Characters of Odd Degree of Symmetric Groups
Eugenio Giannelli
Trinity Hall, University of Cambridge, CB21TJ, UK.
Abstract
We construct a natural bijection between odd-degree irreducible characters of S
2
n
and linear
characters of its Sylow 2-subgroup P
2
n
. We show that such a bijection is nicely induced by the
restriction functor. We conclude with a characterization of the irreducible characters χ of S
n
such that the restriction of χ to P
n
has a unique linear constituent.
1. Introduction
J. McKay conjectured in his landmark paper [14] that the number n
2
(G) of odd-degree
irreducible characters of a finite group G equals n
2
(N
G
(P )), where P is a Sylow 2-subgroup
of G. This is now known as the p = 2 case of the McKay Conjecture and has been recently
proved after more than 40 years in [13].
Sometimes, we do not only have that n
p
(G) = n
p
(N
G
(P )), but it is also possible to determine
a natural bijection between Irr
p
0
(G) and Irr
p
0
(N
G
(P )), where Irr
p
0
(G) is the set of complex
irreducible characters of G of degree not divisible by p. This occurs for solvable groups and
p = 2 (see [5]), for groups with a normal p-complement, using the Glauberman correspondence
and for groups having self-normalising Sylow p-subgroup for p odd (see [16] and [17]).
In this article we fix p = 2 and we focus our attention on the symmetric groups. It is well
known that in this case, if P is a Sylow 2-subgroup of the symmetric group S
n
then N
S
n
(P ) =
P . Our main result is the following theorem.
Theorem 1.1. Let n ∈ N and let P be a Sylow 2-subgroup of the symmetric group S
2
n
.
If χ ∈ Irr
2
0
(S
2
n
), then χ
y
P
has a unique linear constituent Φ
n
(χ). Moreover, the map
Φ
n
: Irr
2
0
(S
2
n
) −→ Irr
2
0
(P ),
defined by χ 7→ Φ
n
(χ) is a bijection.
2010 Mathematics Subject Classification 20C30 (primary), 20C15 (secondary).
Page 2 of 18 EUGENIO GIANNELLI
McKay conjecture for symmetric groups was first proved with a beautiful counting argument
by J. Olsson in [19]. Later, bijections between Irr
p
0
(S
n
) and Irr
p
0
(N
S
n
(P )) were given by
P. Fong in [3], when checking the Isaacs-Navarro conjecture (first stated in [7]). These bijections
are not choice-free (canonical), in fact their definition depends on some arbitrarily chosen
labelling of the characters of P (see [3, Section 2] for precise details).
Theorem 1.1 has a particular story. G. Navarro brought to our attention that in [8] it is
explained that J. Alperin privately communicated to M. Isaacs and G. Navarro that if χ is an
irreducible odd-degree character of S
2
n
, then χ
y
P
has a unique linear constituent. We could
not find a proof of this fact anywhere in the literature. With the kind permission of J. Alperin
we take the opportunity to present here our own proof of this correspondence.
We would like to point out that while the present article was under review, the result obtained
in Theorem 1.1 played a fundamental role in the construction of a canonical bijection between
Irr
2
0
(S
n
) and Irr
2
0
(P
n
), for any natural number n (here P
n
denotes a Sylow 2-subgroup of S
n
).
This was done in [4, Section 4]. At the end of Section 3 below, we will take the opportunity to
describe the main ideas involved in the construction of such bijection.
We also remark that, building on the bijection described in Theorem 1.1 for the basic case n =
2
k
, it is possible to construct an explicit canonical bijection between Irr
2
0
(S
n
) and Irr
2
0
(M),
where M is any maximal subgroup of S
n
of odd index. This is done in [4, Theorem D]. Similarly,
the result obtained in Theorem 1.1 is fundamental to determine a natural correspondence
between Irr
2
0
(G) and Irr
2
0
(N
G
(P )), where G = GL
n
(q), q is an odd prime power and P is a
Sylow 2-subgroup of G. This is done in [4, Theorem E].
In Section 4, we study the number of linear constituents in the restriction to a Sylow 2-
subgroup of any irreducible character of S
n
. In particular, we prove the following result.
Theorem 1.2. Let χ ∈ Irr(S
n
), and let P
n
be a Sylow 2-subgroup of S
n
.
(i) The restriction of χ to P
n
has a linear constituent.
(ii) If χ(1) > 1, then the restriction of χ to P
n
has a unique linear constituent if and only
if χ(1) is odd and n is a power of 2.
We remark that Theorem 1.2 shows that the only non-linear irreducible characters of
symmetric groups having a unique linear constituent in restriction to a Sylow 2-subgroup
are exactly the odd-degree irreducible characters of S
2
n
, for some natural number n. In this
sense Theorem 1.2 is a strong converse of Theorem 1.1.
CHARACTERS OF ODD DEGREE OF SYMMETRIC GROUPS Page 3 of 18
2. Preliminaries
In this section we fix the notation and we recall the main representation theoretic facts that
we will need throughout the article. We refer the reader to [1], [6] and [15] for a complete
account of the theory.
2.1. Representations of wreath products
Let G be a finite group and let M be a CG-module. Let M M be the two-fold tensor
product of M with itself. For all (g
1
, g
2
; h) ∈ G o C
2
and for all m
1
⊗ m
2
∈ M M, we let
(g
1
, g
2
; h) · m
1
⊗ m
2
= g
1
· m
h(1)
⊗ g
2
· m
h(2)
.
This yields a C(G o C
2
)-module structure on M M. We denote by
^
M M the C(G o C
2
)-
module obtained this way. It is clear that Res
G×G
(
^
M M)
∼
=
M M. If χ is the character
afforded by M , then we denote by χ × χ and
^
χ × χ the characters afforded by M M and
^
M M respectively. For J a CC
2
-module we denote by
b
J the C(G o C
2
)-module obtained by
inflation of J. Similarly, if ψ is the character afforded by J, then we denote by
b
ψ the character
afforded by
b
J.
The following theorem characterizes the irreducible representations of G o C
2
. It is a special
case of the more general Theorem 4.4.3 of [11].
Theorem 2.1. A C(G o C
2
)-module V is simple if and only if one of the following happens.
(i) V
∼
=
(
^
M M)
N
b
J, for some simple CG-module M and for some simple CC
2
-module
J, or
(ii) V
∼
=
Ind
GoC
2
G×G
(L M)
∼
=
Ind
GoC
2
G×G
(M L), for some non-isomorphic simple CG-modules
L and M.
In particular all these possibilities are mutually exclusive (no two of those representations are
isomorphic).
