# An Indicator Formula for the Hopf Algebra $k^{S_{n-1}}\\#kC_{n}$

Kayla Orlinsky
University of Southern California
Los Angeles, CA 90089-1113

An Indicator Formula for the Hopf Algebra $k^{S_{n-1}}\\#kC_{n}$

Kayla Orlinsky University of Southern California, Los Angeles, CA 90089-1113
Email<EMAIL_ADDRESS>Phone: (949) 579-0796

Abstract  The semisimple bismash product Hopf algebra
$J_{n}=k^{S_{n-1}}\\#kC_{n}$ for an algebraically closed field $k$ is
constructed using the matched pair actions of $C_{n}$ and $S_{n-1}$ on each
other. In this work, we reinterpret these actions and use an understanding of
the involutions of $S_{n-1}$ to derive a new Froebnius-Schur indicator formula
for irreps of $J_{n}$ and show that for $n$ odd, all indicators of $J_{n}$ are
nonnegative. We also derive a variety of counting formulas including Theorem
6.2.2 which fully describes the indicators of all $2$-dimensional irreps of
$J_{n}$ and Theorem 6.1.2 which fully describes the indicators of all odd-
dimensional irreps of $J_{n}$ and use these formulas to show that nonzero
indicators become rare for large $n$.

Keywords  Hopf Algebras $\cdot$ Frobenius Schur Indicator $\cdot$
Representations of Hopf Algebras $\cdot$ Bismash Product $\cdot$ Symmetric
Group

Mathematics Subject Classification (2000)  16G99 $\cdot$ 16T30 $\cdot$ 20B35

##  Introduction

In this paper, we study the indicators of the simple modules (irreps) of a
particular semisimple Hopf algebra, specifically the bismash product
$J_{n}=\mathbb{C}^{S_{n-1}}\\#\mathbb{C}C_{n}$ where $S_{n}$ is the symmetric
group. We rely especially on the works of [12] which served as a general
resource on Hopf Algebras as well as [5] which derived an indicator formula
for irreps of bismash product Hopf algebras. Additionally, a key result of [7]
(Theorem 3.2) showed that there is a one-to-one correspondence between
irreducible representations (irreps) of $J_{n}$ and group representations of
$F_{x}\subset C_{n}$ where $F_{x}$ is the stabilizer of a permutation $x\in
S_{n-1}$ under the matched pair action of $C_{n}$ on $S_{n-1}$ (see Section
2). Using this correspondence, we adapt the formulas of [5] and find new ways
to explicitly describe and count irreps of $J_{n}$ with certain indicators.
Finally, we use code developed in Python to provide insight into negative
indicators.

The idea that indicators give insight into the structure of representations
started with groups. Given an irreducible character $\chi$ of a finite group
$G$, averaging the values outputted by the character evaluated at the square
of each group element is called the Frobenius-Schur [4] indicator

$\nu(\chi)=\frac{1}{|G|}\sum_{g\in G}\chi(g^{2}).$

First discovered by its namesakes in 1906, the magic of the Frobenius-Schur
indicator is that it always takes the value $0$, $1,$ or $-1$. These three
values determine whether the associated representation–thought of as
$kG$-module–admits a nondegenerate $G$-invariant bilinear form
($\nu(\chi)\not=0$) and if so, if that form is symmetric or skew-symmetric
($\nu(\chi)=1$ or $-1$ respectively).

Although [3] extended the indicator to Kac algebras in 1999, explicitly
constructing an analogue of the Frobenius-Schur indicator for semisimple Hopf
algebras was first done by Linchenko and Montgomery in 2000 [10]. The key idea
of [10] came from the fact that a semisimple Hopf algebra $H$ over an
algebraically closed field admits a unique two-sided integral element
$\Lambda$ satisfying $h\Lambda=\varepsilon(h)\Lambda=\Lambda h$ for all $h\in
H.$ It was then found that by computing the (irreducible) character of the
second Sweedler power of this integral

$\nu(\chi)=\chi((m\circ\Delta)(\Lambda))=\chi\left(\sum\Lambda_{1}\Lambda_{2}\right),$

that the resulting sum would also only return $0,1,$ or $-1$. Here $m$ and
$\Delta$ are the multiplication and comultiplication on the Hopf algebra.
Furthermore, these three values given by $\nu(\chi)$ determine whether the
associated $H$-module admits a nondegenerate $H$-invariant bilinear form
($\nu(\chi)\not=0$) and if so, if that form is symmetric or skew-symmetric
($\nu(\chi)=1$ or $-1$ respectively). It is a perfect analogue of its group-
theoretic counterpart for semisimple Hopf algebras.

Frobenius-Schur indicators on Hopf algebras are of interest for a variety of
reasons, three of which stand out. First, and most obviously, they determine
if the associated $H$-module (representation) has an $H$-invariant, bilinear
form and further if it is symmetric or skew-symmetric. In fact, it was shown
by Linchenko and Montgomery [10] that $\nu(\chi)\not=0$ if and only if the
associated simple module (with character $\chi$) is self dual.

Secondly, the indicator has been found to aid in the classification of finite
dimensional Hopf algebras. This problem of describing all Hopf algebras of a
certain dimension is in general very difficult and tedious. For example, in
2000, Kashina [6] classified all semisimple Hopf alebgras of dimension $16$
and then in 2001, Kashina, Mason, and Montgomery [7] showed that the
computations of all Hopf Algebras of dimension 16 could be improved using the
(second) indicator $\nu$ [7]. Also [8] used the indicator to put restrictions
on the dimension of a Hopf algebra based on the dimension of its simple
modules. For example, they showed that if $H$ has an even dimensional simple
module, then the dimension of $H$ itself must be even [8].

Thirdly, and most generally, Frobenius-Schur indicators can be adapted to have
broader applications to category theory. If $H$ is a semisimple Hopf algebra
over $\mathbb{C}$, then we can define Rep$(H)$, which is the category of
finite dimensional representation of $H$. In fact, Rep$(H)$ is a pivotal
fusion tensor category meaning that it is a category with finitely many
isomorphism classes of simple objects (fusion) with a notion of dualization
(pivotal) and tensor product (tensor). It was shown by Mason and Ng in 2005
[13] that the Frobenius-Schur indicator is a categorical invariant. That is,
if $H$ and $H^{\prime}$ are two semisimple Hopf algebras over an algebraically
closed field of characteristic $0$ such that Rep$(H)$ is equivalent to
Rep$(H^{\prime})$ as $k$-linear tensor categories, then
$\\{\nu(\chi)\\}_{\chi\in Irr(H)}$ and $\\{\nu(\chi^{\prime})\\}_{\chi\in
Irr(H^{\prime})}$ are identical.

Although we do not discuss them in this paper, there are abundant applications
if one were to expand out to higher indicators. That is, by computing the
$n^{\text{th}}$-Sweedler power of $\Lambda$ (as described above). For example,
in the same paper where the idea is introduced, [9] prove a version of
Cauchy’s Theorem for semisimple Hopf alebras using the regular representation.

In this paper, we focus on the semisimple Hopf algebra
$J_{n}=\mathbb{C}^{S_{n-1}}\\#\mathbb{C}C_{n}$ which is a bismash product of
the two groups $S_{n-1}$ and $C_{n}$. Matched pairs of groups have been known
to group theorists as factorizable groups since the 1950s, but the
construction of Hopf algebras from a mathced pair of groups was first proposed
by Takeuchi [17] in 1981. In fact, it was shown by Masuoka [11] that the Hopf
algebra of interest in this paper, $J_{n}$, is one of only two Hopf algebras
that arise from the decomposition of $S_{n}$ as a matched pair of $S_{n-1}$
and $C_{n}$. The other Hopf alebgra being $H_{n}=(J_{n})^{*}$ the dual of
$J_{n}$ for which [5] have already shown that all indicators of simple modules
are $+1$.

Similar work has been done by Timmer [18] in 2015 using different
factorizations of the symmetric group $S_{n}=S_{n-r}G$ and considering
$H=k^{G}\\#kS_{n-r}$. Timmer [18] shows that the indicator is nonnegative, a
result which we find to be false for $J_{n}$ for certain $n.$

The goal of this work was to fully describe all irreps of $J_{n}$ and state
their second indicators $\nu(\chi)=\nu_{2}(\chi)$. When $n$ is odd, this can
be done explicitly using an understanding of involutions in the symmetric
group $S_{n-1}$ and some computation. However, when $n$ is even, more work is
needed to completely describe all irreps and their indicators. Specifically,
when $n$ is odd, we prove the indicator is nonnegative but for $n\geq 12$ such
that $4$ divides $n$, there exists representations of $J_{n}$ with negative
indicator which has complicated the problem of classifcation. The organization
of the paper is as follows.

In Section 2, we discuss the group actions which determine the underlying
structure of our Hopf algebra. This involves describing facorizable groups,
which are groups $L$ which contain two subgroups $F,G$ such that $F\cap G=1$
and $L=GF=FG$. Note that this allows every element in $L$ to be written
uniquely as a product of elements in $F$ and $G$, but does not imply that $L$
can be written as a direct or even semi-direct product of $F$ and $G$. Since
we can write every element $l\in L$ as a product $xa\in GF=L$ with $a\in F$
and $x\in G$, two natural group actions arise of these subgroups on each other
which describe how to write $xa$ uniquely as an element of $FG$. Note that
each of these actions are group actions in the sense that $G$ acts on the set
$F$ and $F$ acts on the set $G$. However, neither respects the group operation
of the other. The factorizable group of interest to us will be the symmetric
group of order $n$, $S_{n}$ which we will show can be written as a product of
$C_{n}$ and $S_{n-1}$, where $C_{n}$ here is the cyclic group generated by the
standard $n$-cycle $a=(1\,\,\,2\,\,\,\cdots\,\,\,n)$. We will discuss the
orbits, stabilizers, and other useful sets and give some lemmas which we will
need in order to prove our indicator counting formulas later on.

