . . . ×
×
×
...
×
• . . . •
Figure 12. M(λR
)and M(˜λR
)
Case 2, let a ≥ p, b p, with prime p. Similar to Case 1, because the
length of a has no critical effect on the Mullineux map. The Mullineux
Map shows if we add p-partition to the first of row of the original map,
we will get the new Mullineux map M(˜λR
).

15. MULLINEUX MAP 15
M(˜λR
) =




b + p
n times
p...p a − (np)
b + 1
n times
1...1 1



 −→




b + p
n times
p...p a − (np)
p
n times
p − 1...p − 1 a − (np) − 1



 .
where n is a nonnegative integer.
Case 3, let a ≤ p, b ≥ p, with prime p. Since b ≥ p, first we need
to apply ladder algorithm to p regularize λ. We get a Mullineux map
satisfies,




p + 1
n times
p...p a + b − (p + 1) − (np)
p
n times
p − 1...p − 1 a + b − (p + 1) − (np)




−→ 



p + 1
n times
p...p a + b − (p + 1)) − (np)
2
n times
1...1 1



 .
We obtain ˜λR
by adding p nodes to the foot of λ . our M(˜λR
) denote
to be,




p + 1
n + 1 times
p...p a + b − (p + 1) − ((n + 1)p)
p
n times
p − 1...p − 1 a + b − (p + 1) − ((n + 1)p)




−→




p + 1
n + 1 times
p...p a + b − (p + 1) − ((n + 1)p)
2
n + 1 times
1...1 1



 .
Comparing M(˜λR
) and M(λR
), we observe there is an extra p nodes
hanging in the first row of M(˜λR
).

16. MULLINEUX MAP 16
4. Combinatorial proof for M(λR
) ≥ (λT
)R
We are seeking for a combinatorial proof for hooks by using the
relations we have proved in the previous section. First we handle the
base case of our induction, when the hook has both a small arm and
leg length.
Theorem 4.1. λ = (a, 1b
) with a ≤ p, b p.
1. M(λR
) ≥ (λT
)R
with equality p (a + b)
We begin with λ = (a, 1b
),it is easy to check λT
= (b + 1, 1a−1
).
We denote the Mullineux symbol as,
M(λR
) :
a + b
b + 1
−→
a + b
a
.
We can represent our Mullineux map as M(λR
)=(b + 1, 1a−1
).
It turns out M(λR
) = (λT
)R
as we desired.
2.Let a ≤ p, b p and p | (a + b).
We have λ = (a, 1b
), we can derive that λT
= (b + 1, 1a−1
)
the Mullineux map for λ is
a + b
b + 1
−→
a + b
a − 1
.
We show that the total length of (λT
)R
and M(λR
) are the same,which
is (a + b) nodes. On the other hand, the leg length of M(λR
) is
(a − 2),which is one row less than (λT
)R
. That implies we have an
extra ”×” hanging in the first row of M(λR
). By the definition of
partition, we conclude M(λR
) (λT
)R
.
Theorem 4.2. Let λ = (a, 1b
) be arbitrary, then M(λR
) ≥ (λT
)R
,
equality p (a + b).
Proof. Let’s prove by induction.
Base case: Let us fix b, where b ≤ p, and a ≤ p, with prime p.
We can easily compute λT
= (b + 1, 1a−1
) by the method we stated
previously.
The Mullineux map for λR
, where p (a + b) satisfies,
a + b
b + 1
−→
a + b
a
.
for p (a + b). It is trivial to see the formula of M(λR
) = (b + 1, 1a−1
).
Finally, we proved λT
= M(λR
) while p (a + b).
Let us suppose M(λR
) ≥ (λT
)R
is true for every a and we are trying
to prove the inequality also holds for a + 1. If a + 1 p, we are done,

17. MULLINEUX MAP 17
by Theorem 4.1. If a + 1 ≥ p, we begin by breaking a + 1 down as
l + p = a + 1 where 0 l p.
We denote λl as the partition of l, M(λl)R
as the Mullineux symbol
of l. For M(λR
), we add a rim p-hook to M(λl)R
. According to the
assumptions of induction, we have M(λR
l ) ≥ (λT
l )R
. Now by Lemma
1,we can simply add a rim p-hook to (λl)T
in order to get λT
. Similarly,
in order to obtain M(λR
), we only need to add a rim p-hook to M(λl)R
.
If we add a rim p-hook with the same length to both (λl)T
and M(λl)R
,
it directly leads to M(λR
) ≥ (λT
)R
. We can conclude the inequality
holds for the case a + 1. In result, it suffices to show M(λR
) ≥ (λT
)R
for any partitions.
Let us fix a, where a p, b p with prime p. We take a transpose
of λ, then λT
= (b+1, 1a−1
). The Mullineux Symbol for M(λR
)satisfies
a + b
b + 1
−→
a + b
a
.
We can represent M(λR
) = (b+1, 1a−1
). It is showed in our previous
proof that M(λR
) ≥ (λT
)R
is true for the base case.
Let’s suppose the inequality M(λR
) ≥ (λT
)R
holds for every b. Now,
we are trying to prove the inequality is also true for b + 1. If b + 1 p,
it is trivial M(λR
) ≥ (λT
)R
by Theorem 4.1. If b + 1 ≥ p, we can begin
by breaking (b + 1) down to (d + p), where 0 d p.
We let λd represent the partition of d, M(λd)R
stand for the Mullineux
symbol of d.
By Lemma 3, we can acquire λT
by adding p nodes to the first row
of (λd)T
. Additionally, By Lemma 4, we obtain M(λR
) by adding p
nodes to the first row of M(λd)R
If we add a same rim p-hook to both
(λl)T
and M(λl)R
, it leads to M(λR
) ≥ (λT
)R
. So,the inequality also
holds for the case b + 1.
After all, we proved M(λR
) ≥ (λT
)R
by induction.
References
[1] C. Bessenrodt and J.B. Olsson, “On residue symbols and Mullineux conjec-
ture,” J. Alg. Combinatorics 7 (1998), 227-251.
[2] C. Bessenrodt, J.B. Olsson and M. Xu, “On properties of the Mullineux map
with an application to Schur modules,” Math. Proc. Camb. Phil. Soc. 126
(1999), 443-459.
[3] G.D. James, “On the decomposition matrices of symmetric groups, II. J. Al-
gebra. 43 (1976), 45-54.
[4] G.D. James and A. Kerber. The representation theory of the symmetric group
(Addison-Wesley, London, 1981).

18. MULLINEUX MAP 18
[5] G. Mullineux. “Bijections of p-regular partitions and p-modular irreducibles of
the symmetric groups. J. London Math. Soc. (2) 20 (1979), no. 1, 60-66.
Department of Mathematics, University at Buffalo, SUNY, 244 Math-
ematics Building, Buffalo, NY 14260, USA
E-mail address: ftan@buffalo.edu