Denote by K
1
the cyclic group C
2
and define K
n
:= K
n−1
o C
2
for all natural numbers n ≥ 2.
Moreover, denote by B
n
the base group of K
n
. Clearly B
n
∼
=
K
n−1
× K
n−1
. The following
lemma is easily deduced from Theorem 2.1.
Lemma 2.2. Let χ ∈ Irr(K
n
). Then χ
y
B
n
has a linear constituent if and only if χ(1) = 1
or χ = (λ × µ)
x
K
n
B
n
, for some linear characters λ, µ ∈ Irr(K
n−1
) such that λ 6= µ.
Moreover, χ is linear if and only if there exists a linear character λ ∈ Irr(K
n−1
) such that
λ × λ is a constituent of χ
y
B
n
.
Page 4 of 18 EUGENIO GIANNELLI
2.2. Partitions and characters of symmetric groups
In this section we summarize the main facts about the representation theory of symmetric
groups that will be used in the rest of the paper. For a comprehensive account of this theory
we refer the reader to [10],[11] and [18].
The combinatorics of partitions. Let n be a natural number. A composition ρ of n is a
finite sequence of non-negative integers ρ = (ρ
1
, ρ
2
, . . . , ρ
s
), such that ρ
1
+ ρ
2
+ · · · + ρ
s
= n.
If ρ
i
≥ ρ
i+1
for all i ∈ {1, . . . , s − 1} then ρ is a partition of n and we define the Young diagram
of ρ as the set
[ρ] = {(i, j) ∈ N × N | 1 ≤ i ≤ s, 1 ≤ j ≤ ρ
i
}.
Here the set N × N is oriented with the x-axis pointing right and the y-axis pointing down, in
accordance with the anglo-american tradition.
Let P(n) be the set of partitions of n. Given m < n ∈ N and λ = (λ
1
, . . . , λ
k
) ∈ P(n), µ =
(µ
1
, . . . , µ
t
) ∈ P(m), we say that µ is a subpartition of λ if t ≤ k and µ
j
≤ λ
j
for all j ∈
{1, . . . , t}. In this case we denote by λ r µ the composition of n − m defined by
λ r µ = (λ
1
− µ
1
, λ
2
− µ
2
, . . . , λ
t
− µ
t
, λ
t+1
, . . . , λ
k
).
We will refer to λ r µ as a skew partition. Moreover, we will denote by [λ r µ] the subset of
[λ] defined by
[λ r µ] = {(i, j) ∈ N × N | 1 ≤ i ≤ k, µ
i
< j ≤ λ
i
}.
We will call [λ r µ] the skew Young diagram of λ r µ. It is important to remark that the
definition of the skew Young diagram [λ r µ] depends on both λ and µ and not only on
the composition λ r µ. For example, we have that (4, 4) r (2, 1) = (2, 3) = (4, 3) r (2), but
[(4, 4) r (2, 1)] 6= [(4, 3) r (2)].
We denote by R(λ) the subset of [λ] defined by
R(λ) = {(i, j) ∈ [λ] | (i + 1, j + 1) 6∈ [λ]}.
This is known as the rim of λ.
Given a node (r, c) ∈ [λ], the associated rim-hook h(r, c) is defined as
h(r, c) = {(i, j) ∈ R(λ) | r ≤ i, c ≤ j}.
If h = h(r, c) contains e nodes, then we refer to h as an e-hook of λ. Removing h from [λ]
gives the Young diagram of a partition denoted λ − h. In particular |λ − h| = |λ| − e and
[h] = [λ r (λ − h)] is a skew Young diagram. If k(h) + 1 ∈ N is the number of rows of [h], then
we sometimes say that
ˆ
h := (e − k(h), 1
k(h)
) is the hook partition corresponding to h.
For any fixed positive integer e, we have that a partition is an e-core if it has no e-hooks.
Moreover, for any λ ∈ P(n) the e-core of λ is the unique partition obtained by successively
removing e-hooks from λ. The e-core of λ is usually denoted by C
e
(λ).
CHARACTERS OF ODD DEGREE OF SYMMETRIC GROUPS Page 5 of 18
Given λ ∈ P(n) we say that j ∈ {1, . . . , n} has multiplicity k in λ if λ has exactly k parts of
size j. We equivalently denote a partition λ = (λ
1
, . . . , λ
`(λ)
) by
λ = (r
α
1
1
, . . . , r
α
k
k
),
if α
1
+ · · · + α
k
= `(λ), λ has exactly α
j
parts of size r
j
for all j ∈ {1, . . . , k} and r
j
< r
j+1
for
all j ∈ {1, . . . , k − 1}. Here (and for the rest of the article) `(λ) denotes the number of parts
of λ.
Characters of symmetric groups. Ordinary irreducible characters of the symmetric group
S
n
are naturally labelled by partitions of n. Given any λ ∈ P(n) we denote by χ
λ
the irreducible
character of S
n
labelled by λ. Let m < n be natural numbers. For any λ and µ partitions of
m and n − m respectively, the Littlewood-Richardson rule (see [10, Chapter 16]) describes the
decomposition into irreducible constituents of the character χ of S
n
defined by
χ = (χ
λ
× χ
µ
)
x
S
n
S
m
×S
n−m
.
Since we will use it repeatedly in Section 4, for the reader’s convenience we restate the rule
below. In order to do this we need to recall a few technical definitions.
Definition 2.3. Let λ = (λ
1
, . . . , λ
k
) ∈ P(n) and let C = (c
1
, . . . , c
n
) be a sequence of
positive integers. We say that C is of type λ if
|{i ∈ {1, 2, . . . , n} : c
i
= j}| = λ
j
,
for all j ∈ {1, . . . , k}. We say that an element c
j
of C is good if c
j
= 1 or if
|{i ∈ {1, 2, . . . , j − 1} : c
i
= c
j
− 1}| > |{i ∈ {1, 2, . . . , j − 1} : c
i
= c
j
}|.
Finally we say that the sequence C is good if c
j
is good for every j ∈ {1, . . . , n}.
Example 2.4. The sequence (1, 2, 3, 1) is a good sequence of type (2, 1, 1). On the other
hand (1, 1, 3, 2) is not a good sequence since the number 3 is not good.
The Littlewood-Richardson rule can now be stated as follows.
Theorem 2.5. Let n and m be natural numbers such that m < n. Let µ ∈ P(m) and
ν ∈ P(n − m). Then
(χ
µ
× χ
ν
)
x
S
n
S
m
×S
n−m
=
X
λ∈P(n)
a
λ
χ
λ
,
where a
λ
equals the number of ways to replace the nodes of [λ r µ] by natural numbers such
that
Page 6 of 18 EUGENIO GIANNELLI
(i) The sequence obtained by reading the natural numbers from right to left, top to bottom
is a good sequence of type ν.
