# Non-linear Sylow Branching Coefficients for symmetric groups

Eugenio Giannelli  Dipartimento di Matematica e Informatica U. Dini, Viale
Morgagni 67/a, Firenze, Italy  and  Giada Volpato  Dipartimento di Matematica
e Informatica U. Dini, Viale Morgagni 67/a, Firenze, Italy
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We study the restriction to Sylow subgroups of irreducible characters of
symmetric groups. In particular, we focus our attention on constituents of
degree greater than $1$. Our main result is a wide generalization of [GN18,
Theorem 3.1].

##  1\. Introduction

The purpose of this article is to investigate the structure of the restriction
to Sylow subgroups of irreducible characters of the symmetric group
$\mathfrak{S}_{n}$. Let $p$ be a prime number and let $P_{n}$ be a fixed Sylow
$p$-subgroup of $\mathfrak{S}_{n}$. The main question studied in this paper is
the following. Given $k\in\mathbb{N}$, which and how many irreducible
characters of $\mathfrak{S}_{n}$ admit a constituent of degree $p^{k}$ in
their restriction to $P_{n}$? More formally, we let $\mathrm{Irr}_{k}(P_{n})$
denote the set consisting of all the irreducible characters of $P_{n}$ of
degree $p^{k}$, and we focus our attention on the subset $\Omega_{n}^{k}$ of
$\mathrm{Irr}(\mathfrak{S}_{n})$ defined as follows:

$\Omega_{n}^{k}=\\{\chi\in\mathrm{Irr}(\mathfrak{S}_{n})\ |\
[\chi_{P_{n}},\phi]\neq 0,\ \text{for some}\
\phi\in\mathrm{Irr}_{k}(P_{n})\\}.$

In [GN18, Theorem 3.1] it is proved that the restriction to $P_{n}$ of any
irreducible character of $\mathfrak{S}_{n}$ admits a linear constituent. In
other words, $\Omega_{n}^{0}=\mathrm{Irr}(\mathfrak{S}_{n})$. This result was
improved (for odd primes) in [GL21] where, for every linear character $\phi$
of $P_{n}$, the authors classify those irreducible characters $\chi$ of
$\mathfrak{S}_{n}$ such that $\phi$ appears as an irreducible constituent of
$\chi_{P_{n}}$.

In this article we largely extend in a new direction the result obtained in
[GN18] mentioned above. More precisely, for any odd prime number $p$, we are
able to describe the set $\Omega_{n}^{k}$, for any $k\in\mathbb{N}$.
Surprisingly enough, these sets possess quite a regular structure. In order to
describe it we recall that irreducible characters of $\mathfrak{S}_{n}$ are
naturally in bijection with $\mathcal{P}(n)$, the set of partitions of $n$.
With this in mind, we find it useful to think of $\Omega_{n}^{k}$ as a subset
of $\mathcal{P}(n)$ instead of $\mathrm{Irr}(\mathfrak{S}_{n})$. For any
$t\in\mathbb{N}$, we let $\mathcal{B}_{n}(t)$ be the subset of
$\mathcal{P}(n)$ consisting of partitions whose Young diagram fits into a
$t\times t$ grid (i.e. having first row and first column of size at most $t$).

The first main result of the article is Theorem 5.1, where we show that for
any $k\in\mathbb{N}$ there exists a certain $T_{n}^{k}\in\\{1,\ldots,n\\}$
such that $\Omega_{n}^{k}=\mathcal{B}_{n}(T_{n}^{k})$. In other words, Theorem
5.1 shows that the set of partitions of $n$ whose corresponding irreducible
characters admit a constituent of degree $p^{k}$ on restriction to a Sylow
$p$-subgroup of $\mathfrak{S}_{n}$ coincide with the set of partitions of $n$
which fit inside a square, whose size depend on both $n$ and $t$. As mentioned
above, this statement highlights the nice and well-behaved combinatorial
structure of the sets $\Omega_{n}^{k}$.

The description given by Theorem 5.1 is sharpened in Theorem 5.3, where we
explicitly compute the value of $T_{n}^{k}$ for all $n,k\in\mathbb{N}$. We
avoid the precise description of these values here, as it requires the
introduction of some technical definitions. Nevertheless, we refer the reader
to Tables 1 and 2 for several specific and concrete instances of our second
main result.

An intriguing consequence of Theorem 5.3 is that
$\Omega_{n}^{j}\subseteq\Omega_{n}^{k}$, for all $k\leq j$. This means that
whenever $\chi_{P_{n}}$ admits an irreducible constituent of degree $p^{j}$,
then it also admits constituents of degree $p^{k}$, for all $0\leq k\leq j$.

We conclude our article by studying the second part of the question we
proposed above. Namely, we give an estimate for how many characters of
$\mathfrak{S}_{n}$ are contained in $\Omega_{n}^{k}$. In Corollary 5.5 we show
that the restriction to $P_{n}$ of almost all irreducible characters of
$\mathfrak{S}_{n}$ admits an irreducible constituent of degree $p^{k}$, for
all admissible $k\in\mathbb{N}$. More precisely, we prove that

$\lim_{n\rightarrow\infty}\frac{|\Omega_{n}|}{|\mathcal{P}(n)|}=1,$

where $\Omega_{n}$ is the intersection of all of the sets $\Omega_{n}^{k}$,
where $k$ runs among all those natural numbers such that $p^{k}$ is the degree
of an irreducible character of $P_{n}$.

######  Remark 1.1.

As mentioned above, this article treats the case of odd primes. When $p=2$,
linear constituents of the restriction to Sylow $2$-subgroups of odd degree
characters of $\mathfrak{S}_{n}$ were studied in [INOT17], mainly in
connection with the McKay Conjecture [N18]. Despite this, the object of our
study seems to be particularly difficult when $p=2$. For instance, we
immediately notice in this case that the set
$\Omega_{4}^{1}=\\{(3,1),(2,1,1)\\}$ and therefore is not of the form
$\mathcal{B}_{4}(T)$, for any $T\in\\{1,2,3,4\\}$. This shows that the main
theorems of the present article do not hold for the prime $2$. Even if this
irregularity might disappear for larger natural numbers, more serious
obstacles arise in this setting. For example, Lemma 3.3 below asserts that the
restriction to $P_{n}$ of every non-linear irreducible character of
$\mathfrak{S}_{n}$ cannot admit a unique irreducible constituent of a certain
degree. This is a crucial ingredient in the proofs of our main results.
Unfortunately, this is plainly false when the prime is $2$. For instance, in
[G17] it is shown that if $\lambda=(2^{n}-x,1^{x})$ then
$(\chi^{\lambda})_{P_{2^{n}}}$ admits a unique constituent of degree $1$.
Things can go even worse: if $\lambda=(2^{n}-1,1)$ then it is not difficult to
see that $(\chi^{\lambda})_{P_{2^{n}}}$ admits a unique constituent of degree
$2^{k}$, for all $k\in\\{0,1,\ldots,n-1\\}$.

Acknowledgments. We thank Stacey Law for carefully reading an earlier version
of this article and for providing us with many helpful comments and
suggestions. We are also grateful to the referees for several useful comments
that improved our article.

##  2\. Notation and Background

Throughout this article, $p$ denotes an odd prime. Given integers $n\leq m$,
we denote by $[n,m]$ the set $\\{n,n+1,\ldots,m\\}$. If $n<m$ then $[m,n]$ is
regarded as the empty set. We let $\mathcal{C}(n)$ be the set of compositions
of $n$, i.e. the set consisting of all the finite sequences
$(a_{1},a_{2},\ldots,a_{z})$ such that $a_{i}$ is a non-negative integer for
all $i\in[1,z]$ and such that $a_{1}+\cdots+a_{z}=n$. Given
$\lambda=(\lambda_{1},\dots,\lambda_{z})\in\mathcal{C}(n)$, we sometimes
denote by $l(\lambda)=z$ the number of parts of $\lambda$. As already
mentioned in the introduction, $P_{n}$ denotes a Sylow $p$-subgroup of the
symmetric group $\mathfrak{S}_{n}$. As usual, given a finite group $G$, we
denote by $\mathrm{Irr}(G)$ the set of irreducible complex characters of $G$,
and by $\mathrm{Lin}(G)$ the subset of linear characters of $G$. Finally,
$\mathrm{cd}(G)=\\{\chi(1)\ |\ \chi\in\mathrm{Irr}(G)\\}$ is the set of
irreducible character degrees.

###  2.1. Wreath products

Here we fix the notation for characters of wreath products. For more details
see [JK81, Chapter 4]. Let $G$ be a finite group and let $H$ be a subgroup of
$\mathfrak{S}_{n}$. We denote by $G^{\times n}$ the direct product of $n$
copies of $G$. The natural action of $\mathfrak{S}_{n}$ on the direct factors
of $G^{\times n}$ induces an action of $\mathfrak{S}_{n}$ (and therefore of
$H\leq\mathfrak{S}_{n}$) via automorphisms of $G^{\times n}$, giving the
wreath product $G\wr H:=G^{\times n}\rtimes H$. We refer to $G^{\times n}$ as
the base group of the wreath product $G\wr H$. We denote the elements of $G\wr
H$ by $(g_{1},\dots,g_{n};h)$ for $g_{i}\in G$ and $h\in H$. Let $V$ be a
$\mathbb{C}G$-module and suppose it affords the character $\phi$. We let
$V^{\otimes n}:=V\otimes\cdots\otimes V$ ($n$ copies) be the corresponding
$\mathbb{C}G^{\times n}$-module. The left action of $G\wr H$ on $V^{\otimes
n}$ defined by linearly extending

$(g_{1},\dots,g_{n};h):v_{1}\otimes\cdots\otimes v_{n}\mapsto
g_{1}v_{h^{-1}(1)}\otimes\cdots\otimes g_{n}v_{h^{-1}(n)},$

turns $V^{\otimes n}$ into a $\mathbb{C}(G\wr H)$-module, which we denote by
$\tilde{V}^{\otimes n}$. We denote by $\tilde{\phi}$ the character afforded by
the $\mathbb{C}(G\wr H)$-module $\tilde{V}^{\otimes n}$. For any character
$\psi$ of $H$, we let $\psi$ also denote its inflation to $G\wr H$ and let
$\mathcal{X}(\phi;\psi):=\tilde{\phi}\cdot\psi$ be the character of $G\wr H$
obtained as the product of $\tilde{\phi}$ and $\psi$. Let $\phi\in{\rm
Irr}(G)$ and let $\phi^{\times n}:=\phi\times\cdots\times\phi$ be the
corresponding irreducible character of $G^{\times n}$. Observe that
$\tilde{\phi}\in{\rm Irr}(G\wr H)$ is an extension of $\phi^{\times n}$. Given
$K\leq G$, we denote by ${\rm Irr}(G|\psi)$ the set of characters $\chi\in{\rm
Irr}(G)$ such that $\psi$ is an irreducible constituent of the restriction
$\chi_{K}$. Hence, by Gallagher’s Theorem [I76, Corollary 6.17] we have

${\rm Irr}(G\wr H|\phi^{\times n})=\Set{\mathcal{X}(\phi;\psi)}{\psi\in{\rm
Irr}(H)}.$

If $H=C_{p}$ is a cyclic group of prime order $p$, every
$\psi\in{\mathrm{Irr}}(G\wr C_{p})$ is either of the form

  * (i)

$\psi=\phi_{1}\times\cdots\times\phi_{p}\big{\uparrow}^{G\wr C_{p}}_{G^{\times
p}}$, where $\phi_{1},\dots\phi_{p}\in{\mathrm{Irr}}(G)$ are not all equal; or

  * (ii)

$\psi=\mathcal{X}(\phi;\theta)$ for some $\phi\in{\mathrm{Irr}}(G)$ and
$\theta\in{\mathrm{Irr}}(C_{p})$.

We remark that in case (i) we have that ${\rm Irr}(G\wr
C_{p}|\phi_{1}\times\cdots\times\phi_{p})=\\{\psi\\}$.

###  2.2. Sylow subgroups of $\mathfrak{S}_{n}$

We record some facts about Sylow subgroups of symmetric group and we refer to
[JK81, Chapter 4] or to [O76] for more details.

We let $P_{n}$ denote a Sylow $p$-subgroup of $\mathfrak{S}_{n}$. Clearly
$P_{1}$ is the trivial group while $P_{p}\cong C_{p}$ is cyclic of order $p$.
If $i\geq 2$, then $P_{p^{i}}=\big{(}P_{p^{i-1}}\big{)}^{\times p}\rtimes
P_{p}=P_{p^{i-1}}\wr P_{p}\cong P_{p}\wr\cdots\wr P_{p}$ ($i$-fold wreath
product). Let $n=\sum_{i=1}^{t}a_{i}p^{n_{i}}$ be the $p$-adic expansion of
$n$. Then $P_{n}\cong P_{p^{n_{1}}}^{\times a_{1}}\times P_{p^{n_{2}}}^{\times
a_{2}}\times\cdots\times P_{p^{n_{t}}}^{\times a_{t}}$.