In Section 3, we explicitly describe the Hopf algebra $J_{n}$ and summarize
relevant results in [5], which include a discussion of the classification of
the indicators of $H_{n}=(J_{n})^{*}$ the dual of $J_{n}$. In [5], a complete
classification of indicators is given for $J_{p}$ where $p$ is prime, and also
for $H_{n}$ for all $n.$ While we have not been able to fully describe all
indicators of $J_{n}$ for all $n$, in this section we discuss the methods used
for when $p$ is prime and provide some useful results which apply for general
$n.$

In Section 4, we give a rewritten indicator formula which describes how to
compute indicators from irreps induced from permutations using minimal
information about the permutation. However, to obtain it, we must first
completely re-interpret the actions of $C_{n}$ on $S_{n-1}$. We describe how
viewing each permutation in the symmetric groups as a bijection on
$\mathbb{Z}/n\mathbb{Z}$ and then using some modular arithmetic, we uncover
new relationships within the orbits and stabilizers of the group actions
described in Section 3. For example, when we consider the action of
$C_{n}=\langle a\rangle$ on $S_{n-1}$, we show that if a permutation $x\in
S_{n-1}$ (viewed as a permutation in $S_{n}$ which fixes $n$) has stabilizer
$F_{x}=\langle a^{t}\rangle$, then $x$ exhibits a number of remarkable
properties including a sort of linearity in $t$. We use these properties to
derive Theorem 4.3.2 which is an improved indicator formula which drastically
reduces–but does not eliminate–the computation requirement. Using this, we
show that for any $n$ not divisible by $4$, the indicator of any irrep of
$J_{n}$ is always nonnegative.

In Section 5, we cover majority of key results for this paper. Due to the
computational nature of the approach we take, we have broken this section into
subsections. Throughout this section, we use the relations discovered in
Section 4 to fully construct (and therefore count) many sets of interest. We
show that any permutation $x\in S_{n-1}$ with stabilizer $\langle
a^{t}\rangle\subset C_{n}$ can in fact be constructed from some permutation in
$S_{t-1}$.

In Section 4.3, we use the formulas from the previous section to show that if
$n$ is odd, then counting irreducible representation of $J_{n}$ is equivalent
to counting involutions of $S_{n-1}$ with certain properties and, given an odd
integer $t$, provide a formula for counting all $t$-dimensional irreps with
indicator $+1$ (Theorem 6.1.2). Although we cannot describe all indicators of
$t$-dimensional irreps of $J_{n}$ when $t$ is even, we are able to fully
describe the indicators of all $2$-dimensional irreps of $J_{n}$ (Theorem
6.2.2).

In Section 7, we discuss limiting behavior using the formulas we developed
previously. We show that the ratio of nonzero irreps of dimension $t$ (odd or
$t=2$) with indicator $+1$ to the total number of irreps of dimension $t$
tends to $0$ as $n\to\infty$ (Theorem 7.4 and Theorem 7.6).

##  Group Action Results

In this section, we describe the underlying group actions which make up the
structure of the bismash product Hopf algebra $J_{n}$. Throughout, $L,F,G$
will be finite groups. Elements $a,b\in F$ and $x,y\in G$ and $l\in L$.

Consider a (finite) group $L$ with two subgroups $F$ and $G$ which intersect
trivially and whose product is the entire group $L$. We call such a group
factorizable and such objects have been studied for decades due to their
unique decomposition.

######  Definition 2.1.

A group $L$ is factorizable if it has two subgroups $F$ and $G$ such that
$F\cap G=1$ and $L=GF=FG$.

Being able to write that $L=GF$ and knowing that $F\cap G=1$ implies that
$L=FG$ and it also implies that every element $l\in L$ can be written uniquely
as a product of something in $F$ and something in $G$.

Note we cannot say that $L$ is a direct (or even semi-direct) product of $F$
and $G$ because nowhere are we assuming normality of either subgroup.

######  Definition 2.2 ([17]).

Let $F$ and $G$ be groups. Let $\lhd:G\times F\to G$ and $\rhd:G\times F\to F$
be group actions. Then $(F,G,\rhd,\lhd)$ is a matched pair if for all $x,y\in
G$ and all $a,b,\in F$,

$\displaystyle x\rhd ab$ $\displaystyle=(x\rhd a)((x\lhd a)\rhd c)$
$\displaystyle xy\lhd a$ $\displaystyle=(x\lhd(y\rhd a))(y\lhd a)$

While this definition turns out to be equivalent to that of a factorizable
group, [17] did not know that when he coined this term in the ’80s.

######  Lemma 2.3.

Let $L=GF$ be a factorizable group. Then $(F,G,\rhd,\lhd)$ is a matched pair
where for $x\in G$ and $b\in F$, the decomposition below is unique:

$xb=(x\rhd b)(x\lhd b).$

Conversely, if $(F,G,\rhd,\lhd)$ is a matched pair, then $L$ which has
underlying set $G\times F$ and multiplication defined by

$(a,x)(b,y)=(a(x\rhd b),(x\lhd b)y)$

is a factorizable groups

We use the same definitions as from [5]. Many of the results given next proved
in [5] or follow immediately from the definitions.

######  Proposition 2.4.

Let $L=GF=FG$ be a factorizable group where the actions $\lhd$ and $\rhd$ are
defined such that $xb=(x\rhd b)(x\lhd b)$ is unique for all $x\in G$ and $b\in
F$. Then the following properties hold:

  1. (1)

$x\rhd bc=(x\rhd b)((x\lhd b)\rhd c)$

  2. (2)

$xy\lhd b=(x\lhd(y\rhd b))(y\lhd b)$

  3. (3)

$xy\rhd b=x\rhd(y\rhd b)$

  4. (4)

$x\lhd bc=(x\lhd b)\lhd c$

  5. (5)

$1\rhd b=b$

  1. (6)

$1\lhd b=1$

  2. (7)

$x\lhd 1=x$

  3. (8)

$x\rhd 1=1$

  4. (9)

$(x\rhd b)^{-1}=(x\lhd b)\rhd b^{-1}$

  5. (10)

$(x\lhd b)^{-1}=x^{-1}\lhd(x\rhd b)$

Both Lemma 2.3 and Proposition 2.4 are proved from the definitions using
direct computation.

######  Definition 2.5.

Let $L=FG$ be a factorizable group. Then for $x\in G$ and $b\in F$, we define
the following:

  1. (1)

The orbits $\mathcal{O}_{x}=\\{x\lhd b\,|\,b\in F\\}\subset G$

  2. (2)

the stabilizers $F_{x}=\\{b\in F\,|\,x\lhd b=x\\}\subset F$

  3. (3)

The sets $F_{x^{-1},x}=\\{b\in F\,|\,x^{-1}\lhd b=x\\}$

######  Lemma 2.6 ([5], Corollary 4.3).

Let $L=FG$. Let $x\in G$, then the following are equivalent:

  1. (1)

$x^{-1}\in\mathcal{O}_{x}$

  2. (2)

$y^{-1}\in\mathcal{O}_{x}$ for all $y\in\mathcal{O}_{x}$

  3. (3)

$F_{x^{-1},x}\not=\varnothing$

  4. (4)

$F_{y^{-1},y}\not=\varnothing$ for all $y\in\mathcal{O}_{x}$

In fact, it is always true that $|F_{x^{-1},x}|=|F_{y^{-1},y}|$ for all
$y\in\mathcal{O}_{x}$.

As we will see in the next section, the heart of our problem of computing
second indicators will become a problem of understanding which $x\in S_{n-1}$
satisfy that $x^{-1}\in\mathcal{O}_{x}$.

######  Lemma 2.7.

Let $x^{-1}\in\mathcal{O}_{x}$. Then $|F_{y^{-1},y}|=|F_{x}|$ for every
$y\in\mathcal{O}_{x}$.

Proof. Since $x^{-1}\in\mathcal{O}_{x}$, there exists a $b\in F$ such that
$x^{-1}=x\lhd b$.

Define a function $\Lambda:F_{x^{-1},x}\to F_{x}$ such that $\Lambda(c)=bc$
for all$c\in F_{x^{-1},x}$.

This function is well defined since

$x\lhd bc=(x\lhd b)\lhd c=x^{-1}\lhd c=x.$

This map is also clearly a bijective with inverse $\Lambda^{-1}(d)=b^{-1}d$
for all $d\in F_{x}$. ∎

######  Theorem 2.8 ([5], Lemma 4.2 (4)).

Let $F$ be an abelian group and let $x^{-1}\in\mathcal{O}_{x}.$ Then for all
$y\in\mathcal{O}_{x}$ and all $b\in F_{y^{-1},y}$, $(y^{-1}\rhd b)b\in F_{x}$

The specific factorizable group of interest to us is $L=S_{n}$. Letting
$a=(1\,\,\,2\,\,\,\cdots\,\,\,n)$ be the standard $n$-cycle and considering
$F=C_{n}=\langle a\rangle$ and $G=S_{n-1}$ we can write $L=FG$ as a
factorizable group. Note that clearly $F\cap G=1$ and since $|L|=|FG|=|F||G|$
we have that $L=GF=FG$. To describe the action explicitly, we use the
following theorem.

######  Proposition 2.9.

For any $\lambda\in S_{n},$ let $a$ be the standard $n$-cycle and write
$\lambda=a^{r}x$ where $x=a^{-r}\lambda\in S_{n-1}$ and $r=\lambda(n)$. This
is the unique way of representing $\lambda$ as a product of elements in
$C_{n}S_{n-1}$.

Proof. Clearly $C_{n}\cap S_{n-1}=1$ since aside from the identity,
permutations in $S_{n-1}$ fix $n$ and permutations in $C_{n}$ do not.
Furthermore, $C_{n}S_{n-1}\subset S_{n}$ and
$|C_{n}||S_{n-1}|=n(n-1)!=n!=|S_{n}|$, we have that $S_{n}$ is a factorizable
with $F=C_{n}$ and $G=S_{n-1}$. ∎

In Section 4, we will use the properties described in this section to better
understand and reinterpret the group actions of $F$ and $G$ on each other. But
before we can derive anything useful, we must describe the Hopf structure
which is at the heart of this paper.