(ii) The numbers are weakly increasing along rows.
(iii) The numbers are strictly increasing down the columns.
For k ∈ {0, 1, . . . , n − 1} we say that the partition (n − k, 1
k
) is a hook partition. Moreover,
we will denote by χ
k
n
the irreducible character of S
n
labelled by (n − k, 1
k
). In the next sections
we will be interested in studying irreducible characters of S
n
labelled by hook partitions. In
particular, we will need the following special case of the Littlewood-Richardson rule.
Lemma 2.6. Let n be a positive natural number and let k ∈ {0, 1, . . . , 2n − 1}. If k ≤ n − 1,
then
(χ
k
2n
)
y
S
n
×S
n
=
k−1
X
i=0
(χ
i
n
× χ
k−1−i
n
) +
k
X
i=0
(χ
i
n
× χ
k−i
n
).
If k = j + n for some j ∈ {0, 1, . . . , n − 1}, then
(χ
k
2n
)
y
S
n
×S
n
=
n−1
X
i=j
(χ
i
n
× χ
n+j−i−1
n
) +
n−1
X
i=j+1
(χ
i
n
× χ
n+j−i
n
).
Proof. We sketch the proof for the case k ∈ {0, 1, . . . , n − 1}. An easy application of the
Littlewood-Richardson rule (Theorem 2.5), shows that
Ξ
k
:=
k−1
X
i=0
(χ
i
n
× χ
k−1−i
n
) +
k
X
i=0
(χ
i
n
× χ
k−i
n
)
is a constituent of (χ
k
2n
)
y
S
n
×S
n
. Since χ
j
n
(1) =
n−1
j
for all j ∈ {0, 1, . . . , n − 1}, it follows by
direct computation that χ
k
2n
(1) = Ξ
k
(1). Hence Ξ
k
= (χ
k
2n
)
y
S
n
×S
n
.
A completely similar argument suffices to prove the case where k = j + n for some j ∈
{0, 1, . . . , n − 1}.
2.3. Sylow 2-subgroups of S
n
Let P
2
be the cyclic group h(1, 2)i ≤ S
2
. Let further P
1
:= {1} and, for d ≥ 1, we set
P
2
d+1
:= P
2
d
o P
2
:= {(σ
1
, σ
2
; π) : σ
1
, σ
2
∈ P
2
d
, π ∈ P
2
} ,
so that P
2
d+1
∼
=
K
d+1
.
We shall always identify P
2
d
with a Sylow 2-subgroup of S
2
d
in the usual way. That is,
(σ
1
, σ
2
; π) ∈ P
2
d
is identified with the element (σ
1
, σ
2
; π) ∈ S
2
d
that is defined as follows: if
j ∈ {1, . . . , 2
d
} is such that j = b + 2
d−1
(a − 1), for some a ∈ {1, 2} and some b ∈ {1, . . . , 2
d−1
}
CHARACTERS OF ODD DEGREE OF SYMMETRIC GROUPS Page 7 of 18
then
(σ
1
, σ
2
; π)(j) := 2
d−1
(π(a) − 1) + σ
π(a)
(b).
Now let n ∈ N. Let k
1
> k
2
> · · · > k
t
≥ 0 be integers such that n = 2
k
1
+ · · · + 2
k
t
is the
2-adic expansion of n. By [11, 4.1.22, 4.1.24], the Sylow 2-subgroups of S
n
are isomorphic to
the direct product
Q
t
i=1
(P
2
k
i
). For subsequent computations it will be useful to fix a particular
Sylow 2-subgroup P
n
of S
n
as follows: for i ∈ {1, . . . , t} let m(i) :=
P
i−1
l=1
2
k
l
and
P
2
k
i
,m(i)
:= (1, 1 + m(i)) · · · (2
k
i
, 2
k
i
+ m(i)) · P
2
k
i
· (1, 1 + m(i)) · · · (2
k
i
, 2
k
i
+ m(i)) .
Now set
P
n
:= P
2
k
1
,m(1)
× P
2
k
2
,m(2)
× · · · × P
2
k
t
,m(t)
.
It is very easy to check that N
S
n
(P
n
) = P
n
.
3. Canonical bijections
In this section we will prove Theorem 1.1. Afterwards we will briefly describe how this result
was applied to construct a canonical McKay bijection for arbitrary symmetric groups in [4]. In
order to do this, we start by restricting our attention to symmetric groups S
2
n
, for all n ∈ N.
In the following lemma we characterize the odd-degree irreducible representations of S
2
n
.
Lemma 3.1. Let χ ∈ Irr(S
2
n
). Then χ ∈ Irr
2
0
(S
2
n
) if and only if χ = χ
λ
, for some hook
partition λ.
Proof. There are many ways to prove this. It is well known that |Irr
2
0
(S
2
n
)| = 2
n
(a proof
can be found in [12, Corollary 1.3]). Moreover if λ = (2
n
− k, 1
k
) then 2 does not divide χ
λ
(1).
This follows from the hook Formula for dimensions (see [10, Page 77]), since
χ
λ
(1) =
(2
n
)!
2
n
· k!(2
n
− 1 − k)!
=
(2
n
− 1)(2
n
− 2) · · · (2
n
− k)
k!
,
and this is clearly an odd integer.
For the reader’s convenience, we modify the notation previously introduced in Section 2.2.
For n ∈ N and k ∈ {0, 1, . . . , 2
n
− 1}, we now denote by χ
k
n
the irreducible character of S
2
n
labelled by the partition (2
n
− k, 1
k
).
We are now ready to prove Theorem 1.1. We restate it below.
Theorem 1.1. Let χ ∈ Irr
2
0
(S
2
n
). Then χ
y
P
2
n
has a unique linear constituent Φ
n
(χ).
Moreover, the map
Φ
n
: Irr
2
0
(S
2
n
) −→ Irr
2
0
(P
2
n
),
Page 8 of 18 EUGENIO GIANNELLI
defined by χ 7→ Φ
n
(χ) is a bijection.
Proof. We proceed by induction on n. For n = 1 we have that S
2
= P
2
, and the statement
holds trivially. Let n ≥ 2 and assume that the statement holds for n − 1. Let χ ∈ Irr
2
0
(S
2
n
).