For $n\in\mathbb{N}$, the normalizer of a Sylow $p$-subgroup of
$\mathfrak{S}_{p^{n}}$ is
$N_{\mathfrak{S}_{p^{n}}}(P_{p^{n}})=P_{p^{n}}\rtimes H$, where
$H\cong(C_{p-1})^{\times n}$. More generally, if
$n=\sum_{i=1}^{t}a_{i}p^{n_{i}}$, $n_{t}>\cdots>n_{1}\geq 0$, then
$N_{\mathfrak{S}_{n}}(P_{n})=N_{1}\wr\mathfrak{S}_{a_{1}}\times\cdots\times
N_{t}\wr\mathfrak{S}_{a_{t}}$, where
$N_{i}:=N_{\mathfrak{S}_{p^{n_{i}}}}(P_{p^{n_{i}}})$ for every $i\in[1,t]$. We
refer the reader to [G21, Section 2] for more details about the structure of
the normaliser of a Sylow $p$-subgroup. The following fact is certainly well
known. We state it here as we will need it in the following section of the
article.

######  Lemma 2.1.

Let $p$ be an odd prime, let $n\in\mathbb{N}$ and let $H$ be a complement of
$P_{p^{n}}$ in $N_{\mathfrak{S}_{p^{n}}}(P_{p^{n}})$. There are no non trivial
elements of $P_{p^{n}}$ that are centralized by $H$.

###### Proof.

This follows directly from the discussion in [G21, Section 2.2]. ∎

We remark that Lemma 2.1 is equivalent to say that
$C_{\mathfrak{S}_{p^{n}}}(P_{p^{n}})=Z(P_{p^{n}})$, for any $n\in\mathbb{N}$.

###  2.3. The Littlewood-Richardson coefficients

For each $n\in\mathbb{N}$, ${\mathrm{Irr}}(\mathfrak{S}_{n})$ is naturally in
bijection with $\mathcal{P}(n)$, the set of all partitions of $n$. For
$\lambda\in\mathcal{P}(n)$, the corresponding irreducible character is denoted
by $\chi^{\lambda}$. Let $m,n\in\mathbb{N}$ with $m<n$. Given
$\chi^{\mu}\times\chi^{\nu}\in\mathrm{Irr}(\mathfrak{S}_{m}\times\mathfrak{S}_{n-m})$,
the decomposition into irreducible constituents of the induction

$\left(\chi^{\mu}\times\chi^{\nu}\right)^{\mathfrak{S}_{n}}=\sum_{\lambda\in\mathcal{P}(n)}\mathcal{LR}(\lambda;\mu,\nu)\chi^{\lambda}$

is described by the Littlewood-Richardson rule (see [F97, Chapter 5] or [J78,
Chapter 16]). Here the natural numbers $\mathcal{LR}(\lambda;\mu,\nu)$ are
called _Littlewood-Richardson coefficients_. Given
$(n_{1},\ldots,n_{k})\in\mathcal{C}(n)$, $\lambda\in\mathcal{P}(n)$ and
$\mu_{j}\in\mathcal{P}(n_{j})$ for all $j\in[1,k]$, we let
$\mathcal{LR}(\lambda;\mu_{1},\ldots,\mu_{k})$ be the multiplicity of
$\chi^{\lambda}$ as an irreducible constituent of
$(\chi^{\mu_{1}}\times\cdots\times\chi^{\mu_{k}})^{\mathfrak{S}_{n}}_{Y}$.
Here $Y$ denotes the Young subgroup
$\mathfrak{S}_{n_{1}}\times\mathfrak{S}_{n_{2}}\times\cdots\times\mathfrak{S}_{n_{k}}$
of $\mathfrak{S}_{n}$. The following lemma describes the behavior of the first
parts of the partitions involved in a non-zero Littlewood-Richardson
coefficient. This will be used several times in the following sections.

######  Lemma 2.2.

Let $\mathcal{LR}(\lambda;\mu_{1},\ldots,\mu_{k})\neq 0$ then
$\lambda_{1}\leq\sum_{j=1}^{k}(\mu_{j})_{1}$.

###### Proof.

When $k=2$, the statement is a straightforward consequence of the
combinatorial description of the Littlewood-Richardson coefficient
$\mathcal{LR}(\lambda;\mu_{1},\mu_{2})$, as given in [F97, Section 5.2]. The
lemma is then proved by iteration. ∎

As in [GL21], we define $\mathcal{B}_{n}(t)$ as the set of those partitions of
$n$ whose Young diagram fits inside a $t\times t$ square grid, i.e. for
$n,t\in\mathbb{N}$, we set

$\mathcal{B}_{n}(t):=\Set{\lambda\in\mathcal{P}(n)}{\lambda_{1}\leq t,\
l(\lambda)\leq t}.$

Moreover, for $(n_{1},\ldots,n_{k})\in\mathcal{C}(n)$ and
$A_{j}\subseteq\mathcal{P}(n_{j})$ for all $j\in[1,k]$, we let

$A_{1}\star A_{2}\star\cdots\star
A_{k}:=\Set{\lambda\in\mathcal{P}(n)}{\mathcal{LR}(\lambda;\mu_{1},\ldots,\mu_{k})>0,\
\text{for some}\ \mu_{1}\in A_{1},\ldots,\mu_{k}\in A_{k}}.$

It is easy to check that $\star$ is both commutative and associative. The
following lemma was first proved in [GL21, Proposition 3.3].

######  Lemma 2.3.

Let $n,n^{\prime},t,t^{\prime}\in\mathbb{N}$ be such that $\tfrac{n}{2}<t\leq
n$ and $\tfrac{n^{\prime}}{2}<t^{\prime}\leq n^{\prime}$. Then

$\mathcal{B}_{n}(t)\star\mathcal{B}_{n^{\prime}}(t^{\prime})=\mathcal{B}_{n+n^{\prime}}(t+t^{\prime}).$

##  3\. Preliminary results

In this section we start collecting some results on restriction of characters
to Sylow $p$-subgroups. These will be used to prove our main theorems in the
second part of the paper.

Unless otherwise stated, from now on $p$ will always denote a fixed odd prime
number. Let $n\in\mathbb{N}$ and let $n=\sum_{i=1}^{t}p^{n_{i}}$ be its
$p$-adic expansion, where $n_{1}\geq n_{2}\geq\cdots\geq n_{t}\geq 0$. We
define the integer $\alpha_{n}$ as follows. For powers of $p$ we set
$\alpha_{1}=\alpha_{p}=0$ and $\alpha_{p^{k}}=(p^{k-1}-1)/(p-1)$, for $k\geq
2$. For general $n=\sum_{i=1}^{t}p^{n_{i}}$, we set
$\alpha_{n}=\sum_{i=1}^{t}\alpha_{p^{n_{i}}}$. As shown in Lemma 3.2 below,
$p^{\alpha_{n}}$ is the greatest degree of an irreducible character of
$P_{n}$. It is interesting to note that
$\alpha_{n}=\nu(\lfloor\frac{n}{p}\rfloor!)$, where $\nu(n)$ denotes the
highest power of $p$ dividing $n$. We omit the proof of this statement and we
refer the reader to [V23] for the complete calculations.

As mentioned in the Introduction, we let
$\mathrm{Irr}_{k}(P_{n})=\\{\theta\in\mathrm{Irr}(P_{n})\ |\
\theta(1)=p^{k}\\}$. This notation will be kept throughout the article. In the
following lemma we give a lower bound for the size of the set
$\mathrm{Irr}_{k}(P_{n})$. This is certainly far from being attained (in
general), but it will be sufficient for our purposes.

######  Lemma 3.1.

Let $k,t\in\mathbb{N}$ be such that $p^{k}\in\mathrm{cd}(P_{p^{t}})$. Then
$|\mathrm{Irr}_{k}(P_{p^{t}})|\geq p$.

###### Proof.

We proceed by induction on $t$. If $t=1$ then we know that the statement holds
as ${\mathrm{Irr}}(P_{p})=\mathrm{Irr}_{0}(P_{p})$ has size $p$. The elements
of ${\mathrm{Irr}}(P_{p})$ are denoted by
$\phi_{0},\phi_{1},\ldots,\phi_{p-1}$, where we conventionally set $\phi_{0}$
to be the trivial character. Let $t\geq 2$, and let
$\psi\in{\mathrm{Irr}}_{k}(P_{p^{t}})$. If $\psi=\mathcal{X}(\theta;\phi_{i})$
for some $\theta\in{\mathrm{Irr}}_{k}(P_{p^{t-1}})$ and $i\in[0,p-1]$, then
$\mathcal{X}(\theta;\phi_{j})\in\mathrm{Irr}_{k}(P_{p^{t}})$ for all
$j\in[0,p-1]$. Hence $|\mathrm{Irr}_{k}(P_{p^{t}})|\geq p$. Otherwise
$\psi=(\theta_{1}\times\cdots\times\theta_{p})^{P_{p^{t}}}$ where
$\theta_{1},\dots,\theta_{p}\in{\mathrm{Irr}}(P_{p^{t-1}})$ are not all equal.
If there exists $x\in[1,p]$ such that $\theta_{1}(1)\neq\theta_{x}(1)$ then we
define $\eta_{1},\ldots,\eta_{p}\in\mathrm{Irr}_{k}(P_{p^{t}})$ as follows.
For any $j\in[1,p]$ we let

$\eta_{j}=(\tau_{j}\times\theta_{2}\times\cdots\times\theta_{p})^{P_{p^{t}}},$

where $\tau_{1},\tau_{2},\ldots,\tau_{p}$ are $p$ distinct irreducible
characters of $P_{p^{t-1}}$ of degree $\theta_{1}(1)$. These exist by
inductive hypothesis. On the other hand, if $\theta_{1}(1)=\theta_{x}(1)$ for
all $x\in[1,p]$ then we let

$\eta_{1}=(\tau_{2}\times\cdots\times\tau_{2}\times\tau_{1})^{P_{p^{t}}},\
\text{and}\
\eta_{j}=(\tau_{1}\times\cdots\times\tau_{1}\times\tau_{j})^{P_{p^{t}}},\
\text{for all}\ j\in[2,p].$

As before, here we chose $\tau_{1},\tau_{2},\ldots,\tau_{p}$ to be $p$
distinct irreducible characters of $P_{p^{t-1}}$ of degree $\theta_{1}(1)$.
These exist by inductive hypothesis. In both cases $\eta_{1},\ldots,\eta_{p}$
are $p$ distinct elements of $\mathrm{Irr}_{k}(P_{p^{t}})$. Hence
$|\mathrm{Irr}_{k}(P_{p^{t}})|\geq p$. ∎

The next lemma shows that $P_{n}$ has irreducible characters of each degree
$1,p,p^{2},\ldots,p^{\alpha_{n}}$.

######  Lemma 3.2.

Let $n\in\mathbb{N}$. Then $\mathrm{cd}(P_{n})=\\{p^{k}\ |\
k\in[0,\alpha_{n}]\\}$.

###### Proof.

Let us first suppose that $n=p^{t}$ is a power of $p$ and proceed by induction
on $t$. The case $t=1$ is trivial, since $P_{p}$ is cyclic and $\alpha_{p}=0$.
If $t\geq 2$, notice that $\alpha_{p^{t}}=1+p\alpha_{p^{t-1}}$. Let
$k\in[0,\alpha_{p^{t}}-1]$, and let $q\leq\alpha_{p^{t-1}}$ and $r\in[0,p-1]$
be such that $k=qp+r$. If $r=0$, by inductive hypothesis there exists
$\phi\in{\mathrm{Irr}}(P_{p^{t-1}})$ such that $\phi(1)=p^{q}$. Hence for any
$\psi\in{\mathrm{Irr}}(P_{p})$,
$\mathcal{X}(\phi;\psi)\in{\mathrm{Irr}}(P_{p^{t}})$ has degree $p^{k}$. If
$r>0$, then $q<\alpha_{p^{t-1}}$. By inductive hypothesis, there exist
$\phi_{1},\dots,\phi_{p}\in{\mathrm{Irr}}(P_{p^{t-1}})$ such that
$\phi_{i}(1)=p^{q+1}$ for every $i\in[1,r]$, and $\phi_{j}(1)=p^{q}$ for every
$j\in[r+1,p]$. Hence
$(\phi_{1}\times\cdots\times\phi_{r}\times\phi_{r+1}\times\cdots\times\phi_{p})^{P_{p^{t}}}\in{\mathrm{Irr}}(P_{p^{t}})$
has degree $p^{k}$. Finally, let $k=\alpha_{p^{t}}$. By inductive hypothesis
and by Lemma 3.1, there exist
$\phi_{1},\dots,\phi_{p}\in{\mathrm{Irr}}(P_{p^{t-1}})$ not all equal and such
that $\phi_{i}(1)=p^{\alpha_{p^{t-1}}}$ for all $i\in[1,p]$. Hence
$(\phi_{1}\times\cdots\times\phi_{p})^{P_{p^{t}}}\in{\mathrm{Irr}}(P_{p^{t}})$
has degree $p^{k}$. This concludes the proof in the case $n=p^{t}$, for
$t\in\mathbb{N}$.