##  Hopf and Representation Results

Here we construct the Hopf Algebra that will be the object of study for the
remainder of the paper. Throughout this section, we will take $k$ to be an
algebraically closed field of characteristic $0$. When doing explicit
computations, we will often take $k=\mathbb{C}$. A general resource on Hopf
Algebras, [12], was consulted often for background definitions and facts about
Semisimple Hopf Algebras.

The first half of this section is largely a summary of key results of [5]
since they provided the indicator formula for all semisimple Hopf algebras
which are bismash products and they also classified all indicators of our main
example in the prime case. We conclude that further simplifications to their
formula are possible in the case where $F$ is abelian.

######  Definition 3.1 ([17]).

Let $(F,G,\lhd,\rhd)$ be a matched pair of finite groups. Then the bismash
product $J=k^{G}\\#kF$ has underlying set $k^{G}\otimes_{k}kF$. Let $x,y\in G$
and $a,b\in F$ and define $p_{x}:G\to k$ such that

$p_{x}(y)=\delta_{x,y}=\begin{cases}1&\text{ if }y=x\\\ 0&\text{
otherwise}\end{cases}$

which is basis of $k^{G}$ dual to $kG$. We will typically write $p_{x}\otimes
a=p_{x}\\#a$ for elements of $J$.

  1. (1)

unit: $1\\#1$

  2. (2)

multiplication: $(p_{x}\\#a)(p_{y}\\#b)=\delta_{y,x\lhd a}p_{x}\\#ab$

  3. (3)

counit: $\varepsilon(p_{x}\\#a)=\delta_{x,1}$

  4. (4)

comultiplication:

$\Delta(p_{x}\\#a)=\sum_{y\in G}(p_{xy^{-1}}\\#(y\rhd a))\otimes(p_{y}\\#a)$

  5. (5)

antipode: $S(p_{x}\\#a)=p_{(x\lhd a)^{-1}}\\#(x\rhd a)^{-1}$

We will now summarize a variety of general results in representation theory
which we will be using as a starting point for computing the second indicators
of $J$. Additionally, we will provide results on bismash products from [5]
that are on significance.

######  Theorem 3.2 ([7]).

Let $J=k^{G}\\#kF$. Let $x\in G$ have orbit $\mathcal{O}_{x}$ and stabilizer
$F_{x}$. Let $V$ be an irreducible representation of $F_{x}$, thought of as a
simple $kF_{x}$-module. Then

$\hat{V}=kF\otimes_{kF_{x}}V$

is an irreducible $J$-module under the action of

$(p_{y}\\#a)\cdot(b\otimes v)=\delta_{y\lhd(ab),x}(ab\otimes v)$

for all $x\in G$, $a,b\in F$ and $v\in V$. Furthermore, every irreducible
representation of $J$ is induced from an irreducible representation of
subgroups of $F$ in this way.

Explicitly, if we want to construct an irreducible representation $\hat{V}$ of
the Hopf algebra $J$ we follow the following steps.

  1. (1)

choose an element $x\in G$

  2. (2)

under the matched pair action of $F$ on $G$, compute the stabilizer of $x$
which is $F_{x}$

  3. (3)

$F_{x}$ is a subgroup of $F$ and so has irreducible (group) representations
$V$

  4. (4)

for each of these $F_{x}$-group-representations $V$, there is an induced
$J$-module $\hat{V}$ whose structure is described by Theorem 3.2.

Since all irreps of $J$ are constructed in this way, the first step in
understanding the irreps of $J$ (which is ultimately how we will understand
the indicators) is to understand the orbits and stabilizers of the action of
$F$ on $G.$

######  Lemma 3.3 ([7]).

If $F$ is abelian, the dimension of $\hat{V}$ as described in Theorem 3.2 is
$[F:F_{x}]$.

In fact, in the original proof of Theorem 3.2, we know that
dim$(\hat{V})=[F:F_{x}]$dim$(V)$. So, if $F$ is abelian, so is $F_{x}$ for all
$x$ so $\dim(V)=1$ for all irreducible representations of $F_{x}$ so
$\dim(\hat{V})=[F:F_{x}].$

We now introduce a crucial formula, the indicator formula for bimash product
Hopf Algebras.

######  Theorem 3.4 ([5]).

For $J=k^{G}\\#kF$ and $x\in G$, the indicator of $\hat{\chi}$ is given by

$\nu(\hat{\chi})=\frac{1}{|F|}\sum_{y\in\mathcal{O}_{x}}\sum_{a\in
F_{y^{-1},y}}\hat{\chi}(p_{y}\\#(y^{-1}\rhd a)a)$ (1)

where $F_{y^{-1},y}=\\{a\in F\,|\,y^{-1}\lhd a=y\\}$. Furthermore, if
$\hat{V}$ is an irreducible representation, then $\nu(\hat{\chi})$ takes
values in $\\{1,0,-1\\}$

This formula was adapted from a formula derived by [[14], Proposition 5.5]
from [[7], p 898]. Additionally, it was shown in [10] that the formula
below–as with its group analogue–always outputs either $1,0,$ or $-1$ and
these values determine if the associated module admits a nondegenerate,
$J$-invariant, bilinear form (iff $\nu(\hat{\chi})\not=0)$, and if it does
admit such a form, if this form is symmetric (iff $\nu(\hat{\chi})=1$) or
skew-symmetric (iff $\nu(\hat{\chi})=-1$).

######  Theorem 3.5 ([5]).

Let $L=GF=FG$ be a factorizable group and $J=k^{G}\\#kF$. Let $x\in G$ and
$F_{x}=\\{1\\}.$ Then there is a unique simple $F_{x}$-module $V\cong k$ and
the induced simple $J$-module $\hat{V}$ has Schur indicator equal to $1$ if
and only if $x^{-1}\in\mathcal{O}_{x}$. Otherwise the indicator is $0$.

######  Theorem 3.6 ([5]).

Let $x\in G$ and $F_{x}=F$. Then any simple module $V$ of $kF_{x}$ is already
an $F$-module and so $\hat{V}=V$ becomes a simple $J$-module. Then:

  1. (1)

if $x=1$ then the value of the Frobenius-Schur indicator of $\hat{V}$ is the
same as the indicator of $V$ as a simple $kF$-module;

  2. (2)

if $x^{2}\not=1$ then the value of the Frobenius-Schur indicator of $\hat{V}$
is $0$;

  3. (3)

if $x^{2}=1$ (for $x\not=1$) and additionally if $F=C_{p^{r}}$ for $p$ and odd
prime, then the Frobenius-Schur indicator of $\hat{V}$ is $1.$

Now we describe a main result proved in [5]. This is a complete classification
of the indicators of the Hopf algebra $J_{p}$ where $p$ is a prime.

######  Definition 3.7.

Let $S_{n}=S_{n-1}C_{n}=C_{n}S_{n-1}$ be a matched pair of finite groups. Then
the bismash product $J_{n}=k^{S_{n-1}}\\#kC_{n}$ where $F=C_{n}$ and
$G=S_{n-1}$.

######  Theorem 3.8 ([5]).

Consider $J_{p}=k^{G}\\#kC_{p}$ where $p>2$ is prime. There are two cases:

I: $\hat{V}$ has dimension $1$. Here $F_{x}=F$, so that $x\in G^{F}=\\{x\in
G\,|\,x\lhd a=x$ for all $a\in F\\}$. Then either,

  1. (a)

$x=1$, in which case we have the trivial module $\hat{V}_{0}$ with
$\nu(\hat{V}_{0})=1$, and $p-1$ other $1$-dim simple modules $\hat{V}$, all
with indicator $0$;

  2. (b)

$x$ has order $2$, in which case we have $p$ simple modules with
$\nu(\hat{V})=1$;

  3. (c)

$x$ has order $>2$, in which case we have $p$ simple modules with
$\nu(\hat{V})=0.$

II: $\hat{V}$ has dimension $p$. Here $F_{x}=\\{1\\}$ and either

  1. (e)

$\mathcal{O}_{x}$ contains an element of order $2$. Then there is one simple
module with $\nu(\hat{V})=1;$

  2. (f)

$\mathcal{O}_{x}$ contains no elements of order $2$. Then there is one simple
module with $\nu(\hat{V})=0.$

We now cite a slightly revised theorem of [5].

######  Theorem 3.9 ([5]).

Consider $J_{p}=k^{G}\\#kC_{p}$ for $p>2$ prime. Let $m_{p,1}$ be the number
of orbits of size $p$ (so $F_{x}=\\{1\\}$) containing at least one involution.
Let $m_{p,0}$ be the number of orbits of size $p$ containing no involutions.
Then

  1. (1)

$\displaystyle m_{p,1}=\frac{i_{p}-1}{p}-1\qquad i_{p}=\\#\\{\sigma\in
S_{n}\,|\,\sigma^{2}=1\\}$

  2. (2)

$\displaystyle m_{p,1}+m_{p,0}=(p-1)\frac{(p-2)!-1}{p}$

  3. (3)

There are $pm_{p,1}$ irreps of dimension $p$ with indicator $+1$

  4. (4)

There are $p(p-1)$ irreps of dimension $1$ and of these $p+1$ have indicator
$+1$

Let us reiterate. In their paper, Jedwab and Montgomery fully describe all
indicators for the Hopf algebra $J_{p}=k^{S_{p-1}}\\#kC_{p}$ where $p$ is an
odd prime. Specifically, they describe completely all $x\in G=S_{p-1}$ for
which an irrep $\hat{V}$ of $J_{p}$ induced from the (group) representation
$V$ of the group $F_{x}=(C_{p})_{x}$ has nonzero indicator and further show
that all indicators are either $0$ or $1$. In Section 6.1 and Section 6.2, we
do the same for $J_{n}$ for $n$ odd and $n=2$ respectively.

Note that [5] show that the algebraic dual
$(J_{n})^{*}=H_{n}=\mathbb{C}^{C_{n}}\\#\mathbb{C}S_{n-1}$ is totally
orthogonal. That is, every simple $H_{n}$-module admits an $H_{n}$-invariant,
symmetric, nondegenerate, bilinear form. However, we will see in the next
section that the only $n$ for which $J_{n}$ is totally orthogonal is $n=2.$

Thus, the question of describing indicators becomes much more interesting for
$J_{n}$.