By Lemma 3.1, there exists k ∈ {0, 1, . . . , 2
n
− 1} such that χ = χ
k
n
. From Section 2.3, it is
easy to see that P
2
n
= B
n
o hgi, where
g =
2
n−1
Y
i=1
(i, i + 2
n−1
),
and B
n
= P
2
n−1
× P
g
2
n−1
∼
=
P
2
n−1
× P
2
n−1
. Let T
n
be the maximal Young subgroup of S
2
n
containing B
n
. Clearly T
n
= S
2
n−1
× S
2
n−1
.
By Lemma 2.6 we have that χ
k
n
y
T
n
equals
P
k−1
j=0
(χ
j
n−1
× χ
k−1−j
n−1
) +
P
k
j=0
(χ
j
n−1
× χ
k−j
n−1
) if 0 ≤ k ≤ 2
n−1
− 1 ,
P
2
n−1
−1
i=k−2
n−1
(χ
i
n−1
× χ
k−1−i
n−1
) +
P
2
n−1
−1
i=k−2
n−1
+1
(χ
i
n−1
× χ
k−i
n−1
) if 2
n−1
≤ k ≤ 2
n
− 1 .
Therefore, using the inductive hypothesis we deduce that χ
k
n
y
B
n
is equal to
P
k−1
j=0
(φ
j
n−1
× φ
k−1−j
n−1
) +
P
k
j=0
(φ
j
n−1
× φ
k−j
n−1
) + ∆
k
if 0 ≤ k ≤ 2
n−1
− 1 ,
P
2
n−1
−1
i=k−2
n−1
(φ
i
n−1
× φ
k−1−i
n−1
) +
P
2
n−1
−1
i=k−2
n−1
+1
(φ
i
n−1
× φ
k−i
n−1
) + ∆
k
if 2
n−1
≤ k ≤ 2
n
− 1 ,
where φ
j
n−1
= Φ
n−1
(χ
j
n−1
) for all j ∈ {0, 1, . . . , 2
n−1
− 1} and ∆
k
is a (possibly empty) sum of
irreducible characters of P
2
n−1
× P
2
n−1
of even degree, for all k ∈ {0, . . . 2
n
− 1}.
For k ∈ {0, 1, . . . , 2
n
− 1}, let γ(k) be
k
2
(respectively
k−1
2
) if k is even (respectively odd).
Then
(χ
k
n
y
P
2
n
)
y
B
n
= (χ
k
n
y
T
n
)
y
B
n
= (φ
γ(k)
n−1
× φ
γ(k)
n−1
) + Γ
k
+ ∆
k
,
where Γ
k
is a sum of linear characters of B
n
of the form α × β, where α and β are non-
isomorphic linear characters of S
2
n−1
. By Lemma 2.2, we deduce that χ
k
n
y
P
2
n
has a unique
linear constituent Φ
n
(χ
k
n
). Moreover,
Φ
n
(χ
k
n
)
y
B
n
= φ
γ(k)
n−1
× φ
γ(k)
n−1
.
To conclude the proof, we just need to show that whenever k
1
< k
2
are such that γ(k
1
) =
γ(k
2
), then Φ
n
(χ
k
1
n
) 6= Φ
n
(χ
k
2
n
). Clearly γ(k
1
) = γ(k
2
) if and only if k
1
= 2j and k
2
= 2j + 1,
for some j ∈ {0, 1, . . . , 2
n−1
− 1}. Let g ∈ P
2
n
be a 2
n
-cycle. By the Murnaghan-Nakayama rule
CHARACTERS OF ODD DEGREE OF SYMMETRIC GROUPS Page 9 of 18
(see [10, Page 79]), we have that
χ
k
1
n
(g) = 1 and χ
k
2
n
(g) = −1.
Moreover, it is very easy to show that θ(g) = 0 for all θ ∈ Irr(P
2
n
) such that θ(1) is even. Hence
Φ
n
(χ
k
1
n
)(g) = χ
k
1
n
(g) = 1 6= −1 = χ
k
2
n
(g) = Φ
n
(χ
k
2
n
)(g).
The proof is concluded.
As explained in the introduction, the results obtained in Theorem 1.1 were applied in [4]
to construct various types of character correspondences. To conclude this section we briefly
present the main ideas involved in the construction of a natural McKay bijection between
Irr
2
0
(S
n
) and Irr
2
0
(P
n
) for any arbitrary natural number n different from a power of 2. This
is done in full details in [4, Theorem 4.3]. Exactly as in [5], the word natural in the previous
sentence is intended to mean that an algorithm is given for constructing the correspondence
and that the result is independent of any choices made in application of the algorithm.
The following lemma is a reformulation of the characterization of the irreducible characters
of odd degree of S
n
, first proved in [12].
Lemma 3.2. Let n, k ∈ N be such that 2
k
≤ n < 2
k+1
. Let λ ∈ P(n). Then χ
λ
∈ Irr
2
0
(S
n
)
if and only if the following hold:
(i) There exists a unique 2
k
-hook h in [λ], and
(ii) if we let µ := C
2
(λ) = λ − h then χ
µ
∈ Irr
2
0
(S
n−2
k
).
Proof. We refer the reader to [2, Lemma 1] for an elegant proof of this lemma.
Let now n ∈ N be such that n = 2
n
1
+ 2
n
2
+ · · · + 2
n
t
is the binary expansion of n, for some
n
1
> n
2
> · · · > n
t
≥ 0. If χ
λ
∈ Irr
2
0
(S
n
) then we deduce from Lemma 3.2 that there exists a
hook partition
ˆ
h
1
of 2
n
1
corresponding to the unique removable 2
n
1
-hook h
1
of λ. Moreover
if we set λ
2
= λ − h
1
, we have that χ
λ
2
is an irreducible character of odd degree of S
n−2
n
1
.
Using again Lemma 3.2 we uniquely determine a hook partition
ˆ
h
2
of 2
n
2
and a partition λ
3
labelling an irreducible character of odd degree of S
n−2
n
1
−2
n
2
. After t iterations of this process
we are able to associate a sequence λ
∗
of t hook-partitions to χ
λ
. In particular we have
λ
∗
:= (
ˆ
h
1
,
ˆ
h
2
, . . . ,
ˆ
h
t
) ∈ H(2
n
1
) × · · · × H(2
n
t
).
(Here for m ∈ N, we denoted by H(m) the subset of P(m) consisting of hook partitions).