The case where $n$ is not a power of $p$ follows easily. Indeed, if
$n=\sum_{i=1}^{t}p^{n_{i}}$ is the $p$-adic expansion of $n$ then $P_{n}\cong
P_{p^{n_{1}}}\times P_{p^{n_{2}}}\times\cdots\times P_{p^{n_{t}}}$. ∎

Let $n,k\in\mathbb{N}$. As we mentioned in the introduction, it is convenient
to think of the set $\Omega_{n}^{k}$ as a subset of $\mathcal{P}(n)$. More
precisely, for $\lambda\in\mathcal{P}(n)$, we will sometimes write
$\lambda\in\Omega_{n}^{k}$ instead of $\chi^{\lambda}\in\Omega_{n}^{k}$.

The following is an important ingredient when proving statements by induction.
For an odd prime $p$ let $\chi$ be a non-linear character of
$\mathfrak{S}_{n}$ and suppose that $\chi_{P_{n}}$ has an irreducible
constituent of degree $p^{k}$. Then it has at least two distinct irreducible
constituents.

######  Lemma 3.3.

Let $n\in\mathbb{N}$ be such that $n\geq p$ and let
$\lambda\in\Omega_{n}^{k}\smallsetminus\\{(n),(1^{n})\\}$ for some
$k\in[0,\alpha_{n}]$. Then there are at least two distinct irreducible
constituents of $(\chi^{\lambda})_{P_{n}}$ of degree $p^{k}$.

###### Proof.

Let us first suppose that $n=p^{t}$, and let us set $P=P_{p^{t}}$ and
$N=N_{\mathfrak{S}_{p^{t}}}(P_{p^{t}})$. We observe that the only
$N$-invariant irreducible character of $P$ is the trivial one. To show this,
we let $\mathrm{Irr}_{K}(P)$ denote the set of $K$-invariant irreducible
characters of $P$, for any $K\leq N$. Let $H$ be a $p^{\prime}$-complement of
$P$ in $N$. Clearly $\mathrm{Irr}_{N}(P)=\mathrm{Irr}_{H}(P)$. On the other
hand the set $C_{P}(H)=\\{x\in P\ |\ x^{h}=x,\ \text{for all}\ h\in H\\}$
consists of the only identity element, by Lemma 2.1. Using the Glauberman
correspondence [I76, Theorem 13.1], we get that
$|\mathrm{Irr}_{H}(P)|=|\mathrm{Irr}(C_{P}(H))|=1$. It follows that
$\mathrm{Irr}_{N}(P)=\\{1_{P}\\}$, as claimed.

Since $\lambda\notin\\{(n),(1^{n})\\}$, by [GL21, Lemma 4.3] we know that
$(\chi^{\lambda})_{P}$ necessarily admits a non-trivial linear constituent
(direct computations show that this holds also in the case
$(p,n,\lambda)=(3,9,(3,3,3))$, which is not covered by the lemma). It follows
that for any $k\in\mathbb{N}$ such that $\lambda\in\Omega_{n}^{k}$, we can
find a non-trivial $\theta\in\mathrm{Irr}_{k}(P)$ such that $\theta$ is a
constituent of $(\chi^{\lambda})_{P}$. Since $\chi^{\lambda}$ is $N$-invariant
we deduce that every $N$-conjugate of $\theta$ is a constituent of
$\chi^{\lambda}$. The statement follows. Recalling the structure of $P_{n}$
described in Section 2.2, we observe that the case where $n$ is not a prime
power is an easy consequence of the prime power case. ∎

######  Definition 3.4.

Let $G$ be a finite group and let $H$ be a $p$-subgroup of $G$. Given a
character $\theta$ of $G$, we let $\mathrm{cd}(\theta_{H})$ be the set of
degrees of the irreducible constituents of $\theta_{H}$. Moreover, we let
$\partial_{H}(\theta)$ be the non-negative integer defined as follows:

$\partial_{H}(\theta)=\mathrm{max}\\{k\in\mathbb{N}\ |\
p^{k}\in\mathrm{cd}(\theta_{H})\\}.$

######  Proposition 3.5.

Let $n\in\mathbb{N}$ and let $Y=(\mathfrak{S}_{p^{n-1}})^{\times
p}\leq\mathfrak{S}_{p^{n}}$ be such that $B\leq Y$, where
$B=(P_{p^{n-1}})^{\times p}$ is the base group of $P_{p^{n}}$. Let
$\lambda\in\mathcal{B}_{p^{n}}(p^{n}-1)$. Then

$\partial_{P_{p^{n}}}(\chi^{\lambda})=1+\mathrm{max}\\{\partial_{B}(\chi^{\nu_{1}}\times\cdots\times\chi^{\nu_{p}})\
|\ \chi^{\nu_{1}}\times\cdots\times\chi^{\nu_{p}}\in\mathrm{Irr}(Y)\
\text{and}\ \mathcal{LR}(\lambda;\nu_{1},\ldots,\nu_{p})\neq 0\\}.$

###### Proof.

Let
$M=1+\mathrm{max}\\{\partial_{B}(\chi^{\nu_{1}}\times\cdots\times\chi^{\nu_{p}})\
|\ \chi^{\nu_{1}}\times\cdots\times\chi^{\nu_{p}}\in\mathrm{Irr}(Y)\
\text{and}\ \mathcal{LR}(\lambda;\nu_{1},\ldots,\nu_{p})\neq 0\\}.$ Let
$\mu_{1},\dots,\mu_{p}\in\mathcal{P}(p^{n-1})$ be such that
$\mathcal{LR}(\lambda;\mu_{1},\ldots,\mu_{p})\neq 0$ and
$M=1+\partial_{B}(\chi^{\mu_{1}}\times\cdots\times\chi^{\mu_{p}})$. Since
$\lambda\notin\\{(n),(1^{n})\\}$, we can assume that $\mu_{1},\ldots,\mu_{p}$
are not all in $\\{(p^{n-1}),(1^{p^{n-1}})\\}$. Moreover, let $\phi$ be an
irreducible constituent of
$(\chi^{\mu_{1}}\times\cdots\times\chi^{\mu_{p}})_{B}$ such that
$\phi(1)=p^{M-1}$. By Lemma 3.3, we can take
$\phi=\phi_{1}\times\cdots\times\phi_{p}$ with
$\phi_{1},\dots,\phi_{p}\in{\mathrm{Irr}}(P_{p^{n-1}})$ not all equal. Hence
$\phi^{P_{p^{n}}}\in{\mathrm{Irr}}(P_{p^{n}})$, it has degree $p^{M}$ and
$\left[(\chi^{\lambda})_{P_{p^{n}}},\phi^{P_{p^{n}}}\right]\neq 0$. Thus
$p^{M}\in\mathrm{cd}((\chi^{\lambda})_{P_{p^{n}}})$.

Now suppose for a contradiction that there exists an integer $N>M$ such that
$p^{N}\in\mathrm{cd}((\chi^{\lambda})_{P_{p^{n}}})$. Then there exists
$\varphi\in{\mathrm{Irr}}(P_{p^{n}})$ such that
$\left[(\chi^{\lambda})_{P_{p^{n}}},\varphi\right]\neq 0$ and
$\varphi(1)=p^{N}$. Let $\phi_{1}\times\cdots\times\phi_{p}$ be an irreducible
constituent of $\varphi_{B}$. Hence there exist
$\mu_{1},\dots,\mu_{p}\in\mathcal{P}(p^{n-1})$ such that
$\mathcal{LR}(\lambda;\mu_{1},\ldots,\mu_{p})\neq 0$ and
$\left[\phi_{1}\times\cdots\times\phi_{p},(\chi^{\mu_{1}}\times\cdots\times\chi^{\mu_{p}})_{B}\right]\neq
0$. We have that

$M>\partial_{B}(\chi^{\mu_{1}}\times\cdots\times\chi^{\mu_{p}})\geq N-1,$

since the degree of $\phi_{1}\times\cdots\times\phi_{p}$ is either $p^{N}$ or
$p^{N-1}$. Hence $M<N<M+1$, which is a contradiction. ∎

######  Proposition 3.6.

Let $n$ be a natural number and let $n=\sum_{i=1}^{t}p^{n_{i}}$ be the
$p$-adic expansion of $n$, where $n_{1}\geq n_{2}\geq\cdots\geq n_{t}\geq 0$.
Let
$Y=\mathfrak{S}_{p^{n_{1}}}\times\mathfrak{S}_{p^{n_{2}}}\times\cdots\times\mathfrak{S}_{p^{n_{t}}}$
be such that $P_{n}\leq Y\leq\mathfrak{S}_{n}$, and let $\lambda$ be a
partition of $n$. Then

$\partial_{P_{n}}(\chi^{\lambda})=\mathrm{max}\\{\partial_{P_{n}}(\chi^{\mu_{1}}\times\cdots\times\chi^{\mu_{t}})\
|\ \chi^{\mu_{1}}\times\cdots\times\chi^{\mu_{t}}\in\mathrm{Irr}(Y)\
\text{and}\ \mathcal{LR}(\lambda;\mu_{1},\ldots,\mu_{t})\neq 0\\}.$

###### Proof.

Since $P_{n}=P_{p^{n_{1}}}\times P_{p^{n_{2}}}\times\cdots\times
P_{p^{n_{t}}}\leq Y$, the statement follows. ∎

######  Lemma 3.7.

Let $n\in\mathbb{N}_{\geq 2}$, let $k\in[2,\alpha_{p^{n}}]$ and let
$(a_{1},\ldots,a_{p})\in\mathcal{C}(k-1)$ be such that
$a_{i}\in[0,\alpha_{p^{n-1}}]$, for all $i\in[1,p]$. Then

$\Omega_{p^{n-1}}^{a_{1}}\star\Omega_{p^{n-1}}^{a_{2}}\star\cdots\star\Omega_{p^{n-1}}^{a_{p}}\subseteq\Omega_{p^{n}}^{k}.$

###### Proof.

To ease the notation we let $q=p^{n-1}$. If
$\lambda\in\Omega_{q}^{a_{1}}\star\Omega_{q}^{a_{2}}\star\cdots\star\Omega_{q}^{a_{p}}$,
by definition there exists an irreducible constituent
$\chi^{\mu_{1}}\times\dots\times\chi^{\mu_{p}}$ of
$(\chi^{\lambda})_{(\mathfrak{S}_{q})^{\times p}}$ such that
$\mu_{i}\in\Omega_{q}^{a_{i}}$ for all $i\in[1,p]$. Hence for every
$i\in[1,p]$ there exists an irreducible constituent $\phi_{i}$ of
$(\chi^{\mu_{i}})_{P_{q}}$ such that $\phi_{i}(1)=p^{a_{i}}$. Since $k\geq 2$,
there exists $j\in[1,p]$ such that $a_{j}\geq 1$. Hence
$\mu_{j}\notin\\{(q),(1^{q})\\}$. Thus, by Lemma 3.3 we can assume that
$\phi_{1},\dots,\phi_{p}$ are not all equal. It follows that
$(\phi_{1}\times\cdots\times\phi_{p})^{P_{p^{n}}}$ is an irreducible
constituent of $(\chi^{\lambda})_{P_{p^{n}}}$ of degree equal to $p^{k}$.
Hence $\lambda\in\Omega_{p^{n}}^{k}.$ ∎

######  Lemma 3.8.

Let $n\in\mathbb{N}_{\geq 2}$ and let $n=\sum_{i=1}^{t}p^{n_{i}}$ be its
$p$-adic expansion, where $n_{1}\geq n_{2}\geq\cdots\geq n_{t}\geq 0$. Let
$k\in[1,\alpha_{n}]$ and let $(a_{1},\ldots,a_{t})\in\mathcal{C}(k)$ be such
that $a_{i}\in[0,\alpha_{p^{n_{i}}}]$, for all $i\in[1,t]$. Then

$\Omega_{p^{n_{1}}}^{a_{1}}\star\Omega_{p^{n_{2}}}^{a_{2}}\star\cdots\star\Omega_{p^{n_{t}}}^{a_{t}}\subseteq\Omega_{p^{n}}^{k}.$

###### Proof.

Recall that $P_{n}\cong P_{p^{n_{1}}}\times P_{p^{n_{2}}}\times\cdots\times
P_{p^{n_{t}}}$ and let
$\lambda\in\Omega_{p^{n_{1}}}^{a_{1}}\star\Omega_{p^{n_{2}}}^{a_{2}}\star\cdots\star\Omega_{p^{n_{t}}}^{a_{t}}$.
By definition, for every $i\in[1,t]$ there exists
$\phi_{i}\in{\mathrm{Irr}}(P_{p^{n_{i}}})$ with $\phi_{i}(1)=p^{a_{i}}$, such
that $\phi_{1}\times\cdots\times\phi_{t}$ is an irreducible constituent of
$(\chi^{\lambda})_{P_{n}}$ of degree $p^{k}$. Hence
$\lambda\in\Omega_{p^{n}}^{k}.$ ∎

##  4\. The prime power case

The aim of this section is to completely describe the sets
$\Omega_{p^{n}}^{k}$ for all odd primes $p$, all natural numbers $n$ and all
$k\in[0,\alpha_{p^{n}}]$. We remind the reader that from [GN18, Theorem 3.1],
we know that $\Omega_{p^{n}}^{0}=\mathcal{B}_{p^{n}}(p^{n})$, for all
$n\in\mathbb{N}$. Equivalently, every irreducible character of
$\mathfrak{S}_{p^{n}}$ admits a linear constituent on restriction to a Sylow
$p$-subgroup. This result will be used frequently, with no further reference.
We start by analysing the cases where $k\in\\{1,2\\}$. In the next lemma we
show that for $j=1,2,$ every non-linear character of $\mathfrak{S}_{p^{n}}$
affords an irreducible constituent of degree $p^{j}$ on restriction to a Sylow
$p$-subgroup $P_{p^{n}}$, as long as $P_{p^{n}}$ has an irreducible character
of degree $p^{j}$.