######  Theorem 3.10 ([5]).

Let $H_{n}=\mathbb{C}^{C_{n}}\\#\mathbb{C}S_{n-1}$ be the algebraic dual of
$J_{n}$. Then if $\hat{V}$ is an irreducible representation of $H_{n}$ with
character $\hat{\chi}$, $\nu(\hat{\chi})=+1.$

######  Lemma 3.11.

Let $x\in S_{n-1}$. If $x^{-1}\notin\mathcal{O}_{x}$, then $\hat{V}$ has
indicator $0$ for any representation $V$ of $F_{x}$.

Proof. From Theorem 3.4, if $F_{y^{-1},y}=\varnothing$, then
$\nu(\hat{\chi})=0$. From Lemma 2.6, if $x^{-1}\notin\mathcal{O}_{x}$, then
$F_{y^{-1},y}=\varnothing$ for all $y\in\mathcal{O}_{x}$ and so
$\nu(\hat{\chi})=0$. ∎

Lemma 3.11 is a more general version of Theorem 3.5.

######  Proposition 3.12 ([5]).

Let $L=FG$ as usual and let $V$ be an irreudcible representation of $F_{x}$
with character $\chi$. Then for $x\in G$, the induced $J=k^{G}\\#kF$-module
$\hat{V}$ has character $\hat{\chi}$ which is defined by

$\hat{\chi}(p_{y}\\#a)=\sum_{\begin{subarray}{c}b\in T\\\ \text{ where
}b^{-1}ab\in F_{x}\end{subarray}}\delta_{y\lhd b}(x)\chi(b^{-1}ab)$

where, the sum is taken over a set of representatives $T$ for the right cosets
of $F/F_{x}$.

######  Lemma 3.13.

If $F$ is abelian, then the formula for $\hat{\chi}$ From Proposition 3.12
reduces to

$\hat{\chi}(p_{y}\\#a)=\begin{cases}\chi(a)&\text{ if
}y\in\mathcal{O}_{x}\text{ and }a\in F_{x}\\\ 0&\text{ otherwise}\end{cases}$

Proof. Fix the element $p_{y}\\#a$. Consider the following cases.

  1. (1)

First, if $y\notin\mathcal{O}_{x}$, then there does not exist any $b\in F$ for
which $y\lhd b=x$. Namely, $\delta_{y\lhd b}(x)=0$ for all $b\in F$ and so
$\hat{\chi}(p_{y}\\#a)=0$.

  2. (2)

Second, if $y\in\mathcal{O}_{x}$ but $a\notin F_{x}$, then there is no $b\in
F$ for which $bab^{-1}=bb^{-1}a=a\notin F_{x}$ ($F$ is abelian), so there is
no sum and thus $\hat{\chi}(p_{y}\\#a)=0$.

  3. (3)

Now, if $y\in\mathcal{O}_{x}$ and $a\in F_{x}$, then by Proposition 3.12, we
claim that there is only one coset of $F/F_{x}$ for which $y\rhd b=x$.

First, since $y\in\mathcal{O}_{x}=\\{x\rhd b\,|\,b\in F\\}$ there exists a
$b\in F$ so $x\lhd b=y$. Namely, $x=y\lhd b^{-1}$ and so there is at least one
coset ($b^{-1}F_{x}$) of $F/F_{x}$ whose representative satisfies that
$\delta_{y\lhd b^{-1}}(x)=1$.

Now, assume $c\in F$ is another element such that $y\lhd c=x$ (and so
equivalently $x\lhd c^{-1}=y$). Then

$y=x\lhd c^{-1}=x\lhd b\implies x=x\lhd bc.$

Thus, $bc\in F_{x}$ so $c\in b^{-1}F_{x}$ (keep in mind $F$ is abelian).

Thus, there is exactly one coset for which $\delta_{y\lhd b^{-1}}(x)\not=0$.

Finally, we obtain the result

$\hat{\chi}(p_{y}\\#a)=\begin{cases}\chi(a)&\text{ if
}y\in\mathcal{O}_{x}\text{ and }a\in F_{x}\\\ 0&\text{ otherwise}\end{cases}$
∎

######  Theorem 3.14.

Let $C_{n}=\langle a\rangle$. Let $k=\mathbb{C}$ and $\zeta_{n}\in k$ be a
primitive $n^{\text{th}}$-root of unity. Let $V_{i}$ be a simple
$kC_{n}$-module equivalent to the irreducible (group) representation
$\rho_{i}$ of $C_{n}$ defined by

$\displaystyle\rho_{i}:C_{n}$ $\displaystyle\to\mathbb{C}^{\times}$
$\displaystyle a$ $\displaystyle\mapsto\zeta_{n}^{i}.$

Then, for any $0\leq i<n$,

$\nu(V_{i})=\begin{cases}+1&\text{ if }n|2i\\\ 0&\text{ otherwise}\end{cases}$

Finally, this gives us that for $n$ even, all the representations have
indicator $0$ except for exactly the trivial representation and the
representation defined by $\rho_{i}(a)=(\zeta_{n})^{\frac{n}{2}}$ both of
which have indicator $+1$.

And for $n$ odd, only the trivial representation has indicator $+1$ and the
rest have indicator $0.$

Proof. Let $\chi_{i}$ be the character of $V_{i}$. First, note that
$V_{i}\cong k$ and so writing $\rho_{i}(a)=\zeta_{n}^{i}$ makes sense and so
Tr$(\rho_{i}(a))=\zeta_{n}^{i}$.

Since $C_{n}=\langle a\rangle$, we can write

$\displaystyle\nu(V_{i})$ $\displaystyle=\frac{1}{|C_{n}|}\sum_{b\in
C_{n}}\chi_{i}(b^{2})=\frac{1}{|C_{n}|}\sum_{t=0}^{n-1}\chi_{i}((a^{t})^{2})$
$\displaystyle=\frac{1}{n}\sum_{t=0}^{n-1}\text{Tr}(\rho_{i}(a^{2t}))=\frac{1}{n}\sum_{t=0}^{n-1}\rho_{i}(a)^{2t}$
$\displaystyle=\frac{1}{n}\sum_{t=0}^{n-1}(\zeta_{n}^{i})^{2t}=\frac{1}{n}\sum_{t=0}^{n-1}\zeta_{n}^{2it}$
$\displaystyle=\begin{cases}1&\text{ if }\zeta_{n}^{2i}=1\\\
\frac{1}{n}\cdot\frac{\zeta_{n}^{2ni}-1}{\zeta_{n}^{2i}-1}&\text{ if
}\zeta_{n}^{2i}\not=1\end{cases}$ $\displaystyle=\begin{cases}1&\text{ if
}\zeta_{n}^{2i}=1\\\ 0&\text{ if }\zeta_{n}^{2i}\not=1\end{cases}$

Recall that for any $w\not=1$,
$\displaystyle\frac{w^{n}-1}{w-1}=\sum_{t=0}^{n-1}w^{t}$. Since $\zeta_{n}$ is
a primitive $n^{\text{th}}$-root of unity, $\zeta_{n}^{2i}=1$ if and only if
$n|2i$.

Note that since $i<n$, $2i<2n$ and so $n|2i$ if and only if $n=2i$ (so
$i=\frac{n}{2}$) or $i=0$. Namely, the trivial representation always has
indicator $+1$, and there is a non-trivial representation with nonzero
indicator if and only if $n$ is even. ∎

##  Reinterpreting the Action

Consider $x\in S_{n}$ such that $x(n)=n$. Then, letting $C_{n}=\langle
a\rangle$ where $a=(1\,\,\,2\,\,\,\cdots\,\,\,n)$ is the standard $n$-cycle,
we can multiply powers of $a$ with $x$ without issue.

Now, we note that $S_{n}\cong Bij(\mathbb{Z}/n\mathbb{Z})$ which is the group
of bijections of $\mathbb{Z}/n\mathbb{Z}$ where permutation multiplication is
now viewed as function composition. By viewing $a$ as a shifting map which
sends $i$ to $i+1$, we can consider how the action of $C_{n}$ on $S_{n-1}$
affects each permutation $x$ pointwise.

Using this notation, we will often describe addition and multiplication of the
outputs of $x$ and its shifts. This makes sense using the ring structure of
$\mathbb{Z}/n\mathbb{Z}$. Unless otherwise stated, all arithmetic in this
section will be done modulo $n$.

###  Reinterpreting Orbits and Stabilizers

######  Proposition 4.1.1.

Let $L=FG=C_{n}S_{n-1}$ where $F=C_{n}=\langle a\rangle$ and $G=S_{n-1}$.
consider the actions $\lhd$ and $\rhd$ of $F$ and $G$ on each other as from
Section 2 and let $x\in S_{n-1}$. Then,

  1. (1)

Let $a^{r}\in C_{n}$. Then $x\lhd a^{r}=a^{-x(r)}xa^{r}$ and $x\rhd
a^{r}=a^{x(r)}$;

  2. (2)

$F_{x}=\\{a^{t}\,|\,a^{-x(t)}xa^{t}=x\\}=\\{a^{t}\,|\,x(u+t)-x(t)=x(u)$ for
$u=1,...,n\\}$;

  3. (3)

$F_{x^{-1},x}=\\{a^{s}\,|\,x^{-1}(u+s)-x^{-1}(s)=x(u)\text{ for }i=1,...,n\\}$

  4. (4)

$\mathcal{O}_{x}=\\{a^{-x(l)}xa^{l}\,|\,a^{l}\in C_{n}\\}$;

###### Proof.

Note that all arithmetic is done modulo $n$. That is, when we write
$x(u+t)-x(t)=x(u)$ we really mean $(x(u+t\mod n)-x(t))\mod n=x(u)$.

  1. (1)

Proposition 2.9 tells us that $S_{n}$ is factorizable with subgroups $F=C_{n}$
and $G=S_{n-1}$ and Lemma 2.3 tells us how we can obtain a matched pair from
$F$ and $G$.