In [4] it is shown that the process described above defines a bijection α
n
between Irr
2
0
(S
n
)
and H := H(2
n
1
) × · · · × H(2
n
t
). The description of the structure of a Sylow 2-subgroup of S
n
given in Section 2.3, together with Theorem 1.1 imply the existence of a bijection β
n
between
Page 10 of 18 EUGENIO GIANNELLI
H and Irr
2
0
(P
n
). This is simply defined by
β
n
((h
1
, . . . , h
t
)) = Φ
n
1
(χ
h
1
) × · · · × Φ
n
t
(χ
h
t
), for all (h
1
, . . . , h
t
) ∈ H.
The composition γ
n
:= β
n
◦ α
n
of the two bijections described above provides us with the
desired correspondence between Irr
2
0
(S
n
) and Irr
2
0
(P
n
). A remarkable fact, proved in [4,
Theorem 4.3] is that for all χ ∈ Irr
2
0
(S
n
) we have that γ
n
(χ) is always an irreducible constituent
of χ ↓
P
n
. We conclude by mentioning that the bijection γ
n
was very recently described in
character theoretical terms only (avoiding the combinatorics of hook partitions used in [4] and
here) by M. Isaacs, G. Navarro, J. Olsson and P. Tiep in [9, Section 5].
In the following example we concretely construct the bijection γ
6
.
Example 3.3. Let n = 6 and let P
6
be a Sylow 2-subgroup of S
6
. Then P
6
∼
=
P
4
× P
2
∼
=
C
2
o C
2
× C
2
. Let
K
2
= {(2), (1
2
)} ⊆ P(2),
K
4
= {(4), (3, 1), (2, 1
2
), (1
4
)} ⊆ P(4), and
K
6
= {(6), (5, 1), (4, 2), (3
2
), (2
3
), (2
2
, 1
2
), (2, 1
4
), (1
6
)} ⊆ P(6).
Then Irr
2
0
(S
n
) = {χ
λ
: λ ∈ K
n
} for all n ∈ {2, 4, 6}. Let H be the subset of P(4) × P(2)
defined by
H = {((4), (2)), ((4), (1
2
)), ((3, 1), (2)), . . . , ((1
4
), (2)), ((1
4
), (1
2
))}.
For α = (α(1), α(2)) ∈ H, let ψ
α
be the linear character of P
6
defined by
ψ
α
= Φ
2
(χ
α(1)
) × Φ
1
(χ
α(2)
).
Then Irr
2
0
(P
6
) = {ψ
α
: α ∈ H}.
For n = 6 the bijection γ
6
constructed in [4] and described above is as follows:
χ
(6)
7−→ ψ
((4),(2))
,
χ
(5,1)
7−→ ψ
((4),(1
2
))
,
χ
(4,2)
7−→ ψ
((3,1),(1
2
))
,
χ
(3
2
)
7−→ ψ
((3,1),(2))
,
χ
(2
3
)
7−→ ψ
((2,1
2
),(1
2
))
,
χ
(2
2
,1
2
)
7−→ ψ
((2,1
2
),(2))
,
χ
(2,1
4
)
7−→ ψ
((1
4
),(2))
,
χ
(1
6
)
7−→ ψ
((1
4
),(1
2
))
.
CHARACTERS OF ODD DEGREE OF SYMMETRIC GROUPS Page 11 of 18
Let n = 2
k
and let χ
λ
∈ Irr
2
0
(S
2
k
). By Theorem 1.1 we have that Φ
k
(χ
λ
) is the unique linear
constituent of the restriction of the character χ
λ
to a Sylow 2-subgroup of S
2
k
. This extremely
special fact does not always occur. For example, direct computations show that when n = 6
we have that
χ
(3,3)
y
P
6
= ψ
((3,1),(2))
+ ψ
((4),(2))
+ ψ
((1
4
),(1
2
))
+ ∆,
where ∆ is an irreducible character of P
6
, such that ∆(1) = 2. Hence we see that χ
(3,3)
y
P
6
has
not one, but three linear constituents (as predicted by Theorem 1.2).
4. More on linear constituents
As already mentioned in the introduction, this section is concerned with the study of the
number of linear constituents in the restriction to a Sylow 2-subgroup of any irreducible
character of S
n
. In particular, the section is entirely devoted to the proof of Theorem 1.2.
Adopting the notation introduced in Section 2.2, let λ ∈ P(2n) be such that
λ = (λ
1
, . . . , λ
`(λ)
) = (r
α
1
1
, . . . , r
α
t
t
).
Definition 4.1. Let λ ∈ P(2n) be as above. Call each r
α
i
i
a cluster of λ, and call r
α
i
i
an
odd cluster if r
i
and α
i
are odd. Since λ ∈ P(2n) we have that λ has an even number of odd
clusters. Let ∆(λ) be the partition of n obtained from λ as follows:
Replace the cluster r
α
i
i
in λ by
(
r
i
2
)
α
i
if r
i
is even ,
(
r
i
+1
2
)
α
i
2
, (
r
i
−1
2
)
α
i
2
if r
i
is odd and α
i
is even ,
(
r
i
+1
2
)
α
i
−1
2
,
r
i
+ε
i
2
, (
r
i
−1
2
)
α
i
−1
2
if r
i
and α
i
are odd ,
where ε
i
= 1 if r
α
i
i
is the 1st, 3rd, 5th, ... odd cluster in λ; and ε
i
= −1 if r
α
i
i
is the 2nd, 4th,
6th, ... odd cluster in λ.
Moreover denote by S(λ) the skew partition defined by S(λ) = λ r ∆(λ) and let [S(λ)] be
the corresponding skew Young diagram.
The following example should clarify the idea behind the above definition.
Example 4.2. We will now construct ∆(λ) and S(λ) in a specific situation. This could be
very helpful to follow the two steps in the proof of Proposition 4.3 below.
Let λ ∈ P(44) be the partition defined by
λ = (7, 7, 7, 6, 5, 4, 3, 3, 1, 1) = (7
3
, 6, 5, 4, 3
2
, 1
2
).
Page 12 of 18 EUGENIO GIANNELLI
Then the odd clusters of λ are 7
3
and 5
1
. Moreover we have that
∆(λ) = (4, 4, 3
|{z}
7
, 3
|{z}
6
, 2
|{z}
5
, 2
|{z}
4
, 2, 1
|{z}
3
, 1, 0
|{z}
1
) ∈ P(22),
where the labelling below each part of ∆(λ) records the corresponding cluster of λ. In this
situation we have
S(λ) = (3, 3, 4, 3, 3, 2, 1, 2, 0, 1).
Figure 1 below highlights the relation between λ, ∆(λ) and S(λ).
1 1 1
2 2 2
3 3 3
4 44
55
6 6
7
8
1
2
7
9
• • • •
• • • •
•
•
•
•
•
•
•
•
•
•
•
•
•
•
[λ] =
Figure 1: The Young diagram [∆(λ)] consists of those boxes containing a black dot. The skew Young
diagram [S(λ)] of the skew partition S(λ) consists of the remaining boxes, containing numbers.