######  Lemma 4.1.

Let $p$ be an odd prime, $n\in\mathbb{N}$ and
$k\in[1,\alpha_{p^{n}}]\cap\\{1,2\\}$. Then
$\Omega_{p^{n}}^{k}=\mathcal{B}_{p^{n}}(p^{n}-1)$.

###### Proof.

Let $k=1$. Then necessarily $n\geq 2$. We first observe that clearly
$\Omega_{p^{n}}^{1}\subseteq\mathcal{B}_{p^{n}}(p^{n}-1)$. On the other hand,
if $\lambda\in\mathcal{B}_{p^{n}}(p^{n}-1)$, then there exist
$\mu_{1}\in\mathcal{B}_{p^{n-1}}(p^{n-1}-1)$ and
$\mu_{2},\ldots,\mu_{p}\in\mathcal{P}(p^{n-1})$ such that
$\mathcal{LR}(\lambda;\mu_{1},\ldots,\mu_{p})\neq 0$. Using Lemma 3.3 we
deduce that $(\chi^{\mu_{1}})_{P_{p^{n-1}}}$ admits two distinct linear
constituents. Therefore, there exists
$\phi_{1},\ldots,\phi_{p}\in\mathrm{Lin}(P_{p^{n-1}})$ not all equal and such
that $\phi_{i}$ is a constituent of $(\chi^{\mu_{i}})_{P_{p^{n-1}}}$, for all
$i\in[1,p]$. It follows that
$(\phi_{1}\times\cdots\times\phi_{p})^{P_{p^{n}}}$ is an irreducible
constituent of $(\chi^{\lambda})_{P_{p^{n}}}$ of degree $p$. We conclude that
$\lambda\in\Omega_{p^{n}}^{1}$ and hence that
$\Omega_{p^{n}}^{1}=\mathcal{B}_{p^{n}}(p^{n}-1)$.

Let $k=2$. Then necessarily $n\geq 3$. It is clear that
$\Omega_{p^{n}}^{2}\subseteq\mathcal{B}_{p^{n}}(p^{n}-1)$. On the other hand,
if $\lambda\in\mathcal{B}_{p^{n}}(p^{n}-1)$, then there exist
$\mu_{1}\in\mathcal{B}_{p^{n-1}}(p^{n-1}-1)$ and
$\mu_{2},\ldots,\mu_{p}\in\mathcal{P}(p^{n-1})$ such that
$\mathcal{LR}(\lambda;\mu_{1},\ldots,\mu_{p})\neq 0$. We can now argue exactly
as above to deduce that $(\chi^{\lambda})_{P_{p^{n}}}$ admits an irreducible
constituent $\theta$ of the form
$\theta=(\psi\times\phi_{1}\times\cdots\times\phi_{p-1})^{P_{p^{n}}}$, where
$\psi\in\mathrm{Irr}_{1}(P_{p^{n-1}})$ and
$\phi_{1},\ldots,\phi_{p-1}\in\mathrm{Lin}(P_{p^{n-1}})$. Hence
$\theta(1)=p^{2}$, $\lambda\in\Omega_{p^{n}}^{2}$ and therefore we have that
$\Omega_{p^{n}}^{2}=\mathcal{B}_{p^{n}}(p^{n}-1)$. ∎

Lemma 4.1 is a special case of the following more general result.

######  Theorem 4.2.

Let $n\in\mathbb{N}$ and let $k\in[0,\alpha_{p^{n}}]$. Then there exists
$t_{n}^{k}\in[\frac{p^{n}+1}{2},p^{n}]$ such that
$\Omega_{p^{n}}^{k}=\mathcal{B}_{p^{n}}(t_{n}^{k}).$ Moreover, if
$k\in[0,\alpha_{p^{n}}-1]$, then $t_{n}^{k+1}\in\\{t_{n}^{k}-1,t_{n}^{k}\\}.$

###### Proof.

We proceed by induction on $n$. If $n=1$, then $\alpha_{p}=0$ and
$\Omega_{p}^{0}=\mathcal{P}(p)$. If $n\geq 2$, we assume that the statement
holds for $n-1$. If $k=0$ then by [GN18, Theorem 3.1],
$\Omega_{p^{n}}^{0}=\mathcal{B}_{p^{n}}(p^{n})$, and $t_{n}^{0}=p^{n}$.
Moreover, by Lemma 4.1 we know that $t_{n}^{1}=p^{n}-1=t_{n}^{0}-1$, as
required. The case $k=1$ is completely treated by Lemma 4.1. In fact, we know
that $\Omega_{p^{n}}^{1}=\mathcal{B}_{p^{n}}(p^{n}-1)$ and that
$t_{n}^{2}=p^{n}-1=t_{n}^{1}$, as required. We can now suppose that $k\geq 2$.
We define

$\mathcal{L}(k-1)=\\{(j_{1},\dots,j_{p})\in\mathcal{C}(k-1)\ |\
j_{i}\in[0,\alpha_{p^{n-1}}]\text{ for all }i\in[1,p]\\}.$

Moreover, we set

$M=\max\Set{t_{n-1}^{j_{1}}+\cdots+t_{n-1}^{j_{p}}}{(j_{1},\dots,j_{p})\in\mathcal{L}(k-1)}.$

Notice that for any $j\in[0,\alpha_{p^{n-1}}]$, the value $t_{n-1}^{j}$ is
well-defined by induction as the integer such that
$\Omega_{p^{n-1}}^{j}=\mathcal{B}_{p^{n-1}}(t_{n-1}^{j})$. We claim that
$M=t_{n}^{k}$. In other words, we want to prove that
$\Omega_{p^{n}}^{k}=\mathcal{B}_{p^{n}}(M)$. Let
$(j_{1},\dots,j_{p})\in\mathcal{L}(k-1)$ be such that
$M=t_{n-1}^{j_{1}}+\cdots+t_{n-1}^{j_{p}}$. By inductive hypothesis and by
Lemmas 2.3 and 3.7, we have that

$\mathcal{B}_{p^{n}}(M)=\mathcal{B}_{p^{n-1}}(t_{n-1}^{j_{1}})\star\cdots\star\mathcal{B}_{p^{n-1}}(t_{n-1}^{j_{p}})=\Omega_{p^{n-1}}^{j_{1}}\star\cdots\star\Omega_{p^{n-1}}^{j_{p}}\subseteq\Omega_{p^{n}}^{k}.$

For the opposite inclusion, suppose for a contradiction that
$\lambda\in\Omega_{p^{n}}^{k}\smallsetminus\mathcal{B}_{p^{n}}(M)$. Since $p$
is odd, we have that $\Omega_{p^{n}}^{k}$ is closed under conjugation of
partitions. Hence, we can assume that $\lambda_{1}\geq M+1$. Since
$\lambda\in\Omega_{p^{n}}^{k}$, there exists an irreducible constituent
$\theta$ of $(\chi^{\lambda})_{P_{p^{n}}}$ with $\theta(1)=p^{k}$.

$\bullet$ If $\theta=(\phi_{1}\times\cdots\times\phi_{p})^{P_{p^{n}}}$ with
$\phi_{1},\ldots,\phi_{p}\in{\mathrm{Irr}}(P_{p^{n-1}})$ not all equal, then
there exists $(j_{1},\ldots,j_{p})\in\mathcal{L}(k-1)$ such that
$\phi_{i}(1)=p^{j_{i}}$ for all $i\in[1,p]$. Then, for every $i\in[1,p]$ there
exists an irreducible constituent $\chi^{\mu_{i}}$ of
$(\phi_{i})^{\mathfrak{S}_{p^{n-1}}}$ such that
$\left[\chi^{\mu_{1}}\times\cdots\times\chi^{\mu_{p}},(\chi^{\lambda})_{(\mathfrak{S}_{p^{n-1}})^{\times
p}}\right]\neq 0$. Hence using the inductive hypothesis, we have that
$\mu_{i}\in\Omega_{p^{n-1}}^{j_{i}}=\mathcal{B}_{p^{n-1}}(t_{n-1}^{j_{i}})$,
for all $i\in[1,p]$. Hence

$M\geq t_{n-1}^{j_{1}}+\cdots+t_{n-1}^{j_{p}}\geq\lambda_{1}\geq M+1,$

where the first inequality holds by definition of $M$ and the second one by
Lemma 2.2. This is a contradiction.

$\bullet$ On the other hand, if $\theta=\mathcal{X}\left(\phi;\psi\right)$ for
some $\phi\in{\mathrm{Irr}}(P_{p^{n-1}})$ and $\psi\in{\mathrm{Irr}}(P_{p})$,
then, $\phi(1)=p^{\frac{k}{p}}$ and there exist
$\mu_{1},\dots,\mu_{p}\in\Omega_{p^{n-1}}^{\frac{k}{p}}$ such that
$\mathcal{LR}(\lambda;\mu_{1},\dots,\mu_{p})\neq 0$. Hence, using the
inductive hypothesis we have that

$\lambda\in\left(\Omega_{p^{n-1}}^{\frac{k}{p}}\right)^{\star
p}=\left(\mathcal{B}_{p^{n-1}}(t_{n-1}^{\frac{k}{p}})\right)^{\star p}.$

Here we denoted by $A^{\star p}$ the $p$-fold $\star$-product
$A\star\cdots\star A$. By inductive hypothesis we also know that
$t_{n-1}^{\frac{k}{p}}\in\Set{t_{n-1}^{\frac{k}{p}-1}-1,t_{n-1}^{\frac{k}{p}-1}}$.
Using Lemma 2.2 we obtain that

$M+1\leq\lambda_{1}\leq
pt_{n-1}^{\frac{k}{p}}\leq(p-1)t_{n-1}^{\frac{k}{p}}+t_{n-1}^{\frac{k}{p}-1}\leq
M.$

This is a contradiction. Notice that the last inequality above follows from
the definition of $M$, as
$(\frac{k}{p},\ldots,\frac{k}{p},\frac{k}{p}-1)\in\mathcal{L}(k-1)$.

For $k\in[2,\alpha_{p^{n}}-1]$, what we have proved so far is summarised here.