  2. (2)

Recall Definition LABEL:*def:orbits and stabs (2) . From (1)

$\displaystyle F_{x}$ $\displaystyle=\\{a^{t}\in C_{n}\,|\,x\lhd a^{t}=x\\}$
$\displaystyle=\\{a^{t}\in C_{n}\,|\,a^{-x(t)}xa^{t}=x\\}$
$\displaystyle=\\{a^{t}\in C_{n}\,|\,(a^{-x(t)}xa^{t})(u)=x(u)\text{ for all
}u=1,...,n\\}$ $\displaystyle=\\{a^{t}\in C_{n}\,|\,x(u+t)-x(t)=x(u)\text{ for
all}u=1,...,n\\}\ $

  3. (3)

Recall Definition LABEL:*def:orbits and stabs (3) . Then, by similar reasoning
as (2),

$\displaystyle F_{x^{-1},x}$ $\displaystyle=\\{a^{s}\in C_{n}\,|\,x^{-1}\lhd
a^{s}=x\\}$ $\displaystyle=\\{a^{s}\in
C_{n}\,|\,a^{-x^{-1}(s)}x^{-1}a^{s}=x\\}$ $\displaystyle=\\{a^{s}\in
C_{n}\,|\,(a^{-x^{-1}(s)}x^{-1}a^{s})(u)=x(u)\text{ for all }u=1,...,n\\}$
$\displaystyle=\\{a^{s}\in C_{n}\,|\,x^{-1}(u+s)-x^{-1}(s)=x(u)\text{ for
all}u=1,...,n\\}\ $

  4. (4)

Recall Definition LABEL:*def:orbits and stabs (1) . Again from (1)

$\displaystyle\mathcal{O}_{x}$ $\displaystyle=\\{x\lhd a^{l}\,|\,a^{l}\in
C_{n}\\}$ $\displaystyle=\\{a^{-x(l)}xa^{l}\,|\,\text{ for }l=1,...,n\\}\qed$

Let us consider an example.

######  Example 4.1.2.

Let $n=6$ and $x=(1\,\,\,2)(3\,\,\,4\,\,\,5)$. Then, using permutation
multiplication we get that $x\lhd a=a^{-x(1)}xa=a^{-2}xa=a^{4}xa$. Since
$a=(1\,\,\,2\,\,\,3\,\,\,4\,\,\,5\,\,\,6)$, direct computation by multiplying
cycles gives that $x\lhd a=(1\,\,\,5\,\,\,4)$.

However, instead of multiplying permutations, this result can also be seen by
understanding $x\lhd a$ pointwise. That is, using ideas from the proof of
Proposition LABEL:*prop:for python easier formulas (1) ,

$(x\lhd a)(i)=(a^{4}xa)(i)=x(i+1)+4.$

This gives us a formula to write out the product as a bijection

$\displaystyle 1$ $\displaystyle\mapsto x(2)+4=5$ $\displaystyle 2$
$\displaystyle\mapsto x(3)+4=2$ $\displaystyle 3$ $\displaystyle\mapsto
x(4)+4=3$

$\displaystyle 4$ $\displaystyle\mapsto x(5)+4=1$ $\displaystyle 5$
$\displaystyle\mapsto x(6)+4=4$ $\displaystyle 6$ $\displaystyle\mapsto
x(1)+4=6$

which is, of course, the same permutaiton we described via direct computation.

######  Proposition 4.1.3.

Let $x\in S_{n-1}.$ Let $F_{x}=\langle a^{t}\rangle$. Then

  1. (1)

$a^{-x(l)}xa^{l}=a^{-x(s)}xa^{s}$ if and only if $x(s-l)=x(s)-x(l)$ and
$s=l\mod t$

  2. (2)

$\mathcal{O}_{x}=\\{a^{-x(l)}xa^{l}\,|\,l=1,...,t\\}$ and so if
$y=a^{-x(l)}xa^{l}\in\mathcal{O}_{x}$, then $y(u)=x(u+l)-x(l)$ for all $u.$
Specifically, $F_{y}=F_{x}$ and $y(t)=x(t).$

  3. (3)

If $x^{-1}=a^{-x(s)}xa^{s}\in\mathcal{O}_{x}$. Then $x(s)+s$ is a fixed point
of $x$ and $x(s)+s=ut$ for some $u$.

  4. (4)

If $x^{-1}=a^{-x(s)}xa^{s}\in\mathcal{O}_{x}$, and we let
$y\in\mathcal{O}_{x}$ with $y=a^{-x(l)}xa^{l}$. Then

$F_{y^{-1},y}=\\{a^{mt+l-s-x(l)}\,|\,m=0,1,...,\frac{n}{t}-1\\}$

Proof. First, note that Proposition LABEL:*prop:for python easier formulas (2)
tells us that if $a^{t}\in F_{x}$, then $x(t+u)=x(t)+x(u)$ for all $u.$

  1. (1)

    1. $\implies$

Let $a^{-x(l)}xa^{l}=a^{-x(s)}xa^{s}$ for some $l$ and $s$. Then
$x=a^{x(l)-x(s)}xa^{s-l}$. Now,

$x(t)=(a^{x(l)-x(s)}xa^{s-l})(t)=x(t+s-l)+x(l)-x(s)=x(t)+x(s-l)+x(l)-x(s)$

and so

$-x(s-l)=x(l)-x(s).$

Thus,

$\displaystyle x$ $\displaystyle=a^{x(l)-x(s)}xa^{s-l}$
$\displaystyle=a^{-x(s-l)}xa^{s-l}$

so $a^{s-l}\in F_{x}$ so there exists a $p$ such that $s-l=pt$ so $s=l\mod t$.

    2. $\impliedby$

Assume there exists $s,l$ so $x(s-l)=x(s)-x(l)$ and $s=l\mod t$. Let $s-l=pt$
for some $p$. Then, since $a^{-pt}=(a^{t})^{-p}\in F_{x}$, Proposition
LABEL:*prop:for python easier formulas (2) gives that $a^{-x(-pt)}xa^{-pt}=x$.
Using the substitution $l=s-pt$, routine computation gives that
$a^{-x(l)}xa^{l}=a^{-x(s)}xa^{s}$ as desired.

  2. (2)

From Proposition 4.1.1 we know that
$\mathcal{O}_{x}=\\{a^{-x(l)}xa^{l}\,|\,l=1,...,n\\}$. Let
$y\in\mathcal{O}_{x}$. Then $y=a^{-x(l)}xa^{l}$ for some $l$. Thus,

$y(t)=(a^{-x(l)}xa^{l})(t)=x(t+l)-x(l)=x(t)+x(l)-x(l)=x(t).$

Furthermore,

$a^{-y(t)}ya^{t}=a^{-x(t)}ya^{t}=a^{-x(t)}(a^{-x(l)}xa^{l})a^{t}=a^{-x(l)}a^{-x(t)}xa^{t}a^{l}=a^{-x(l)}xa^{l}=y.$

Therefore, $a^{t}\in F_{y}$ and so $F_{x}\subset F_{y}$.

However, $\mathcal{O}_{x}=\mathcal{O}_{y}$ by definition of group actions the
choice of representative does not change the orbit and so this shows that
stabilizers are contained within each other. Thus, $F_{x}=F_{y}$ for all
$y\in\mathcal{O}_{x}.$

Finally, let

$\mathcal{O}_{x}=\\{x,y_{1},y_{2},...,y_{n-1}\\}$

where $y_{l}=a^{-x(l)}xa^{l}$. Now, since the orbit stabilizer theorem tells
us that $|\mathcal{O}_{x}|=\frac{|F|}{|F_{x}|}=\frac{n}{\frac{n}{t}}=t$ If
$n=t$, then there is nothing to show. However, if $t<n$, then consider
$p=ut+q$ for some $q<t$. Then inductively, $x(ut+q)=x(ut)+x(q)$ and so
$x(ut+q-q)=x(ut)=x(ut+q)-x(q)$. Clearly $ut+q=q\mod t$ and so (1) gives us
that $y_{p}=y_{q}$. Therefore,

$\mathcal{O}_{x}=\\{x,y_{1},y_{2},...,y_{n-1}\\}=\\{x,y_{1},...,y_{t-1},x,y_{1},...y_{t-1},...\\}=\\{x,y_{1},...,y_{t-1}\\}.$

That is, $\mathcal{O}_{x}=\\{a^{-x(l)}xa^{l}\,|\,l=1,...,t\\}$

  3. (3)

Let $x^{-1}=a^{-x(s)}xa^{s}$ for some $s\leq t$. Then we have that

$x^{-1}(u)=x(u+s)-x(s)\qquad\text{ for all }u.$

Namely, $x(x(l)+s)=x^{-1}(x(l))+x(s)=l+x(s)$ and so Then

$s=x^{-1}(x(s))=x(x(s)+s)-x(s)\implies x(x(s)+s)=x(s)+s$

so $x(s)+s$ is a fixed point of $x.$ Furthermore,

$\displaystyle a^{-x(x(s)+s)}xa^{x(s)+s}$
$\displaystyle=a^{-(x(s)-s)}xa^{x(s)+s}=a^{-s}(a^{-x(s)}xa^{s})a^{x(s)}$
$\displaystyle=a^{-s}x^{-1}a^{x(s)}=(a^{-x(s)}xa^{s})^{-1}$
$\displaystyle=(x^{-1})^{-1}=x$

so $a^{x(s)+s}\in F_{x}$ so $x(s)+s=ut$ for some $1\leq u\leq\frac{n}{t}$.

  4. (4)

Let $x^{-1}=a^{-x(s)}xa^{s}$ and $y\in\mathcal{O}_{x}$ with
$y=a^{-x(l)}xa^{l}$. Then

$\displaystyle y^{-1}$
$\displaystyle=(a^{-x(l)}x^{l})^{-1}=a^{-l}x^{-1}a^{x(l)}$
$\displaystyle=a^{-l}a^{-x(s)}xa^{s}a^{x(l)}=a^{-l-x(s)}xa^{x(l)+s}$

Note that (3) gives that $y^{-1}=a^{-x(x(l)+s)}xa^{x(l)+s}.$

Next, let $a^{r}\in F_{y^{-1},y}$. Note that since we are assuming that
$x^{-1}\in\mathcal{O}_{x}$, we know that $F_{y^{-1},y}$ is nonempty and has
the same size as $F_{x}$ from Lemma 2.6.