Underlined numbers correspond to Step 1 of the algorithm described in the proof of Proposition 4.3.
Non-underlined numbers are assigned to the boxes of [S(λ)] according to Step 2 of the algorithm
describ ed in the proof of Proposition 4.3.
Proposition 4.3. Let λ ∈ P(2n) and let ∆(λ) ∈ P(n) be as in Definition 4.1. Then
χ
∆(λ)
× χ
∆(λ)
is a constituent of χ
λ
↓
S
n
×S
n
.
Proof. By the Littlewood-Richardson rule (Theorem 2.5) and Frobenius reciprocity, it is
enough to show that there exists a way of replacing the nodes of [S(λ)] = [λ r ∆(λ)] by integers
such that conditions (1), (2) and (3) of Theorem 2.5 are satisfied. From the definition of ∆(λ)
CHARACTERS OF ODD DEGREE OF SYMMETRIC GROUPS Page 13 of 18
it is easy to check that if λ
j
is even then S(λ)
j
= ∆(λ)
j
. On the other hand we have that
S(λ)
j
= ∆(λ)
j
± 1, when λ
j
is odd. (Notice that we allow S(λ) to have parts of size 0, in order
to have `(S(λ)) = `(λ)). Below we describe (in two steps) an algorithm to replace the nodes of
[S(λ)] by integers.
Step 1. Let j ∈ {1, 2, . . . , `(λ)}.
(a) If S(λ)
j
= ∆(λ)
j
, then replace every node of row j of [S(λ)] by j.
(b) If S(λ)
j
= ∆(λ)
j
− 1, then replace every node of row j of [S(λ)] by j.
(c) If S(λ)
j
= ∆(λ)
j
+ 1, then leave the most left node of row j of [S(λ)] empty and replace
every other node of row j of [S(λ)] by j.
(In Figure 1 the nodes of [S(λ)] are replaced by underlined numbers according to Step 1 of our
algorithm.)
Remarks on Step 1.
• If ∆(λ)
j
= 1 and S(λ)
j
= 0 then [S(λ)] does not have nodes in row j, hence in (b) we are
doing nothing. Similarly, If ∆(λ)
j
= 0 and S(λ)
j
= 1 then the unique node in row j of [S(λ)]
is the most left one, hence also in this case in (c) we are doing nothing.
• Notice that, once Step 1 is completed, we have that for all j ∈ {1, 2, . . . , `(λ)}, the integer j
appears only in row j of [S(λ)]. Therefore at the moment our numbers are weakly increasing
along rows and obviously strictly increasing down the columns. Moreover the sequence σ
obtained by reading the numbers we distributed in [S(λ)] from right to left, top to bottom
contains exactly n
j
entries equal to j, where
n
j
=
∆(λ)
j
or ,
∆(λ)
j
− 1 .
Since each j appears only in row j, we have that every j appears before any j + 1 in σ.
Moreover, if n
j
= ∆(λ)
j
then clearly n
j
≥ n
j+1
. On the other hand, if n
j
= ∆(λ)
j
− 1 and
n
j+1
= ∆(λ)
j+1
− 1 again we have n
j
≥ n
j+1
. Finally, if n
j
= ∆(λ)
j
− 1 and n
j+1
= ∆(λ)
j+1
then S(λ)
j
= ∆(λ)
j
− 1 = n
j
. Hence λ
j
= 2n
j
+ 1. Similarly
2n
j+1
≤ λ
j+1
≤ 2n
j+1
+ 1,
hence n
j
≥ n
j+1
, because λ is a partition. This proves that σ is a good sequence.
• If λ
j
is even for all j ∈ {1, 2, . . . , `(λ)} then the algorithm ends. In fact, by the above remarks,
it follows that Step 1 is enough to obtain a way to replace the nodes of [S(λ)] with integers
such that: the sequence obtained by reading the natural numbers from right to left, top to
bottom is a good sequence of type ∆(λ); the numbers are weakly increasing along rows; the
numbers are strictly increasing down the columns.
Page 14 of 18 EUGENIO GIANNELLI
Let 2q be the number of odd parts of λ and suppose that q 6= 0 (i.e. λ has odd parts). In
order to obtain a sequence of type ∆(λ) we need to replace the remaining q nodes of [S(λ)]
with integers z
1
< z
2
< · · · < z
q
such that z
i
∈ {1, 2, . . . , `(λ)} for all i ∈ {1, 2, . . . , q}. By Step
1, we have that no two of the remaining q empty nodes lie in the same row of [S(λ)], since
they are the most left nodes of q distinct rows. Let k
1
< k
2
< · · · < k
q
be the rows of [S(λ)]
containing an empty (most left) node. By construction, for all i ∈ {1, 2, . . . , q} we have that
z
i
< k
i
.
Step 2. If q 6= 0, replace the most left node of row k
i
by z
i
, for all i ∈ {1, 2, . . . , q}.
(In Figure 1 the nodes of [S(λ)] are replaced by non-underlined numbers according to Step 2
of our algorithm.)
Remarks on Step 2. The overall configuration obtained after Steps 1 and 2 has the following
properties.
• Let τ = (τ
1
, . . . , τ
n
) be the sequence obtained by reading the numbers from right to left, top
to bottom. Let j ∈ {2, . . . , `(∆(λ))} and suppose that τ
x
= j for some x ∈ {1, 2, . . . , n}. We
want to show that τ
x
is good. Let n
j
denote the number of js lying in row j of [S(λ)]. By
construction this number did not change after Step 2. Hence, from the remarks after Step 1,
we deduce that n
j−1
≥ n
j
. Therefore we have that every j lying in row j of [S(λ)] is good
in τ . Suppose now that j lies in row k of [S(λ)]. Then necessarily k = k
i
> z
i
= j for some
i ∈ {1, . . . , q} and j = z
i
lies in the most left node of row k = k
i
of [S(λ)]. In this case we have
that
|{y ∈ {1, . . . , x − 1} : τ
y
= j}| = ∆(λ)
j
− 1.