$\begin{split}\Omega_{p^{n}}^{k}=\mathcal{B}_{p^{n}}(T)\mbox{, with
}&T=\max\Set{t_{n-1}^{j_{1}}+\cdots+t_{n-1}^{j_{p}}}{(j_{1},\dots,j_{p})\in\mathcal{L}(k-1)}\\\
\Omega_{p^{n}}^{k+1}=\mathcal{B}_{p^{n}}(V)\mbox{, with
}&V=\max\Set{t_{n-1}^{h_{1}}+\cdots+t_{n-1}^{h_{p}}}{(h_{1},\dots,h_{p})\in\mathcal{L}(k)}.\end{split}$

Let $(j_{1},\dots,j_{p})\in\mathcal{L}(k-1)$ be such that
$T=t_{n-1}^{j_{1}}+\cdots+t_{n-1}^{j_{p}}$. Without loss of generality, we can
assume that $j_{1}<\alpha_{p^{n-1}}$. Then
$(j_{1}+1,j_{2},\dots,j_{p})\in\mathcal{L}(k)$. By inductive hypothesis we
know that $t_{n-1}^{j_{1}+1}\in\Set{t_{n-1}^{j_{1}}-1,t_{n-1}^{j_{1}}}$. Hence

$V\geq
t_{n-1}^{j_{1}+1}+t_{n-1}^{j_{2}}+\cdots+t_{n-1}^{j_{p}}\in\Set{T-1,T}.$ (1)

On the other hand, let $(h_{1},\dots,h_{p})\in\mathcal{L}(k)$ be such that
$V=t_{n-1}^{h_{1}}+\cdots+t_{n-1}^{h_{p}}$. Since $k\geq 2$, without loss of
generality we can assume that $h_{1}>0$. Then
$(h_{1}-1,h_{2},\dots,h_{p})\in\mathcal{L}(k-1)$. Thus, as above:

$V=t_{n-1}^{h_{1}}+\cdots+t_{n-1}^{h_{p}}\leq T,$ (2)

since $t_{n-1}^{h_{1}}\in\Set{t_{n-1}^{h_{1}-1}-1,t_{n-1}^{h_{1}-1}}$.
Inequalities (1) and (2) imply that $V\in\Set{T-1,T}$. ∎

We refer the reader to the second part of Example 4.8 for a description of the
key steps of the proof of Theorem 4.2 in a small concrete instance. The
following definitions may seem artificial but are crucial for determining the
exact value of $t_{n}^{k}$ for all $n,k\in\mathbb{N}$.

######  Definition 4.3.

Let $n\in\mathbb{N}_{\geq 2}$ and let $x\in[1,p^{n-2}]$. We define the
integers $m_{x}$ and $\ell(n,x)$ as follows:

$m_{x}=\mathrm{min}\\{m\ |\ x\leq p^{m-2}\\},\ \text{and}\
\ell(n,x)=n-m_{x}+1.$

Notice that $\sum_{x=1}^{p^{n-2}}\ell(n,x)=\alpha_{p^{n}}$ (this is proved in
Lemma 4.4 below). For $x\in[1,p^{n-2}]$ we let

$A_{x}=\left[\sum_{j=1}^{x-1}\ell(n,j)+1,\sum_{j=1}^{x}\ell(n,j)\right].$

We observe that $\\{A_{1},A_{2},\ldots,A_{p^{n-2}}\\}$ is a partition of
$[1,\alpha_{p^{n}}]$ and that $|A_{x}|=\ell(n,x)$ for all $x\in[1,p^{n-2}]$.
We refer the reader to Example 4.8 for a description of these objects in a
specific setting.

For the convenience of the reader we give a more informal explanation of
Definition 4.3 above. For fixed $n\geq 2$, we define an increasing sequence
$0=a_{0}<a_{1}<a_{2}<\cdots<a_{p^{n-2}}=\alpha_{p^{n}}$ as follows. First
$a_{1}=n-1$. Then $a_{i}-a_{i-1}=n-2$, for $i=2,\ldots,p$. Next
$a_{i}-a_{i-1}=n-3$, for $i=p+1,p+2,\ldots,p^{2}$. Continue in this manner, we
find that $a_{i}-a_{i-1}=1$, for $i=p^{n-3}+1,\ldots,p^{n-2}$. Now set
$A_{i}:=(a_{i-1},a_{i}]$, for $i=1,\ldots,p^{n-2}$. Then
$\\{A_{1},A_{2},\ldots,A_{p^{n-2}}\\}$ is clearly a partition of
$[1,p^{n-2}]$.

######  Lemma 4.4.

With the notation introduced in Definition 4.3, we have that
$\sum_{x=1}^{p^{n-2}}\ell(n,x)=\alpha_{p^{n}}$.

###### Proof.

If $n=2$, then $\ell(2,1)=1=\alpha_{p^{2}}$. Let $n\geq 3$ and $i\in[0,n-3]$,
then for every $x\in[p^{i}+1,p^{i+1}]$, $m_{x}=i+3$ and $\ell(n,x)=n-i-2$.
Hence

$\begin{split}\sum_{x=1}^{p^{n-2}}\ell(n,x)&=\ell(n,1)+\sum_{i=0}^{n-3}\sum_{x=p^{i}+1}^{p^{i+1}}\ell(n,x)=(n-1)+\sum_{i=0}^{n-3}p^{i}(p-1)(n-i-2)\\\
&=(n-1)-(n-2)+\left(\sum_{i=1}^{n-3}p^{i}[(n-i-1)-(n-i-2)]\right)+p^{n-2}[n-(n-3)-2]\\\
&=1+\left(\sum_{i=1}^{n-3}p^{i}\right)+p^{n-2}=\alpha_{p^{n}}.\end{split}$

∎

The following technical lemma will be useful to prove Theorem 4.6.

######  Lemma 4.5.

Let $n\in\mathbb{N}\geq 2$ and $p$ be an odd prime. If $x=pa+r$, for some
$r\in[0,p-1]$ and $a\in\mathbb{N}$, then

$p\cdot\sum_{j=1}^{a}\ell(n-1,j)+r\cdot\ell(n-1,a+1)=\sum_{j=1}^{x}\ell(n,j)-1.$

###### Proof.

Notice that $\ell(n,1)=n-1$ and if $y\in[2,p]$, $\ell(n,y)=n-2$. Thus

$\sum_{y=1}^{p}\ell(n,y)=p\ell(n-1,1)+1.$

Moreover, for $j\in\mathbb{N}$ we have that

$\sum_{y=jp+1}^{jp+p}\ell(n,y)=p\ell(n-1,j+1).$

This follows by observing that $\ell(n,y)=\ell(n-1,j+1)$, for all
$y\in[jp+1,jp+p]$.

Using these facts, we deduce that

$\begin{split}\sum_{j=1}^{x}\ell(n,j)&=\sum_{j=1}^{p}\ell(n,j)+\sum_{j=1}^{a-1}\sum_{y=jp+1}^{jp+p}\ell(n,y)+\sum_{i=1}^{r}\ell(n,ap+i)\\\
&=1+p\ell(n-1,1)+p\sum_{j=1}^{a-1}\ell(n-1,j+1)+r\ell(n-1,a+1).\end{split}$

∎

The main result of this section shows that if $p^{k}$ is a character degree of
$P_{p^{n}}$, then the partitions of $p^{n}$ whose corresponding irreducible
character admit a constituent of degree $p^{k}$ on restriction to $P_{p^{n}}$
are precisely those which fit inside a square of length $p^{n}-x$, where $k\in
A_{x}$ determines $x$.

######  Theorem 4.6.

Let $n\geq 2$, $k\in[1,\alpha_{p^{n}}]$ and let $x\in[1,p^{n-2}]$ be such that
$k\in A_{x}$. Then $\Omega_{p^{n}}^{k}=\mathcal{B}_{p^{n}}(p^{n}-x).$

###### Proof.

We proceed by induction on $n$: if $n=2$ then $\alpha_{p^{2}}=1$ and
necessarily $k=1$ as $A_{1}=\\{1\\}$. By Lemma 4.1, we have that
$\Omega_{p^{2}}^{1}=\mathcal{B}_{p^{2}}(p^{2}-1)$, as required. If $n\geq 3$,
we proceed by induction on the parameter $x\in[1,p^{n-2}]$. For $x=1$, we want
to show that for every $k\in A_{1}=[1,\ell(n,1)]$ we have that
$\Omega_{p^{n}}^{k}=\mathcal{B}_{p^{n}}(p^{n}-1).$ Using Theorem 4.2 and Lemma
4.1, we know that

$\Omega_{p^{n}}^{\ell(n,1)}\subseteq\Omega_{p^{n}}^{k}\subseteq\Omega_{p^{n}}^{1}=\mathcal{B}_{p^{n}}(p^{n}-1).$

Hence, it is enough to show that
$\Omega_{p^{n}}^{\ell(n,1)}=\mathcal{B}_{p^{n}}(p^{n}-1)$. Since
$\ell(n,1)=\ell(n-1,1)+1$, we use Lemma 3.7, the inductive hypothesis on $n$
and [GN18, Theorem 3.1], to deduce that

$\Omega_{p^{n}}^{\ell(n,1)}\supseteq\Omega_{p^{n}}^{\ell(n-1,1)}\star\big{(}\Omega_{p^{n}}^{0}\big{)}^{\star
p-1}=\mathcal{B}_{p^{n}}(p^{n}-1)\star\big{(}\mathcal{B}_{p^{n}}(p^{n})\big{)}^{\star
p-1}.$

Using Lemma 2.3 we conclude that
$\mathcal{B}_{p^{n}}(p^{n}-1)\subseteq\Omega_{p^{n}}^{\ell(n,1)}$ and
therefore that $\mathcal{B}_{p^{n}}(p^{n}-1)=\Omega_{p^{n}}^{\ell(n,1)}$.

Let us now suppose that $x\geq 2$ and that $k\in A_{x}$. To ease the notation,
for any $y\in[1,p^{n-2}]$ we let $f_{n}(y)=\sum_{j=1}^{y}\ell(n,j)$. With this
notation we have that $A_{x}=[f_{n}(x-1)+1,f_{n}(x)]$. Using Theorem 4.2 and
arguing exactly as above, we observe that in order to show that
$\Omega_{p^{n}}^{k}=\mathcal{B}_{p^{n}}(p^{n}-x)$, it is enough to prove that

$(1)\ \Omega_{p^{n}}^{f_{n}(x-1)+1}=\mathcal{B}_{p^{n}}(p^{n}-x)\ \text{and
that}\ (2)\ \Omega_{p^{n}}^{f_{n}(x)}=\mathcal{B}_{p^{n}}(p^{n}-x).$

To prove $(1)$, we start by observing that by inductive hypothesis we know
that the statement holds for any $j\in A_{x-1}$. In particular we have that
$\Omega_{p^{n}}^{f_{n}(x-1)}=\mathcal{B}_{p^{n}}(p^{n}-(x-1))$. By Theorem 4.2
it follows that $\Omega_{p^{n}}^{f_{n}(x-1)+1}=\mathcal{B}_{p^{n}}(T)$, for
some $T\in\\{p^{n}-x,p^{n}-(x-1)\\}$. It is therefore enough to show that
$\lambda=(p^{n}-(x-1),x-1)\notin\Omega_{p^{n}}^{f_{n}(x-1)+1}$. Let
$\mu_{1},\ldots,\mu_{p}\in\mathcal{P}(p^{n-1})$ be such that
$\mathcal{LR}(\lambda;\mu_{1},\dots,\mu_{p})\neq 0$. By Lemma 2.2 for every
$i\in[1,p]$, there exists $a_{i}\in\mathbb{N}$ such that
$(\mu_{i})_{1}=p^{n-1}-a_{i}$ and such that $\sum_{j=1}^{p}a_{j}\leq x-1$. In
particular, for every $i\in[1,p]$ we have that

$\mu_{i}\in\mathcal{B}_{p^{n-1}}(p^{n-1}-a_{i})\smallsetminus\mathcal{B}_{p^{n-1}}(p^{n-1}-(a_{i}+1))=\Omega_{p^{n-1}}^{f_{n-1}(a_{i})}\smallsetminus\Omega_{p^{n-1}}^{f_{n-1}(a_{i})+1},$

where the equality is guaranteed by the inductive hypothesis on $n$.

Let $B=\left(P_{p^{n-1}}\right)^{\times p}$ be the base group of $P_{p^{n}}$
and let $Y=(\mathfrak{S}_{p^{n-1}})^{\times p}\leq\mathfrak{S}_{p^{n}}$ be
such that $B\leq Y$. Let
$\eta=\chi^{\mu_{1}}\times\cdots\times\chi^{\mu_{p}}\in\mathrm{Irr}(Y)$ and
let $x-1=ap+r$, for some $a\in\mathbb{N}$ and $r\in[0,p-1]$. We observe that

$\displaystyle\partial_{B}(\eta)$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{p}f_{n-1}(a_{j})\ =\
\sum_{j=1}^{p}\sum_{i=1}^{a_{j}}\ell(n-1,i)$ $\displaystyle\leq$
$\displaystyle
p\cdot\big{(}\sum_{j=1}^{a}\ell(n-1,j)\big{)}+r\cdot\ell(n-1,a+1)$
$\displaystyle=$ $\displaystyle\sum_{j=1}^{x-1}\ell(n,j)-1\ =\ f_{n}(x-1)-1.$

Here, the inequality follows immediately by observing that
$\ell(n-1,s)\geq\ell(n-1,s+1)$ for all $s\in\mathbb{N}$. On the other hand,
the third equality holds by Lemma 4.5. Using Proposition 3.5, we deduce that
$\partial_{P_{p^{n}}}(\chi^{\lambda})\leq f_{n}(x-1)$. It follows that
$\lambda\notin\Omega_{p^{n}}^{f_{n}(x-1)+1}$, as desired.