Then,

$\displaystyle y$ $\displaystyle=a^{-x(l)}xa^{l}=y^{-1}\lhd a^{r}$
$\displaystyle=a^{-y^{-1}(r)}y^{-1}a^{r}=a^{-y^{-1}(r)}(a^{-x(x(l)+s)}xa^{x(l)+s})a^{r}$
$\displaystyle=a^{-x(r+x(l)+s)+x(x(l)+s)}a^{-x(x(l)+s)}xa^{x(l)+s}a^{r}=a^{-x(x(l)+s+r)}xa^{x(l)+s+r}$

Thus, we have that

$a^{-x(l)}xa^{l}=a^{-x(x(l)+s+r)}xa^{x(l)+s+r}$

and so from (1), there exists a $p$ such that $x(l)+s+r-l=pt$. Namely,
$a^{r}=a^{pt+l-x(l)-s}$.

Conversely, fix $1\leq m\leq\frac{n}{t}$ and let $r=mt+l-x(l)-s$. Then

$y^{-1}(r)=y^{-1}(mt+l-x(l)-s)=x(mt+l)-l-x(s)=mx(t)+x(l)-l-x(s)$

and

$\displaystyle y^{-1}\lhd a^{r}$
$\displaystyle=a^{-y^{-1}(r)}y^{-1}a^{r}=a^{-y^{-1}(r)}(a^{-l-x(s)}xa^{x(l)+s})a^{r}$
$\displaystyle=a^{-mx(t)-x(l)+l+x(s)}a^{-l-x(s)}xa^{x(l)+s}a^{r}=a^{-x(l)-x(mt)}xa^{x(l)+s+r}$
$\displaystyle=a^{-x(l)}a^{-x(mt)}xa^{mt+l}=a^{-x(l)}xa^{l}$ $\displaystyle=y$

and so $a^{mt+l-x(l)-s}\in F_{y^{-1},y}$ for all $m.$

Therefore,

$F_{y^{-1},y}=\\{a^{mt+l-x(l)-s}\,|\,m=1,2,...,\frac{n}{t}\\}.$

Since we are taking all integers modulo $n$, we could equivalently consider
$u=0,...,\frac{n}{t}-1$, which will be convenient to us later.∎

###  Uncovering New Permutations Properties

Now, using our new notation, we will explore exactly how much we can obtain
from this shifting. It turns out the answer is quite a lot! First, we dive
into understanding what info we can obtain from the stabilizer of
$F_{x}\subset F=C_{n}$. Because the stabilizer is a subgroup and $C_{n}$ is
cyclic, we can write $F_{x}=\langle a^{t}\rangle$ for some value $t$ dividing
$n$. Throughout, we will refer to $t$ often and it will always be reserved for
the (smallest) power of $a$ which generates the stabilizer of a given $x$.

######  Lemma 4.2.1.

Let $x\in S_{n-1}$ and $F_{x}=\langle a^{t}\rangle$ where
$a=(1\,\,\,2\,\,\,\cdots\,\,\,n)$. Then,

  1. (1)

$x(ut+q)=ux(t)+x(q)$ for all $u\in\mathbb{Z}/n\mathbb{Z}$ and $q<t.$

  2. (2)

$x(t)=jt$ where $1\leq j<\frac{n}{t}$ is coprime to $\frac{n}{t}$

  3. (3)

$F_{x^{-1}}=F_{x}$

  4. (4)

If $x(t)=jt$ then $x^{-1}(t)=j^{-1}t$ where $j^{-1}$ is the multiplicative
inverse of $j$ modulo $\frac{n}{t}$.

  5. (5)

if $x^{-1}\in\mathcal{O}_{x}$ and $x(t)=jt$, then $j^{2}=1\mod\frac{n}{t}$.

Proof.

  1. (1)

This is induction on the statment $x(u+t)=x(u)+x(t)$ for all $t$ from
Proposition LABEL:*prop:for python easier formulas (2) .

  2. (2)

Here we are assuming that $F_{x}=\langle a^{t}\rangle\subset F=\langle
a\rangle$. Since $|F|=n$, and $F_{x}$ is a subgroup, Lagrange tells us that
there exists an $m$ such that $n=mt$. Therefore by (1),

$mt=n=x(n)=x(mt)=mx(t).$

Namely, using rules of modular arithmetic,

$mx(t)=mt\mod n\implies x(t)=t\mod\left(\frac{n}{m}=t\right).$

Thus, $x(t)$ is some multiple of $t$.

Now, for contradiction, assume that $x(t)=jt$ where $j$ is not coprime to
$\frac{n}{t}$. Then, let $u=\frac{n}{t\text{gcd}(j,\frac{n}{t})}<\frac{n}{t}$
and so $ut<n$ but

$x(ut)=ux(t)=ujt=\frac{njt}{t\text{gcd}(j,\frac{n}{t})}=n\frac{j}{\text{gcd}(j,\frac{n}{t})}=n\mod
n$

which is not possible since $x\in S_{n-1}$ and so $x(n)=n$. Namely, $x(t)=jt$
where $j$ is coprime to $\frac{n}{t}$.

  3. (3)

Since Proposition LABEL:*prop:for python easier formulas (1) gives us that
$x(u+t)=x(u)+x(t)$ for all $u$, inductively we get that

$x(mt+u)=x((m-1)t+u+t)=x((m-1)t+u)+x(t)=\cdots=x(u)+mx(t).$

  4. (4)

First, since $x\lhd a^{t}=x$, then inverting both sides and using the
properties of Proposition 2.9 and Proposition LABEL:*prop:for python easier
formulas (1) , we obtain

$(x\lhd a^{t})^{-1}=x^{-1}\lhd(x\rhd a^{t})=x^{-1}\lhd a^{x(t)}=x^{-1}.$

Namely, $a^{x(t)}\in F_{x^{-1}}$. However, $x(t)=jt$ where $j$ is coprime to
$\frac{n}{t}$ by (2). Thus, $a^{x(t)}=(a^{t})^{j}$ is a generator of $F_{x}$
and so $F_{x}\subset F_{x^{-1}}$.

Now, the reverse. Let $F_{x^{-1}}=\langle a^{s}\rangle$, then $x^{-1}\lhd
a^{s}=x^{-1}$ so again using the properties of Proposition 2.9 and Proposition
LABEL:*prop:for python easier formulas (1) ,

$(x^{-1}\lhd a^{s})^{-1}=x\lhd(x^{-1}\rhd a^{s})=x\lhd a^{x^{-1}(s)}=x$

so $a^{x^{-1}(s)}\in F_{x}$. But (2) tells us that $x^{-1}(s)=ls$ where $l$ is
coprime to $\frac{n}{s}$ and so again, $a^{x^{-1}(s)}$ is a power of the
generator which is coprime to $\frac{n}{s}=|F_{x^{-1}}|$ and so it is again a
generator. Thus, $F_{x^{-1}}\subset F_{x}$.

  5. (5)

Note that from (2), we know that $x(t)=jt$ where $j$ is coprime to
$\frac{n}{t}$. Thus, $j$ has a unique multiplicative inverse modulo
$\frac{n}{t}$ which we will call $j^{-1}$. Let $x^{-1}(t)=lt$ where $l$ is
coprime to $\frac{n}{t}$. Then, from (1)

$t=x^{-1}(x(t))=x^{-1}(jt)=jx^{-1}(t)=jlt\implies 1=jl\mod\frac{n}{t}.$

Thus, because inverses (if they exist) are unique, we have that
$l=j^{-1}\mod\frac{n}{t}$.

  6. (6)

If there is an $s$ so $x^{-1}=a^{-x(s)}xa^{s}\in\mathcal{O}_{x}$, then
$x^{-1}(u)=x(u+s)-x(s)$ for all $u=1,...,n-1$. Namely,

$x^{-1}(t)=x(t+s)-x(s)=x(s)+x(t)-x(s)=x(t).$

That is, $t=x^{2}(t)=x(jt)=jx(t)=j^{2}t$ and so

$j^{2}t=t\mod n\implies j^{2}=1\mod\frac{n}{t}.$ ∎

######  Remark 4.2.2.

The converse of Lemma LABEL:*lem:x(t)=kt some k (5) is false. That is, knowing
$x$ satisfies that $x(t)=jt$ where $j^{2}=1\mod\frac{n}{t}$ does not force
$x^{-1}\in\mathcal{O}_{x}$.

######  Corollary 4.2.3.

Let $x\in S_{n-1}$. Then $a^{t}\in F_{x}$ if and only if $x(mt+u)=mx(t)+x(u)$
for all $1\leq m\leq\frac{n}{t}$ and $0\leq u<t$.

Proof.

  1. $\implies$

This follows directly from Lemma LABEL:*lem:x(t)=kt some k (1) .

  2. $\impliedby$

Assume $x(mt+u)=mx(t)+x(u)$ for all $1\leq m\leq\frac{n}{t}$ and $0\leq u<t$.
Let $1\leq q\leq n$ and write $q=mt+u$ for some $0\leq u<t$. Then,

$\displaystyle x(q+t)-x(t)$ $\displaystyle=x(mt+u+t)-x(t)=x((m+1)t+u)-x(t)$
$\displaystyle=(m+1)x(t)+x(u)-x(t)=mx(t)+x(u)$ $\displaystyle=x(mt+u)=x(q)$

So $a^{t}\in F_{x}$ by Proposition 4.1.3.∎

######  Corollary 4.2.4.

Let $x\in S_{n-1}$. Then $F_{x}=F=C_{n}=\langle a\rangle$ if and only if for
all $u=1,...,n$, $x(u)=ux(1)$.