Again, we need to distinguish two possibilities. If all (j − 1)s are in row (j − 1) of [S(λ)] then
|{1 ≤ y < x : τ
y
= j − 1}| = ∆(λ)
j−1
> ∆(λ)
j
− 1 = |{1 ≤ y < x : τ
y
= j}|,
since j − 1 < j < k. Hence τ
x
= j is good in τ . On the other hand, if z
i−1
= j − 1 then we have
exactly ∆(λ)
j−1
− 1 integers equal to (j − 1) lying in row (j − 1) of [S(λ)] and one (j − 1) lying
in row k
i−1
. Since k
i−1
< k
i
, we have again that
|{1 ≤ y < x : τ
y
= j − 1}| = ∆(λ)
j−1
> ∆(λ)
j
− 1 = |{1 ≤ y < x : τ
y
= j}|.
We conclude that τ
x
is a good element of τ and therefore that τ is a good sequence of type
∆(λ).
• The numbers are clearly weakly increasing along the rows, since z
i
< k
i
for all i ∈ {1, . . . , q}.
• The numbers are strictly increasing down the columns. Since ∆(λ) is a partition, this follows
by observing that for all i ∈ {1, . . . , q}, z
i
is either the highest entry in its column, or it lies
below z
h
, for some 1 ≤ h ≤ i − 1.
CHARACTERS OF ODD DEGREE OF SYMMETRIC GROUPS Page 15 of 18
By the Littlewood-Richardson rule (Theorem 2.5), the proof is now complete.
Corollary 4.4. Let λ ∈ P(2
n
). Then χ
λ
y
P
2
n
has a linear constituent.
Proof. Proceed by induction on n. If n = 0, 1 then the result follows easily by direct
computation. Suppose that n ≥ 2. By Proposition 4.3 there exists ∆(λ) ∈ P(2
n−1
) such that
χ
∆(λ)
× χ
∆(λ)
is a constituent of χ
λ
y
S
2
n−1
×S
2
n−1
. By inductive hypothesis there exists a linear
character φ of P
2
n−1
such that φ is a constituent of χ
∆(λ)
y
P
2
n−1
. Therefore φ × φ is a linear
constituent of χ
λ
y
P
2
n−1
×P
2
n−1
. By Lemma 2.2 we deduce that χ
λ
y
P
2
n
has a linear constituent.
In order to prove Theorem 1.2, we will need the following technical lemmas.
Lemma 4.5. Let n = 2
k
+ 2
t
for some natural numbers k and t such that k > t ≥ 0. Let
λ ∈ P(n) be such that χ
λ
(1) is odd and χ
λ
(1) > 1. Then χ
λ
y
S
2
k
×S
2
t
is not irreducible.
Proof. Suppose that there exists µ ∈ P(2
k
) and ν ∈ P(2
t
) such that
χ
λ
y
S
2
k
×S
2
t
= χ
µ
× χ
ν
.
Since χ
λ
(1) is odd, we deduce that also χ
µ
(1) and χ
ν
(1) are odd integers. Therefore by Lemma
3.1 there exists i ∈ {0, 1, . . . , 2
k
− 1} and j ∈ {0, 1, . . . , 2
t
− 1} such that
µ = (2
k
− i, 1
i
) and ν = (2
t
− j, 1
j
).
We split the proof into two different cases.
(1) Suppose that j /∈ {0, 2
t
− 1}. Let g ∈ S
n
be a (2
t
− 1)-cycle such that g(x) = x for all
x ∈ {2
k
+ 1, 2
k
+ 2, . . . , n}, and let h ∈ S
n
be the (2
t
− 1)-cycle defined by
h = (2
k
+ 1, 2
k
+ 2, . . . , n − 1).
Clearly (g, 1), (1, h) ∈ S
2
k
× S
2
t
and
χ
µ
(g)χ
ν
(1) = χ
λ
(g) = χ
λ
(h) = χ
µ
(1)χ
ν
(h),
since g and h are conjugate in S
n
. Moreover, from the Murnaghan-Nakayama rule it is easy
to deduce that χ
ν
(h) = 0, since [ν] does not contain (2
t
− 1)-hooks. Hence we obtain that
χ
µ
(g) = 0. This is a contradiction, since again by the Murnaghan-Nakayama rule it is easy to
check that χ
µ
(g) > 0.
(2) Suppose that j ∈ {0, 2
t
− 1}. Then χ
ν
(1) = 1. Let g ∈ S
n
be a 2
t
-cycle such that g(x) =
x for all x ∈ {2
k
+ 1, 2
k
+ 2, . . . , n}, and let h ∈ S
n
be the 2
t
-cycle defined by
h = (2
k
+ 1, 2
k
+ 2, . . . , n).
Page 16 of 18 EUGENIO GIANNELLI
Exactly as before we have that
χ
µ
(g) = χ
µ
(g)χ
ν
(1) = χ
µ
(1)χ
ν
(h) = ±χ
µ
(1).
By the Murnaghan-Nakayama rule we have that
|χ
µ
(g)| ≤ χ
ρ
(1),
where ρ is a hook partition of 2
k
− 2
t
with leg length equal to either i or i − 2
t
(if this number
is non-negative). In both cases we have that
χ
µ
(1) = |χ
µ
(g)| ≤ χ
ρ
(1) <
2
k
− 1
i
= χ
µ
(1).
This is a contradiction.
Lemma 4.6. Let n ∈ N. Suppose that there exist t ≥ 2 and k
1
> k
2
> · · · > k
t
≥ 0 such
that n = 2
k
1
+ · · · + 2
k
t
is the 2-adic expansion of n. Let λ ∈ P(n) be such that χ
λ
(1) is odd
and χ
λ
(1) > 1. Then χ
λ
y
S
2
k
1
×···×S
2
k
t
is not irreducible.
Proof. We proceed by induction on t. The base case t = 2 is proved in Lemma 4.5. Suppose
that t > 2. Let ρ ∈ P(n) and τ ∈ P(n − 2
k
t
) be the partitions defined by
ρ = (2
k
1
, 2
k
2
, . . . , 2
k
t
) , τ = (2
k
1
, 2
k
2
, . . . , 2
k
t−1
).
Denote by S
ρ
and S
τ
the corresponding Young subgroups of S
n
. Let
H = S
n−2
k
t
× S
2
k
t
and K = S
2
k
1
× S
n−2
k
1
.
Clearly S
ρ
≤ H ∩ K and χ
λ
y
S
ρ
= (χ
λ
y
H
)
y
S
ρ
. If χ
λ
y
H
is reducible, then we are done.