To prove $(2)$, we recall that by $(1)$ above we have that
$\Omega_{p^{n}}^{f_{n}(x-1)+1}=\mathcal{B}_{p^{n}}(p^{n}-x)$. Hence, Theorem
4.2 implies that
$\Omega_{p^{n}}^{f_{n}(x)}\subseteq\mathcal{B}_{p^{n}}(p^{n}-x)$. On the other
hand, writing $x=ap+r$ for some $a\in\mathbb{N}$ and $r\in[0,p-1]$, and using
Lemma 4.5, we have that:

$\displaystyle\Omega_{p^{n}}^{f_{n}(x)}$ $\displaystyle=$
$\displaystyle\Omega_{p^{n}}^{1+p\cdot\big{(}\sum_{j=1}^{a}\ell(n-1,j)\big{)}+r\cdot\ell(n-1,a+1)}$
$\displaystyle\supseteq$
$\displaystyle\big{(}\Omega_{p^{n-1}}^{f_{n-1}(a+1)}\big{)}^{\star
r}\star\big{(}\Omega_{p^{n-1}}^{f_{n-1}(a)}\big{)}^{\star p-r}$
$\displaystyle=$
$\displaystyle\big{(}\mathcal{B}_{p^{n-1}}(p^{n-1}-(a+1))\big{)}^{\star
r}\star\big{(}\mathcal{B}_{p^{n-1}}(p^{n-1}-a)\big{)}^{\star p-r}$
$\displaystyle=$ $\displaystyle\mathcal{B}_{p^{n}}(p^{n}-x).$

Here the first inclusion follows from Lemma 3.7. The second equality holds by
inductive hypothesis. Finally, the last equality is given by Lemma 2.3. The
proof is complete. ∎

In the following corollary we collect a number of facts useful to have a
better understanding of the structure of the sets $\Omega_{p^{n}}^{k}$ for all
$n\in\mathbb{N}$ and all $k\in[0,\alpha_{p^{n}}]$.

######  Corollary 4.7.

Let $n\in\mathbb{N}$ and let $1\leq k<t\leq\alpha_{p^{n}}$. The following
hold.

  * (i)

$\mathcal{B}_{p^{n}}(p^{n}-p^{n-2})=\Omega_{p^{n}}^{\alpha_{p^{n}}}\subseteq\Omega_{p^{n}}^{t}\subseteq\Omega_{p^{n}}^{k}$.

  * (ii)

$\Omega_{p^{n}}^{k}=\Omega_{p^{n}}^{t}$ if, and only if, there exists
$x\in[1,p^{n-2}]$ such that $k,t\in A_{x}$.

  * (iii)

Given $x\in[1,p^{n-2}]$ we have that $|\\{k\in[1,\alpha_{p^{n}}]\ |\
\Omega_{p^{n}}^{k}=\mathcal{B}(p^{n}-x)\\}|=\ell(n,x)$.

###### Proof.

Recalling that $|A_{x}|=\ell(n,x)$ for every $x\in[1,p^{n-2}]$, (i), (ii) and
(iii) follow immediately by Theorem 4.6. ∎

We find particularly surprising that a partition of $p^{n}$ whose character
admits an irreducible constituent of degree $p^{k}$ on restriction to
$P_{p^{n}}$ also admits a constituent of degree $p^{j}$, for any
$j\in\\{0,1,\ldots,k-1\\}$. Moreover, the partitions whose character admit a
constituent of maximal possible degree $p^{\alpha_{p^{n}}}$ are precisely
those which fit inside a square of side $p^{n}-p^{n-2}$.

######  Example 4.8.

Let $p=3$ and fix $n=4$. Following the notation introduced in Definition 4.3,
we have $3^{4-2}=9$ and $\ell(4,1)=3,\ \ell(4,2)=\ell(4,3)=2,\
\ell(4,4)=\cdots=\ell(4,9)=1$. Hence

$A_{1}=\Set{1,2,3},\ A_{2}=\Set{4,5},\ A_{3}=\Set{6,7},\ A_{4}=\Set{8},\
A_{5}=\Set{9},\dots,A_{9}=\Set{13}.$

Observe that $\Set{A_{1},\dots,A_{9}}$ is a partition of
$[1,\alpha_{3^{4}}]=[1,13]$, as required. Using Theorem 4.6, we have a
complete description of $\Omega_{3^{4}}^{k}$, for all $k\in[1,13]$. In
particular, we have

$\Omega_{3^{4}}^{1}=\Omega_{3^{4}}^{2}=\Omega_{3^{4}}^{3}=\mathcal{B}_{3^{4}}(3^{4}-1),\
\Omega_{3^{4}}^{4}=\Omega_{3^{4}}^{5}=\mathcal{B}_{3^{4}}(3^{4}-2),\
\Omega_{3^{4}}^{6}=\Omega_{3^{4}}^{7}=\mathcal{B}_{3^{4}}(3^{4}-3),$
$\Omega_{3^{4}}^{8}=\mathcal{B}_{3^{4}}(3^{4}-4),\
\Omega_{3^{4}}^{9}=\mathcal{B}_{3^{4}}(3^{4}-5),\dots,\Omega_{3^{4}}^{13}=\mathcal{B}_{3^{4}}(3^{4}-9).$

These sets are recorded in the fourth column of Table 1.

We use the second part of this example to illustrate a key step of the proof
of Theorem 4.2. Let $n=k=4$. We wish to compute $t_{4}^{4}$. Following the
notation introduced in the proof of Theorem 4.2 we have that

$\mathcal{L}(3)=\\{(j_{1},j_{2},j_{3})\in\mathcal{C}(3)\ |\
j_{i}\in[0,\alpha_{3^{3}}]=[0,4],\text{ for all
}i\in[1,3]\\}=\\{(3,0,0),(2,1,0),(1,1,1)\\}.$

Working by induction we can assume that we know the values $t_{3}^{j}$ for
every $j\in[0,4]$. This can be comfortably read off the third column of Table
1. We set

$M=\max\Set{t_{3}^{3}+t_{3}^{0}+t_{3}^{0},t_{3}^{2}+t_{3}^{1}+t_{3}^{0},t_{3}^{1}+t_{3}^{1}+t_{3}^{1}}=\max\Set{3^{4}-2,3^{4}-2,3^{4}-3}=3^{4}-2.$

We conclude that $t_{4}^{4}=M=3^{4}-2$.

Table 1. Let $p=3$. According to Theorem 4.6, the structure of $\Omega_{p^{n}}^{k}$ is recorded in the entry corresponding to row $k$ and column $n$. $\Omega_{p^{n}}^{k}$ | $n=1$ | $n=2$ | $n=3$ | $n=4$
---|---|---|---|---
$k=0$ | $\mathcal{B}_{3}(3)$ | $\mathcal{B}_{3^{2}}(3^{2})$ | $\mathcal{B}_{3^{3}}(3^{3})$ | $\mathcal{B}_{3^{4}}(3^{4})$
$k=1$ | $\emptyset$ | $\mathcal{B}_{3^{2}}(3^{2}-1)$ | $\mathcal{B}_{3^{3}}(3^{3}-1)$ | $\mathcal{B}_{3^{4}}(3^{4}-1)$
$k=2$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{3^{3}}(3^{3}-1)$ | $\mathcal{B}_{3^{4}}(3^{4}-1)$
$k=3$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{3^{3}}(3^{3}-2)$ | $\mathcal{B}_{3^{4}}(3^{4}-1)$
$k=4$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{3^{3}}(3^{3}-3)$ | $\mathcal{B}_{3^{4}}(3^{4}-2)$
$k=5$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{3^{4}}(3^{4}-2)$
$k=6$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{3^{4}}(3^{4}-3)$
$k=7$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{3^{4}}(3^{4}-3)$
$k=8$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{3^{4}}(3^{4}-4)$
$k=9$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{3^{4}}(3^{4}-5)$
$k=10$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{3^{4}}(3^{4}-6)$
$k=11$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{3^{4}}(3^{4}-7)$
$k=12$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{3^{4}}(3^{4}-8)$
$k=13$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{3^{4}}(3^{4}-9)$
$k=14$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\emptyset$

##  5\. Arbitrary natural numbers

The aim of this section is to complete our investigation by extending Theorem
4.6 to any arbitrary natural number. In order to do this, we first extend
Theorem 4.2. We recall that $p$ is a fixed odd prime.

######  Theorem 5.1.

Let $n\in\mathbb{N}$ and let $k\in[0,\alpha_{n}]$. There exists
$T_{n}^{k}\in[1,n]$ such that $\Omega_{n}^{k}=\mathcal{B}_{n}(T_{n}^{k})$.
Moreover, $T_{n}^{k+1}\in\\{T_{n}^{k}-1,T_{n}^{k}\\},\ \text{for all}\
k\in[0,\alpha_{n}-1].$

###### Proof.

We proceed by induction on $n\in\mathbb{N}$. If $n=1$, then necessarily $k=0$
and $\Omega_{1}^{0}=\mathcal{B}_{1}(1)$. If $n\geq 2$, let
$n=\sum_{i=1}^{t}p^{n_{i}}$ be the $p$-adic expansion of $n$, with
$n_{1}\geq\dots\geq n_{t}\geq 0$. By Theorem 4.2, for every $i\in[1,t]$ and
every $d_{i}\in[0,\alpha_{p^{n_{i}}}]$, there exists
$t_{n_{i}}^{d_{i}}\in[\frac{p^{n_{i}}}{2}+1,p^{n_{i}}]$ such that
$\Omega_{p^{n_{i}}}^{d_{i}}=\mathcal{B}_{p^{n_{i}}}\left(t_{n_{i}}^{d_{i}}\right)$.
Similarly to the procedure used to prove Theorem 4.2, we define

$\mathcal{J}(k)=\\{(j_{1},\dots,j_{t})\in\mathcal{C}(k)\ |\
j_{i}\in[0,\alpha_{p^{n_{i}}}]\text{ for all }i\in[1,t]\\}.$

Moreover, we set

$M=\max\Set{\sum_{i=1}^{t}t_{n_{i}}^{d_{i}}}{(d_{1},\dots,d_{t})\in\mathcal{J}(k)}.$

We claim that $\Omega_{n}^{k}=\mathcal{B}_{n}(M)$.

Let $(d_{1},\ldots,d_{t})\in\mathcal{J}(k)$ be such that
$M=\sum_{i=1}^{t}t_{n_{i}}^{d_{i}}$. Then using Lemma 2.3, Theorem 4.2 and
Lemma 3.7 we have that

$\mathcal{B}_{n}(M)=\mathcal{B}_{p^{n_{1}}}\left(t_{n_{1}}^{d_{1}}\right)\star\cdots\star\mathcal{B}_{p^{n_{t}}}\left(t_{n_{t}}^{d_{t}}\right)=\Omega_{p^{n_{1}}}^{d_{1}}\star\cdots\star\Omega_{p^{n_{t}}}^{d_{t}}\subseteq\Omega_{n}^{k}.$

Suppose now for a contradiction that
$\lambda\in\Omega_{n}^{k}\smallsetminus\mathcal{B}_{n}(M)$. Without loss of
generality we can assume that $\lambda_{1}\geq M+1$. Let
$\phi=\phi_{1}\times\cdots\times\phi_{t}$ be an irreducible constituent of
$(\chi^{\lambda})_{P_{n}}$ with $\phi_{i}(1)=p^{d_{i}}$ for every $i\in[1,t]$
and $\sum_{i=1}^{t}d_{i}=k$. Observe that
$(d_{1},\ldots,d_{t})\in\mathcal{J}(k)$. For every $i\in[1,t]$, let
$\mu_{i}\in\mathcal{P}(p^{n_{i}})$ be such that
$[(\chi^{\mu_{i}})_{P_{p^{n_{i}}}},\phi_{i}]\neq 0$ and such that
$\chi^{\mu_{1}}\times\cdots\times\chi^{\mu_{t}}$ is an irreducible constituent
of $(\chi^{\lambda})_{Y}$. Here
$Y=\mathfrak{S}_{p^{n_{1}}}\times\cdots\times\mathfrak{S}_{p^{n_{t}}}\leq\mathfrak{S}_{n}$
is chosen so that $P_{n}\leq Y$. Thus by Theorem 4.6,
$\mu_{i}\in\Omega_{p^{n_{i}}}^{d_{i}}=\mathcal{B}_{p^{n_{i}}}\left(t_{n_{i}}^{d_{i}}\right)$
for every $i\in[1,t]$. Hence,

$\lambda\in\mathcal{B}_{p^{n_{1}}}\left(t_{n_{1}}^{d_{1}}\right)\star\cdots\star\mathcal{B}_{p^{n_{t}}}\left(t_{n_{t}}^{d_{t}}\right)=\mathcal{B}_{n}\left(\sum_{i=1}^{t}t_{n_{i}}^{d_{i}}\right).$

By Lemma 2.2 and our assumptions, we have that

$M+1\leq\lambda_{1}\leq\sum_{i=1}^{t}t_{n_{i}}^{d_{i}}\leq M,$

which is a contradiction.