Let us pause here to emphasize the volume of information we obtain from
knowing the size of the stabilizer of $x$. Lemma 4.2.1 tells us that if
$F_{x}=\langle a^{t}\rangle$, then $x$ is in some sense linear in $t$. That
is, $x(q)$ where $q>t$ is determined by where $x$ sends $q\mod t$ and where
$x$ sends $t.$

This puts a heavy restriction on what $x$ can look like. All values $x(q)$ for
$q>t$ are determined from the values $x(t)$ and $x(u)$ for $u<t$. Thus, if $t$
is small relative to $n$, then the number of permutations in $S_{n-1}$ which
have stabilizer $\langle a^{t}\rangle$ is small relative to the total $(n-1)!$
which we formalize in Section 5.

Using our new understanding of permutations $x$ satisfying $F_{x}=\langle
a^{t}\rangle$, we can now better understand the relationship between $x$ and
$y$ where $y\in\mathcal{O}_{x}$.

######  Lemma 4.2.5.

Let $x\in S_{n-1}$, $F_{x}=\langle a^{t}\rangle$, and
$x^{-1}\in\mathcal{O}_{x}$. Write
$\mathcal{O}_{x}=\\{x,y_{1},y_{2},...,y_{t-1}\\}$ where
$y_{l}=a^{-x(l)}xa^{l}$. Then

  1. (1)

If $x^{-1}=y_{s}$ for $1\leq s\leq t$, then $y_{l}^{-1}=y_{x(l)+s\text{ mod
}t}$ for all $l=1,...,t.$ (Here we let $y_{t}=x$).

  2. (2)

for some $1\leq l\leq t$, $l=x(l)+s\mod t$ if and only if $y_{l}$ has order
$2$ (Here we let $y_{t}=x$).

Proof.

  1. (1)

If $x^{-1}=y_{s}=a^{-x(s)}xa^{s}$ for some $s\leq t$, then

$\displaystyle y_{l}^{-1}$
$\displaystyle=(a^{-x(l)}xa^{l})^{-1}=a^{-l}x^{-1}a^{x(l)}$
$\displaystyle=a^{-l}a^{-x(s)}xa^{s}a^{x(l)}=a^{-l-x(s)}xa^{s+x(l)}$

Now note that $x^{-1}=a^{-x(s)}xa^{s}$ implies that for all $j$,
$x^{-1}(j)=x(j+s)-x(s)$ and so

$-x(x(l)+s)=-(x^{-1}(x(l))+x(s))=-(l+x(s))=-l-x(s)$

and so we obtain that

$y_{l}^{-1}=a^{-x(x(l)+s)}xa^{x(l)+s}=y_{x(l)+s\mod t}.$

  2. (2)

This is immediate from (1) since $y_{l}$ has order two if and only if
$y_{l}^{-1}=y_{x(l)+s}=y_{l}$ which happens if and only if $x(l)+s=l\mod t$.∎

###  A Revised Indicator Formula

######  Proposition 4.3.1.

Let $x\in S_{n-1},$ $F_{x}=\langle a^{t}\rangle$,
$x^{-1}=a^{-x(s)}xa^{s}\in\mathcal{O}_{x}$ for some $s$, $x(t)+t=u_{1}t$, and
$x(s)+s=u_{2}t$. Fix an integer $i$. If $iu_{1}=0\mod\frac{n}{t}$ then

  1. (1)

if $\frac{n}{t}$ is odd, $iu_{2}=0\mod\frac{n}{t}$

  2. (2)

if $\frac{n}{t}$ is even, then $iu_{2}=\frac{n}{2t}\text{ or
}0\mod\frac{n}{t}$.

Proof. First, note that since $x^{-1}(i)=x(i+s)-x(s)$, letting $i=x(s)$ we get
that

$x(x(s)+s)=x^{-1}(x(s))+x(s)=s+x(s).$

Now,

$\displaystyle a^{-x(x(s)+s)}xa^{x(s)+s}$
$\displaystyle=a^{-(x(s)+s)}xa^{x(s)+s}=a^{-s}a^{-x(s)}xa^{s}a^{x(s)}$
$\displaystyle=a^{-s}x^{-1}a^{x(s)}=(a^{-x(s)}xa^{s})^{-1}$
$\displaystyle=(x^{-1})^{-1}=x.$

Therefore, $a^{x(s)+s}\in F_{x}=\langle a^{t}\rangle$ and so $x(s)+s=u_{2}t$
is a multiple of $t.$

Thus,

$u_{2}x(t)=x(u_{2}t)=x(x(s)+s)=x(s)+s=u_{2}t$

and so $2u_{2}t=u_{2}t+u_{2}t=u_{2}x(t)+u_{2}t=u_{1}u_{2}t.$ This gives us the
relation that

$2u_{2}t=u_{1}u_{2}t\mod n\implies 2u_{2}=u_{1}u_{2}\mod\frac{n}{t}.$

Therefore,

$2iu_{2}=iu_{1}u_{2}=0\mod\frac{n}{t}.$

  1. (1)

Assume $\frac{n}{t}$ is odd. Then $2$ is invertible $\mod\frac{n}{t}$ so

$iu_{2}=0\mod\frac{n}{t}.$

  2. (2)

Assume $\frac{n}{t}$ is even. Then

$iu_{2}=0\mod\frac{n}{2t}.$ ∎

We are at last ready to give the key theorem so far.

######  Theorem 4.3.2.

Consider $J_{n}=\mathbb{C}^{S_{n-1}}\\#\mathbb{C}C_{n}$. Let $C_{n}=\langle
a\rangle$. Fix $x\in S_{n-1}$ such that $F_{x}=\langle a^{t}\rangle$ and
$x^{-1}=x\lhd a^{s}\in\mathcal{O}_{x}$. Let
$\rho_{x,i}:F_{x}\to\mathbb{C}^{\times}$ be an irreducible (group)
representation of $F_{x}$ defined by
$\rho_{x,i}(a^{t})=\zeta_{\frac{n}{t}}^{i}$ where $\zeta_{\frac{n}{t}}$ is a
primitive $\frac{n}{t}$-root of unity. Let $\chi_{x,i}$ be the character of
$\rho_{x,i}$ Then if $\hat{\chi_{x,i}}$ is the character of $J_{n}$ induced
from $\chi_{x,i}$ (as described in Theorem 3.2) we obtain:

  1. (1)

if $\frac{n}{t}$ is odd, then

$\nu(\hat{\chi}_{x,i})=\begin{cases}1&\text{ if
}(\zeta_{\frac{n}{t}})^{iu_{1}}=1\\\ 0&\text{ if
}(\zeta_{\frac{n}{t}})^{iu_{1}}\not=1\end{cases}$ (2)

  2. (2)

if $\frac{n}{t}$ is even, then

$\nu(\hat{\chi}_{x,i})=\begin{cases}(\zeta_{\frac{n}{t}})^{-iu_{2}}=1\text{ or
}-1&\text{ if }(\zeta_{\frac{n}{t}})^{iu_{1}}=1\\\ 0&\text{ if
}(\zeta_{\frac{n}{t}})^{iu_{1}}\not=1\end{cases}$ (3)

where

$x(t)+t=u_{1}t\qquad\text{ and }\qquad x(s)+s=u_{2}t.$

Proof. Let $\rho_{x,i}:F_{x}\to\mathbb{C}^{\times}$ be an irreducible (group)
representation of $F_{x}$ defined by
$\rho_{x,i}(a^{t})=(\zeta_{\frac{n}{t}})^{i}$ where $\zeta_{\frac{n}{t}}$ is a
primitive $\frac{n}{t}$-root of unity. Let $\chi_{x,i}$ be the character of
$V_{x,i}$ which is the $F_{x}$-module equivalent to $\rho_{x,i}$. Then if
$\hat{\chi_{x,i}}$ is the character of $J_{n}$ induced from $\chi_{x,i}$ (as
described in Theorem 3.2)

First, recall Eq. 1. Note that Theorem 2.8 tells us immediately that assuming
$x^{-1}\in\mathcal{O}_{x}$ gives that
$\hat{\chi}_{x,i}(p_{y}\\#b)=\chi_{x,i}(b)$ and so we get

$\nu(\hat{\chi}_{x,i})=\frac{1}{|F|}\sum_{y\in\mathcal{O}_{x}}\sum_{b\in
F_{y^{-1},y}}\chi_{x,i}((y^{-1}\rhd b)b)$

where $\chi_{x,i}$ is the character of $V_{x,i}$ which is the $F_{x}$-module
which induces $\hat{\chi}_{x,i}$.

Since $x^{-1}\in\mathcal{O}_{x}$, let $x^{-1}=a^{-x(s)}xa^{s}$ for some $s\leq
t$.

Let $y_{l}\in\mathcal{O}_{x}$ with $y_{l}=a^{-x(l)}xa^{l}$. Then Proposition
LABEL:*prop:equivalence of sets (4) gives that we can write $b\in
F_{y^{-1},y}$ as $b=a^{mt+l-s-x(l)}$ for some $1\leq m\leq\frac{n}{t}$.

Note that we have that

$y_{l}^{-1}=(a^{-x(l)}xa^{l})^{-1}=a^{-l}x^{-1}a^{x(l)}=a^{-l}(a^{-x(s)}xa^{s})a^{x(l)}$

and so

$y_{l}^{-1}(mt+l-s-x(l))=x(mt+l)-x(s)-l,$

Thus,

$\displaystyle(y_{l}^{-1}\rhd b)b$
$\displaystyle=a^{y_{l}^{-1}(mt+l-s-x(l))}a^{mt+l-s-x(l)}=a^{x(mt+l)-x(s)-l}a^{mt+l-s-x(l)}$
$\displaystyle=a^{x(mt)+mt-x(s)-s}=a^{m(x(t)+t)-(x(s)+s)}$

Note that $(y^{-1}\lhd b)$ is very much dependent on both $y$ and $b$.
However, $(y^{-1}\lhd b)b$ is dependent on only the choice of the $s$ for
which $x^{-1}=a^{-x(s)}xa^{s}$.

Let $x(t)+t=u_{1}t$ and $x(s)+s=u_{2}t$. Note that both these sums are
necessarily multiplies of $t$ since $x(t)$ is a multiple of $t$ from Lemma
LABEL:*lem:x(t)=kt some k (2) and $x(s)+s$ is a multiple of $t$ from
Proposition LABEL:*prop:equivalence of sets (4) .