Suppose that χ
λ
y
H
= χ
µ
× χ
ν
for some µ ∈ P(n − 2
k
t
) and ν ∈ P(2
k
t
). Obviously χ
µ
(1) and
χ
ν
(1) are both odd. If χ
µ
(1) 6= 1 then χ
µ
y
S
τ
is reducible by inductive hypothesis. Since
S
ρ
= S
τ
× S
2
k
t
, we deduce that χ
λ
y
S
ρ
is reducible. On the other hand if χ
µ
(1) = 1 then
χ
ν
(1) 6= 1, because χ
λ
(1) 6= 1. Therefore, there exists ζ ∈ P(n − 2
k
1
) such that χ
ζ
(1) 6= 1 and
χ
λ
y
K
= φ × χ
ζ
,
for some linear character φ of S
2
k
1
. We can use again the inductive hypothesis to deduce that
χ
ζ
y
S
2
k
2
×···×S
2
k
t
is reducible and therefore that also χ
λ
y
S
ρ
is reducible.
We are now ready to prove Theorem 1.2.
Proof of Theorem 1.2. Let n = 2
k
1
+ 2
k
2
+ · · · + 2
k
t
be the 2-adic expansion of n, where
k
1
> k
2
> · · · > k
t
≥ 0. We proceed by induction on t. If t = 1 then n is a power of 2 and the
first statement follows by Corollary 4.4. Assume that t ≥ 2. Then
P
n
= P
2
k
1
× · · · × P
2
k
t
≤ S
n−2
k
t
× S
2
k
t
=: H.
CHARACTERS OF ODD DEGREE OF SYMMETRIC GROUPS Page 17 of 18
We have that
χ
λ
y
H
=
X
µ,ν
c
λ
µν
(χ
µ
× χ
ν
),
where the sum is taken over all the partitions µ ∈ P(n − 2
k
t
) and ν ∈ P(2
k
t
). Let µ ∈ P(n −
2
k
t
) and ν ∈ P(2
k
t
) be such that c
λ
µν
6= 0. Then by inductive hypothesis we have that χ
µ
y
P
n−2
k
t
has a linear constituent. The same holds for χ
ν
y
P
2
k
t
. Since
P
n
= P
n−2
k
t
× P
2
k
t
,
we conclude that χ
λ
y
P
n
has a linear constituent. This completes the proof of part (i).
To prove (ii), we first observe that if n = 2
k
and χ
λ
(1) is odd then by Theorem 1.1, we have
that χ
λ
y
P
n
has a unique linear constituent. On the other hand, if χ
λ
(1) is even then χ
λ
y
P
n
has
a non-zero, even number of linear constituents. Hence it remains to show that whenever n has
2-adic expansion given by n = 2
k
1
+ 2
k
2
+ · · · + 2
k
t
, for some t ≥ 2 and k
1
> k
2
> · · · > k
t
≥ 0,
then for all λ ∈ P(n) such that χ
λ
(1) > 1 is odd we have that χ
λ
y
P
n
has more than one linear
constituent. Let ρ be the partition defined by
ρ = (2
k
1
, 2
k
2
, . . . , 2
k
t
),
and let S
ρ
be the corresponding Young subgroup. Clearly P
n
is a Sylow 2-subgroup of S
ρ
.
By Corollary 4.4 we deduce that the number of linear constituents of χ
λ
y
P
n
is greater or
equal than the number of irreducible constituents of χ
λ
y
S
ρ
. Since λ /∈ {(n), (1
n
)} we have
that χ
λ
y
S
ρ
is not irreducible, by Lemma 4.6.
Acknowledgements
The author’s research was supported by Gunter Malle’s ERC Advanced Grant 291512 and by
Trinity Hall, University of Cambridge.
This paper grew out of conversations with Gabriel Navarro while I was visiting the University
of Valencia. I am grateful to him for his fundamental advice throughout this work. I thank
all the members of the department of mathematics of the University of Valencia, especially
Carolina Vallejo for many interesting conversations on the McKay conjecture. Also, my thanks
to Paul Fong, Gunter Malle and the anonymous reviewer for many helpful comments and
valuable suggestions on a previous version of this article.
References
1. J. L. Alperin, Local representation theory, Cambridge studies in advanced mathematics, vol. 11, CUP,
1986.
2. A. Ayyer, A. Prasad, and S. Spallone, Odd partitions in Young’s lattice. Sm. Lothar. Combin. 75
(2015), Art. B75g, 13 pp.
Page 18 of 18 CHARACTERS OF ODD DEGREE OF SYMMETRIC GROUPS
3. P. Fong, The Isaacs-Navarro conjecture for symmetric groups. Special issue celebrating the 80th birthday
of Robert Steinberg. J. Algebra 260 (2003), no. 1, 154–161.
4. E. Giannelli, A. Kleshchev, G. Navarro and P. H. Tiep, Restriction of odd degree characters and
natural correspondences. To appear in Int. Math. Res. Not.
5. M. Isaacs, Characters of solvable and symplectic groups. Amer. J. Math 95 (1973), 594–635.
6. M. Isaacs, Character theory of finite groups. AMS Chelsea, 2006.
7. M. Isaacs and G. Navarro, New refinements of the McKay conjecture for arbitrary finite groups. Ann.
of Math. (2) (2002), no. 1, 333–344.
8. M. Isaacs and G. Navarro, Character sums and double cosets. J. Algebra 320 (2008), 3749–3764.
9. M. Isaacs, G. Navarro, J. Olsson and P.H. Tiep, Characters restriction and multiplicities in symmetric
groups. To appear in J. Algebra.
10. G. D. James, The representation theory of the symmetric groups, Lecture Notes in Mathematics, vol.
682, Springer, Berlin, 1978.
11. G. James and A. Kerber, The representation theory of the symmetric group, Encyclopedia of
Mathematics and its Applications, vol. 16, Addison-Wesley Publishing Co., Reading, Mass., 1981.
12. I. G. Macdonald, On the degrees of the irreducible representations of symmetric groups, Bull. London
Math. Soc. 3 1971 189192.
13. G. Malle and B. Sp
¨
ath, Characters of odd degree. Annals of Math. 184 (2016), 869–908.
14. J. McKay, Irreducible representations of odd degree. J. Algebra 20 (1972), 416–418
15. G. Navarro, Characters and blocks of finite groups, vol. 250 of London Mathematical Society Lecture
Note Series. CUP, 1998.
16. G. Navarro, Linear characters of Sylow subgroups. J. Algebra 269 (2003), no. 2, 589–598.
17. G. Navarro, P.H. Tiep and C. Vallejo, McKay natural correspondences on characters. Algebra Number
Theory 8 (2014), 183–1856.
18. J. Olsson, Combinatorics and Representations of Finite Groups, Vorlesungen aus dem Facherbeich
Mathematik der Universitat Essen, Heft 20, 1994.
19. J. Olsson, McKay numbers and heights of characters, Math. Scand. 38 (1976), no. 1, 25–42.