In summary, for $k\in[0,\alpha_{n}-1]$ the following holds:

$\begin{split}\Omega_{n}^{k}=\mathcal{B}_{n}(M),&\text{ where
}M=\max\Set{\sum_{i=1}^{t}t_{n_{i}}^{d_{i}}}{(d_{1},\dots,d_{t})\in\mathcal{J}(k)},\text{
and }\\\ \Omega_{n}^{k+1}=\mathcal{B}_{n}(T),&\text{ where
}T=\max\Set{\sum_{i=1}^{t}t_{n_{i}}^{f_{i}}}{(f_{1},\dots,f_{t})\in\mathcal{J}(k+1)}.\end{split}$

Let $(d_{1},\dots,d_{t})\in\mathcal{J}(k)$ be such that
$M=\sum_{i=1}^{t}t_{n_{i}}^{d_{i}}$. Since $k\leq\alpha_{n}-1$, there exists
$i\in[1,t]$ such that $d_{i}\leq\alpha_{p^{n_{i}}}-1$. Hence
$(d_{1},\dots,d_{i-1},d_{i}+1,d_{i+1},\ldots,d_{t})\in\mathcal{J}(k+1)$ and
$t_{n_{i}}^{d_{i}+1}\in\Set{t_{n_{i}}^{d_{i}}-1,t_{n_{i}}^{d_{i}}}$, by
Theorem 4.2. Thus,

$M-1=-1+\sum_{i=1}^{t}t_{n_{i}}^{d_{i}}\leq
t_{n_{1}}^{d_{1}}+\cdots+t_{n_{i-1}}^{d_{i-1}}+t_{n_{i}}^{d_{i}+1}+t_{n_{i+1}}^{d_{i+1}}+\cdots+t_{n_{t}}^{d_{t}}\leq
T.$

On the other hand, let $(f_{1},\dots,f_{t})\in\mathcal{J}(k+1)$ be such that
$T=\sum_{i=1}^{t}t_{n_{i}}^{f_{i}}$. Without loss of generality we can assume
that $f_{1}\geq 1$. Then $(f_{1}-1,f_{2},\dots,f_{t})\in\mathcal{J}(k)$ and by
Theorem 4.2,
$t_{n_{1}}^{f_{1}}\in\Set{t_{n_{1}}^{f_{1}-1}-1,t_{n_{1}}^{f_{1}-1}}$. Hence

$T=\sum_{i=1}^{t}t_{n_{i}}^{f_{i}}\leq
t_{n_{1}}^{f_{1}-1}+t_{n_{2}}^{f_{2}}+\cdots+t_{n_{t}}^{f_{t}}\leq M.$

It follows that $T=M$ or $T=M-1$. This concludes the proof. ∎

Theorem 5.1 shows that for every $n\in\mathbb{N}$ and $k\in[0,\alpha_{n}]$
there exists an integer, denoted by $T_{n}^{k}$, such that
$\Omega_{n}^{k}=\mathcal{B}_{n}(T_{n}^{k})$. In order to prove our main
result, i.e. to precisely compute the value $T_{n}^{k}$ for all
$n\in\mathbb{N}$ and $k\in[0,\alpha_{n}]$, we start by fixing some notation
that will be kept throughout this section. We remark that for $n<p^{2}$ we
have that $P_{n}$ is abelian and that $\Omega_{n}^{0}=\mathcal{P}(n)$. For
this reason we focus on the case $n\geq p^{2}$.

######  Notation 5.2.

Let $n\geq p^{2}$ be a natural number and let $n=\sum_{i=1}^{t}p^{n_{i}}$ be
the $p$-adic expansion of $n$, where $n_{1}\geq n_{2}\geq\cdots\geq n_{t}\geq
0$. Let $\mathcal{R}:=\\{(i,y)\ |\ i\in[1,t],\ \text{and}\
y\in[1,p^{n_{i}-2}]\\}$. We define a total order $\triangleright$ on
$\mathcal{R}$ as follows. Given $(i,y)$ and $(j,z)$ in $\mathcal{R}$ we say
that $(i,y)\triangleright(j,z)$ if and only if one of the following hold:

  * (i)

$\ell(n_{i},y)>\ell(n_{j},z)$, or

  * (ii)

$\ell(n_{i},y)=\ell(n_{j},z)$ and $i<j$, or

  * (iii)

$\ell(n_{i},y)=\ell(n_{j},z)$ and $i=j$ and $y<z$.

Let $N:=\lfloor\frac{n}{p^{2}}\rfloor$ and notice that $N=|\mathcal{R}|$. Let
$\phi:\mathcal{R}\longrightarrow[1,N]$ be the bijection mapping $(i,y)\mapsto
x$ if and only if the pair $(i,y)$ is the $x$-th greatest element in the
totally ordered set $(\mathcal{R},\triangleright)$. We use this bijection to
relabel the integers $\ell(n_{i},y)$, for all $(i,y)\in\mathcal{R}$. In
particular, we let $\ell(x):=\ell(n_{i},y)$ if $\phi((i,y))=x$. Recalling
Definition 4.3, we observe that the definition of $\triangleright$ implies
that $\ell(1)\geq\ell(2)\geq\cdots\geq\ell(N).$

Finally, for any $\alpha\in[1,N]$ we let
$F_{n}(\alpha)=\sum_{a=1}^{\alpha}\ell(a)$ and
$A_{\alpha}=[\\{F_{n}(\alpha-1)+1,F_{n}(\alpha)\\}]$. We observe that
$\\{A_{1},A_{2},\ldots,A_{N}\\}$ is a partition of $[1,\alpha_{n}]$ (this
follows easily from Lemma 4.4). We refer the reader to Example 5.6 for an
explicit description of these objects in a concrete case.

######  Theorem 5.3.

Let $n\in\mathbb{N}_{\geq p^{2}}$ and $k\in[1,\alpha_{n}]$. Let $x\in[1,N]$ be
such that $k\in A_{x}$. Then

$\Omega_{n}^{k}=\mathcal{B}_{n}(n-x).$

###### Proof.

As in Notation 5.2, let $n=\sum_{i=1}^{t}p^{n_{i}}$ be the $p$-adic expansion
of $n$, where $n_{1}\geq n_{2}\geq\cdots\geq n_{t}\geq 0$. We proceed by
induction on $x$. If $x=1$ then $k\in A_{1}=[1,\ell(1)]=[1,\ell(n_{1},1)]$,
because $\phi((n_{1},1))=1$. By Theorem 4.6 we know that
$\Omega_{p^{n_{1}}}^{k}=\mathcal{B}_{p^{n_{1}}}(p^{n_{1}}-1)$. Moreover,
$\Omega_{p^{m}}^{0}=\mathcal{B}_{p^{m}}(p^{m})$ for all $m\in\mathbb{N}$ by
[GN18, Theorem 3.1]. Thus, using first Lemma 3.8 and then Lemma 2.3, we deduce
that

$\Omega_{n}^{k}\supseteq\Omega_{p^{n_{1}}}^{k}\star\Omega_{p^{n_{2}}}^{0}\star\cdots\star\Omega_{p^{n_{t}}}^{0}=\mathcal{B}_{p^{n_{1}}}(p^{n_{1}}-1)\star\mathcal{B}_{p^{n_{2}}}(p^{n_{2}})\star\cdots\star\mathcal{B}_{p^{n_{t}}}(p^{n_{t}})=\mathcal{B}_{n}(n-1).$

Since $(n)\notin\Omega_{n}^{k}$, we conclude that
$\Omega_{n}^{k}=\mathcal{B}_{n}(n-1)$, as desired. Let us now set $x\geq 2$
and assume that the statement holds for any $s\in
A_{x-1}=[F_{n}(x-1)+1,F_{n}(x)]$. From Theorem 5.1 we know that

$\Omega_{n}^{F_{n}(x-1)+1}\subseteq\Omega_{n}^{k}\subseteq\Omega_{n}^{F_{n}(x)},$

hence it is enough to show that:

$(1)\ \Omega_{n}^{F_{n}(x-1)+1}=\mathcal{B}_{n}(n-x),\ \text{and that}\ (2)\
\Omega_{n}^{F_{n}(x)}=\mathcal{B}_{n}(n-x).$

Here $F_{n}(y)=\sum_{j=1}^{y}\ell(j)$, exactly as explained in Notation 5.2.

To prove (1), we first notice that
$\Omega_{n}^{F_{n}(x-1)}=\mathcal{B}_{n}(n-(x-1))$ by inductive hypothesis.
Hence, Theorem 5.1 implies that
$\Omega_{n}^{F_{n}(x-1)+1}=\mathcal{B}_{n}(T)$, for some
$T\in\\{n-x,n-(x-1)\\}$. Therefore it suffices to prove that
$\lambda=(n-(x-1),x-1)\notin\Omega_{n}^{F_{n}(x-1)+1}$. Let
$\\{G_{1},G_{2},\ldots,G_{t}\\}$ be the partition of $[1,x-1]$ defined by

$G_{i}=\\{y\in[1,x-1]\ |\ \phi^{-1}(y)=(i,z),\ \text{for some}\
z\in[1,p^{n_{i}-2}]\\},\ \text{for all}\ i\in[1,t].$

To ease the notation we let $g_{i}=|G_{i}|$ for all $i\in[1,t]$, and we remark
that $g_{1}+g_{2}+\cdots+g_{t}=x-1$.

Let
$Y=\mathfrak{S}_{p^{n_{1}}}\times\mathfrak{S}_{p^{n_{2}}}\times\cdots\times\mathfrak{S}_{p^{n_{t}}}$
be a Young subgroup of $\mathfrak{S}_{n}$ containing $P_{n}$. For every
$i\in[1,t]$ let $\mu^{i}\in\mathcal{P}(p^{n_{i}})$ be such that
$\mathcal{LR}(\lambda;\mu^{1},\ldots,\mu^{t})\neq 0$. Then Lemma 2.2 implies
that there exist $a_{1},a_{2},\ldots,a_{t}\in\mathbb{Z}$ such that

$(\mu^{i})_{1}=p^{n_{i}}-(g_{i}+a_{i})\ \text{for all}\ i\in[1,t],\ \text{and
such that}\ \sum_{i=1}^{t}a_{i}\leq 0.$

In particular, using Theorem 4.6 we have that for every $i\in[1,t]$,

$\mu^{i}\in\mathcal{B}_{p^{n_{i}}}(p^{n_{i}}-(g_{i}+a_{i}))\smallsetminus\mathcal{B}_{p^{n_{i}}}(p^{n_{i}}-(g_{i}+a_{i}+1))=\Omega_{p^{n_{i}}}^{f_{n_{i}}(g_{i}+a_{i})}\smallsetminus\Omega_{p^{n_{i}}}^{f_{n_{i}}(g_{i}+a_{i})+1}.$

Recycling the notation used in the proof of Theorem 4.6, here
$f_{m}(a):=\sum_{j=1}^{a}\ell(m,j)$. It follows that

$\partial_{P_{p^{n_{i}}}}(\chi^{\mu^{i}})=\sum_{j=1}^{g_{i}+a_{i}}\ell(n_{i},j)=\begin{cases}\sum_{y\in
G_{i}}\ell(y)+\sum_{j=g_{i}+1}^{g_{i}+a_{i}}\ell(n_{i},j)&\mathrm{if}\
a_{i}\geq 0,\\\ \\\ \sum_{y\in
G_{i}}\ell(y)-\sum_{j=g_{i}+a_{i}}^{g_{i}}\ell(n_{i},j)&\mathrm{if}\
a_{i}<0.\end{cases}$

Hence, letting
$\chi=\chi^{\mu^{1}}\times\chi^{\mu^{2}}\times\cdots\times\chi^{\mu^{t}}$, we
have that

$\partial_{P_{n}}(\chi)=\sum_{i=1}^{t}\sum_{y\in G_{i}}\ell(y)+E-F,\
\text{where}\ E=\sum_{\begin{subarray}{c}i=1\\\
a_{i}>0\end{subarray}}^{t}\sum_{j=g_{i}+1}^{g_{i}+a_{i}}\ell(n_{i},j),\
\text{and}\ F=\sum_{\begin{subarray}{c}i=1\\\
a_{i}<0\end{subarray}}^{t}\sum_{j=g_{i}+a_{i}}^{g_{i}}\ell(n_{i},j).$

We claim that $E-F\leq 0$. To see this, we notice that the definition of the
set $G_{i}$ implies that $\phi((i,y))>x-1$ for all $y\geq g_{i}+1$. On the
other hand, for the same reasons, we have that $\phi((j,z))\leq x-1$ for all
$z\leq g_{j}$. Therefore every summand $\ell(n_{i},y)$ appearing in $E$ is
smaller than or equal to any summand $\ell(n_{j},z)$ appearing in $F$. Since
$\sum_{i=1}^{t}a_{i}\leq 0$ we have that $E-F\leq 0$, as desired. Using
Proposition 3.6 we conclude that

$\partial_{P_{n}}(\chi^{\lambda})\leq\sum_{i=1}^{t}\sum_{y\in
G_{i}}\ell(y)=\sum_{y=1}^{x-1}\ell(y)=F_{n}(x-1)<F_{n}(x-1)+1.$

Hence $\lambda\notin\Omega_{n}^{F_{n}(x-1)+1}$ and therefore
$\Omega_{n}^{F_{n}(x-1)+1}=\mathcal{B}_{n}(n-x)$ as required.

To prove (2) we observe that the equality (1) shown above implies that
$\Omega_{n}^{F_{n}(x)}\subseteq\mathcal{B}_{n}(n-x)$, by Theorem 5.1. To show
that the opposite inclusion holds we use an idea that is similar to the one
used to prove (1). In particular, we let $\\{H_{1},H_{2},\ldots,H_{t}\\}$ be
the partition of $[1,x]$ defined by

$H_{i}=\\{y\in[1,x]\ |\ \phi^{-1}(y)=(i,z),\ \text{for some}\
z\in[1,p^{n_{i}-2}]\\},\ \text{for all}\ i\in[1,t].$

To ease the notation we let $h_{i}=|H_{i}|$ for all $i\in[1,t]$, and we remark
that $h_{1}+h_{2}+\cdots+h_{t}=x$. We also introduce the following notation.
For each $i\in[1,t]$, we let

$\Gamma_{i}:=\sum_{y\in
H_{i}}\ell(y)=\sum_{j=1}^{h_{i}}\ell(n_{i},j)=f_{n_{i}}(h_{i}).$

We observe that
$(\Gamma_{1},\Gamma_{2},\ldots,\Gamma_{t})\in\mathcal{C}(F_{n}(x))$ and that
$\Gamma_{i}\in[0,\alpha_{p^{n_{i}}}]$, for all $i\in[1,t]$. We can now use
Lemma 3.8, Theorem 4.6 and Lemma 2.3 (in this order) to deduce that

$\Omega_{n}^{F_{n}(x)}\supseteq\Omega_{p^{n_{1}}}^{\Gamma_{1}}\star\Omega_{p^{n_{2}}}^{\Gamma_{2}}\star\cdots\star\Omega_{p^{n_{t}}}^{\Gamma_{t}}=\mathcal{B}_{p^{n_{1}}}(p^{n_{1}}-h_{1})\star\mathcal{B}_{p^{n_{2}}}(p^{n_{2}}-h_{2})\star\cdots\star\mathcal{B}_{p^{n_{t}}}(p^{n_{t}}-h_{t})=\mathcal{B}_{n}(n-x).$

We obtain that $\Omega_{n}^{F_{n}(x)}=\mathcal{B}_{n}(n-x)$, and the proof is
concluded. ∎

As we have done for the prime power case in Corollary 4.7, we record some
facts to understand better the set $\Omega_{n}^{k}$ for every $n\in\mathbb{N}$
and $k\in[0,\alpha_{n}]$. Keeping the notation introduced in 5.2, we recall
that $N=\lfloor\frac{n}{p^{2}}\rfloor$.

######  Corollary 5.4.

Let $n\in\mathbb{N}$ and $n=\sum_{i=1}^{t}p^{n_{i}}$ its $p$-adic expansion,
where $n_{1}\geq n_{2}\geq\cdots\geq n_{t}\geq 0$. Let $1\leq
k<t\leq\alpha_{n}$. The following hold.

  * (i)

$\mathcal{B}_{n}(n-N)=\Omega_{n}^{\alpha_{n}}\subseteq\Omega_{n}^{t}\subseteq\Omega_{n}^{k}$.

  * (ii)

$\Omega_{n}^{k}=\Omega_{n}^{t}$ if, and only if, there exists $x\in[1,N]$ such
that $k,t\in A_{x}$.

  * (iii)

Given $x\in[1,N]$ we have that
$|\Set{k\in[1,\alpha_{n}]}{\Omega_{n}^{k}=\mathcal{B}_{n}(n-x)}|=\ell(x)$.

###### Proof.

Since $|A_{x}|=\ell(x)$ for every $x\in[1,N]$, (i), (ii) and (iii) hold by
Theorem 5.3. ∎

A second consequence of Theorem 5.3 is the following asymptotic result. This
basically says that when $n$ is arbitrarily large, almost all irreducible
characters of $\mathfrak{S}_{n}$ admit constituents of every possible degree
on restriction to a Sylow $p$-subgroup.

######  Corollary 5.5.

Let $\Omega_{n}=\bigcap_{k}\Omega_{n}^{k}$, where $k$ runs over
$[0,\alpha_{n}]$. Then

$\lim_{n\rightarrow\infty}\frac{|\Omega_{n}|}{|\mathcal{P}(n)|}=1.$

###### Proof.

A result of Erdős and Lehner [EL41, (1.4)] guarantees that given $f(n)$ a
function that diverges as $n$ tends to infinity, then for all but
$o(|\mathcal{P}(n)|)$ partitions $\lambda$ of $n$, the quantities
$\lambda_{1}$ and $l(\lambda)$ lie between $\sqrt{n}\cdot(\tfrac{\log n}{d}\pm
f(n))$ where $d$ is a constant. By Theorem 5.3, we observe that
$\Omega_{n}=\Omega_{n}^{\alpha_{n}}=\mathcal{B}(n-N)$, where
$N=\lfloor\frac{n}{p^{2}}\rfloor$. Since $n-N\geq n/2$, the statement follows.
∎

######  Example 5.6.

Let $p=3$ and $n=3^{3}+3^{2}+3$. Following Notation 5.2, we have $n_{1}=3,\
n_{2}=2$ and $n_{3}=1$. Hence $\mathcal{R}=\Set{(1,1),(1,2),(1,3),(2,1)}$,
since $[1,3^{n_{3}-2}]=\emptyset$. Observe that
$|\mathcal{R}|=4=\lfloor\frac{n}{3^{2}}\rfloor$. Using Definition 4.3, we can
see that $\ell(3,1)=2,\ \ell(3,2)=\ell(3,3)=1$ and $\ell(2,1)=1$. Hence, the
definition of the total order $\triangleright$ on $\mathcal{R}$ implies that
$(1,1)\triangleright(1,2)\triangleright(1,3)\triangleright(2,1)$. Thus
$\ell(1)=2,\ \ell(2)=\ell(3)=\ell(4)=1$ and

$A_{1}=\Set{1,2},\ A_{2}=\Set{3},\ A_{3}=\Set{4},\ A_{4}=\Set{5}.$

Notice that $\Set{A_{1},\dots,A_{4}}$ is a partition of
$[1,\alpha_{n}]=[1,5]$, as required. Moreover by Theorem 5.3 we have
$\Omega_{n}^{1}=\Omega_{n}^{2}=\mathcal{B}_{n}(n-1),\
\Omega_{n}^{3}=\mathcal{B}_{n}(n-2),\ \Omega_{n}^{4}=\mathcal{B}_{n}(n-3),\
\Omega_{n}^{5}=\mathcal{B}_{n}(n-4)$.

Using the notation of Theorem 5.1, the above computation gives that
$T_{n}^{2}=n-1$. Following the proof of Theorem 5.1, we can compute
$T_{n}^{2}$ in a different way. We have

$\mathcal{J}(2)=\Set{(j_{1},j_{2},j_{3})\in\mathcal{C}(2)}{j_{1}\in[0,4],\
j_{2}\in[0,1],\ j_{3}\in\Set{0}}=\Set{(2,0,0),(1,1,0)}.$

Hence
$M=\max\Set{t_{3}^{2}+t_{2}^{0}+t_{1}^{0},t_{3}^{1}+t_{2}^{1}+t_{1}^{0}}=\max\Set{n-1,n-2}=n-1$.
Thus $T_{n}^{2}=M=n-1$, as expected. Notice that $n_{3}=1$ does not contribute
at all to the computations. In fact in $\mathcal{R}$ there are no elements of
the form $(3,y)$, $y\in\mathbb{N}$. Furthermore, by looking at the third
column of Table 2, we can see that $T_{n}^{k}=T_{n-3}^{k}-3$ for every
$k\in\mathbb{N}$. A second example of this fact can be found by observing that
the first two columns of Table 2 are equal.

Table 2. Let $p=3$. According to Theorem 5.3, the structure of $\Omega_{n}^{k}$ is recorded in the entry corresponding to row $k$ and column $n$. $\Omega_{n}^{k}$ | $n=3+3^{3}$ | $n=2\cdot 3+3^{3}$ | $n=3^{2}+3^{3}$ | $n=3^{3}+3^{3}$ | $n=3^{3}+3^{4}$
---|---|---|---|---|---
$k=0$ | $\mathcal{B}_{n}(n)$ | $\mathcal{B}_{n}(n)$ | $\mathcal{B}_{n}(n)$ | $\mathcal{B}_{n}(n)$ | $\mathcal{B}_{n}(n)$
$k=1$ | $\mathcal{B}_{n}(n-1)$ | $\mathcal{B}_{n}(n-1)$ | $\mathcal{B}_{n}(n-1)$ | $\mathcal{B}_{n}(n-1)$ | $\mathcal{B}_{n}(n-1)$
$k=2$ | $\mathcal{B}_{n}(n-1)$ | $\mathcal{B}_{n}(n-1)$ | $\mathcal{B}_{n}(n-1)$ | $\mathcal{B}_{n}(n-1)$ | $\mathcal{B}_{n}(n-1)$
$k=3$ | $\mathcal{B}_{n}(n-2)$ | $\mathcal{B}_{n}(n-2)$ | $\mathcal{B}_{n}(n-2)$ | $\mathcal{B}_{n}(n-2)$ | $\mathcal{B}_{n}(n-1)$
$k=4$ | $\mathcal{B}_{n}(n-3)$ | $\mathcal{B}_{n}(n-3)$ | $\mathcal{B}_{n}(n-3)$ | $\mathcal{B}_{n}(n-2)$ | $\mathcal{B}_{n}(n-2)$
$k=5$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{n}(n-4)$ | $\mathcal{B}_{n}(n-3)$ | $\mathcal{B}_{n}(n-2)$
$k=6$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{n}(n-4)$ | $\mathcal{B}_{n}(n-3)$
$k=7$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{n}(n-5)$ | $\mathcal{B}_{n}(n-3)$
$k=8$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{n}(n-6)$ | $\mathcal{B}_{n}(n-4)$
$k=9$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{n}(n-4)$
$k=10$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{n}(n-5)$
$k=11$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{n}(n-6)$
$k=12$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{n}(n-7)$
$k=13$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{n}(n-8)$
$k=14$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{n}(n-9)$
$k=15$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{n}(n-10)$
$k=16$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{n}(n-11)$
$k=17$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\mathcal{B}_{n}(n-12)$
$k=18$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\emptyset$ | $\emptyset$

## References

  * [EL41] P. Erdős and J. Lehner,  The distribution of the number of summands in the partitions of a positive integer,  _Duke Math. J._ 8 (1941), 335–345.
  * [F97] W. Fulton, _Young tableaux,_ London Mathematical Society Student Texts, vol. 35, Cambridge University Press, Cambridge, 1997, With applications to representation theory and geometry.
  * [G17] E. Giannelli, Characters of odd degree of symmetric groups,  _J. London Math. Soc._ (1), 96 (2017), 1–14.
  * [G21] E. Giannelli, McKay bijections for symmetric and alternating groups,  _Algebra Number Theory_ , 15 (2021), no. 7, 1809–1835.
  * [GL21] E. Giannelli and S. Law, Sylow branching coefficients for symmetric groups,  _Journal of the London Mathematical Society_ , (2) 103 (2021), 697–728.
  * [GN18] E. Giannelli and G. Navarro, Restricting irreducible characters to Sylow $p$-subgroups,  _Proc. Amer. Math. Soc._ 146 (2018), no. 5, 1963–1976.
  * [I76] I. M. Isaacs,  _Character theory of finite groups_ ,  Dover, New York, 1976.
  * [INOT17] I. M. Isaacs, G. Navarro, J. B. Olsson and P. H. Tiep, Character restriction and multiplicities in symmetric groups.  _J. Algebra_ 478 (2017) 271–282.
  * [J78] G. James,  _The representation theory of the symmetric groups_ ,  Lecture Notes in Mathematics, vol. 682, Springer, Berlin, 1978.
  * [JK81] 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.
  * [N18] G. Navarro, Character theory and the McKay conjecture,  Cambridge University Press, 2018.
  * [O76] J. B. Olsson,  McKay numbers and heights of characters,  _Math. Scand._ 38 (1976), no. 1, 25–42.
  * [V23] G. Volpato,  Representations of Symmetric groups and Sylow subgroups,  _PhD Thesis_ , Università degli Studi di Firenze.