Then,

$\displaystyle\nu(\hat{\chi}_{x,i})$
$\displaystyle=\frac{1}{|F|}\sum_{y\in\mathcal{O}_{x}}\sum_{b\in
F_{y^{-1},y}}\chi_{x,i}((y^{-1}\rhd b)b)$
$\displaystyle=\frac{1}{n}\sum_{l=1}^{t}\sum_{m=0}^{\frac{n}{t}-1}\chi_{x,i}((y_{l}^{-1}\rhd
a^{mt+l-s-x(l)})a^{mt+l-s-x(l)})$
$\displaystyle=\frac{1}{n}\sum_{l=1}^{t}\sum_{m=0}^{\frac{n}{t}-1}\chi_{x,i}(a^{m(x(t)+t)-(x(s)+s)})$
$\displaystyle=\frac{t}{n}\sum_{m=0}^{\frac{n}{t}-1}\chi_{x,i}((a^{t})^{mu_{1}-u_{2}})$
$\displaystyle=\frac{t}{n}\sum_{m=0}^{\frac{n}{t}-1}(\zeta_{\frac{n}{t}})^{i(mu_{1}-u_{2})}$
$\displaystyle=\frac{t}{n}\sum_{m=0}^{\frac{n}{t}-1}(\zeta_{\frac{n}{t}})^{imu_{1}}(\zeta_{\frac{n}{t}})^{-iu_{2}}$
$\displaystyle=\frac{t}{n}(\zeta_{\frac{n}{t}})^{-iu_{2}}\begin{cases}\frac{n}{t}&\text{
if }(\zeta_{\frac{n}{t}})^{iu_{1}}=1\\\
\frac{(\zeta_{\frac{n}{t}})^{iu_{1}\frac{n}{t}}-1}{(\zeta_{\frac{n}{t}})^{iu_{1}}-1}&\text{
if }(\zeta_{\frac{n}{t}})^{iu_{1}}\not=1\end{cases}$
$\displaystyle=\begin{cases}(\zeta_{\frac{n}{t}})^{-iu_{2}}&\text{ if
}(\zeta_{\frac{n}{t}})^{iu_{1}}=1\\\
\frac{t}{n}(\zeta_{\frac{n}{t}})^{-iu_{2}}\frac{(\zeta_{\frac{n}{t}})^{iu_{1}\frac{n}{t}}-1}{(\zeta_{\frac{n}{t}})^{iu_{1}}-1}&\text{
if }(\zeta_{\frac{n}{t}})^{iu_{1}}\not=1\end{cases}$
$\displaystyle=\begin{cases}(\zeta_{\frac{n}{t}})^{-iu_{2}}&\text{ if
}(\zeta_{\frac{n}{t}})^{iu_{1}}=1\\\ 0&\text{ if
}(\zeta_{\frac{n}{t}})^{iu_{1}}\not=1\end{cases}$

with the last line reducing since $\zeta_{\frac{n}{t}}$ is a
$\frac{n}{t}^{\text{th}}$-root of unity, so
$(\zeta_{\frac{n}{t}})^{iu_{1}\frac{n}{t}}=((\zeta_{\frac{n}{t}})^{\frac{n}{t}})^{iu_{1}}=1$
so $(\zeta_{\frac{n}{t}})^{iu_{1}\frac{n}{t}}-1=0.$

  1. (1)

Finally, Proposition 4.3.1 gives us that $iu_{2}$ is a multiple of
$\frac{n}{t}$ if $\frac{n}{t}$ is odd, and so
$(\zeta_{\frac{n}{t}})^{iu_{2}}=1$ in this case.

  2. (2)

In the other case, $iu_{2}$ is a multiple of $\frac{n}{2t}$ if $\frac{n}{t}$
is even. Namely, there exists an $m$ so $iu_{2}=m\frac{n}{2t}$. Therefore,
$(\zeta_{\frac{n}{t}})^{iu_{2}}=(-1)^{m}$ which could be $-1$ if $m$ is odd.∎

Note that Theorem 4.3.2 agrees completely with the results of [5]. Namely,
Theorem 3.5 tells us that if $F_{x}=\\{1\\}$ then $\nu(\hat{\chi})=1$ if and
only if $x^{-1}\in\mathcal{O}_{x}$ and is $0$ otherwise. Theorem 4.3.2 says
that if $F_{x}=\\{1\\}$ then $t=n$ so $\frac{n}{t}=1$ which is odd. There is
then one (group) irrep of $F_{x}$, the trivial (group) irrep and so the
indicator will be $1$ if $x^{-1}\in\mathcal{O}_{x}$ and $0$ otherwise.

Theorem LABEL:*thm:JM indicator when Fx=F (2) says that if $F_{x}=F$ and
$x^{2}\not=1$ then all indicators are $0.$ Theorem 4.3.2 says that if
$F_{x}=F$ then $t=1$ and so $\frac{n}{t}=n$. Since $|\mathcal{O}_{x}|=t=1$, if
$x^{2}\not=1$ then $x^{-1}\not\in\mathcal{O}_{x}=\\{x\\}$ and so all
indicators are $0$.

Theorem LABEL:*thm:JM indicator when Fx=F (1) tells us that if $x=1$, then the
indicator of [5] and the Frobenius-Schur group indicator agree. Theorem 4.3.2
states that if $x=1$ then $x^{-1}=x\lhd a^{t}$ and so $s=t$. Namely,
$(\zeta_{n})^{iu_{1}}=(\zeta_{n})^{2i}$ and implies
$(\zeta_{n})^{-iu_{2}}=(\zeta_{n})^{-2i}$. Thus, whether $n$ is even or odd,
if $\zeta_{n}^{2i}=1$ then $\zeta_{n}^{-2i}=1$ and so the indicator
$\nu(\hat{\chi}_{x,i})=1$ if and only if $n$ divides $2i$ and $0$ otherwise.
This of course agrees with the indicator $\nu(\chi_{x,i})$ by Theorem 3.14 and
so the indicators are the same.

Finally, we will see in the later sections Theorem 4.3.2 agrees with the work
of [5] when $n=p^{r}$ for $p$ and odd prime and $r>0$.

######  Corollary 4.3.3.

If $n$ is odd, then all indicators of $J_{n}$ are nonnegative.

An important question is whether the indicator is ever negative. Recall that
for $p$ prime, $J_{p}$ has nonnegative indicator. However, the answer is yes,
there exists even $n$ for which $J_{n}$ has negative indicator.

######  Example 4.3.4.

Let $n=16$, one can show the element

$x=(1\,\,\,5\,\,\,9\,\,\,13)(3\,\,\,7\,\,\,11\,\,\,15)$

has stabilizer $F_{x}=\langle a^{2}\rangle$ and so $t=2.$

Now, let $\zeta_{\frac{n}{t}}=e^{\frac{2\pi i}{8}}\in\mathbb{C}^{\times}$ be a
primitive $8^{\text{th}}$-root of unity.

Then, $\frac{n}{t}=8$ is certainly even and by theorem, Theorem 4.3.2,

$\nu(\hat{\chi})=\begin{cases}\zeta_{8}^{-3i}&\text{ if }\zeta_{8}^{2i}=1\\\
0&\text{ if }\zeta_{8}^{2i}\not=1\end{cases}$

Finally, consider the irreducible (group) representation of $F_{x}$,
$\rho_{4}$ defined by $\rho_{4}(a^{2})=\zeta_{8}^{4}$.

That is, we can let $i=4$. Then $\zeta_{8}^{2i}=\zeta_{8}^{8}=1$ and so we
have that $\zeta_{8}^{4}=e^{4\frac{2\pi i}{8}}=e^{\pi i}=-1$ and so

$\nu(\hat{\chi})=\zeta_{8}^{-3i}=\zeta_{8}^{-12}=(\zeta_{8}^{4})^{-3}=(-1)^{-3}=-1.$

In fact, we will see that there exists irreps of $J_{n}$ which have negative
indicator for any $n\geq 12$ with $4|n$.

Before we conclude this section, we provide two important results that we will
use in later sections.

######  Proposition 4.3.5.

Let $x\in S_{n-1},$ $F_{x}=\langle a^{t}\rangle$,
$x^{-1}=a^{-x(s)}xa^{s}\in\mathcal{O}_{x}$, and $x(t)=jt$ for some $j$ coprime
to $\frac{n}{t}$. Let $\hat{\chi}$ be the character induced from an
irreducible representation of $F_{x}$. Then,

  1. (1)

Permutations in the same orbit have the same indicators. That is, if
$y\in\mathcal{O}_{x}$ then for a given representation, the induced characters
$\hat{\chi}_{x}$ and $\hat{\chi}_{y}$ have the same indicator. (This is true
even if $x^{-1}\not\in\mathcal{O}_{x}$).

  2. (2)

If $\mathcal{O}_{x}$ contains an element of order $2$, then
$\nu(\hat{\chi})\geq 0$

  3. (3)

If $t$ is odd, then $\nu(\hat{\chi})\geq 0$.

  4. (4)

If $n=2m$ for $m$ odd then $\nu(\hat{\chi})\geq 0$.

Proof.

  1. (1)

Note Proposition LABEL:*prop:equivalence of sets (2) tells us that if
$y\in\mathcal{O}_{x}$ then $F_{x}=F_{y}$. Thus, any (group) irrep of $F_{x}$
is equivalent to a group irrep of $F_{y}$. However, Theorem 4.3.2 uses values
dependent on the outputs of $x$ and so we must show that in any case, the
values used to compute the indicator in Theorem 4.3.2 are the same. Let
$y\in\mathcal{O}_{x}$.

    1. (i)

First, Lemma LABEL:*lem:orbit and stab inv equiv (2) tells us that
$x^{-1}\not\in\mathcal{O}_{x}$ if and only if
$y^{-1}\not\in\mathcal{O}_{y}=\mathcal{O}_{x}$ and so if
$x^{-1}\not\in\mathcal{O}_{x}$ then $y^{-1}\not\in\mathcal{O}_{x}$ so all
indicators are $0$ and certainly they are all the same.

    2. (ii)

Now, assume $x^{-1}\in\mathcal{O}_{x}$. Then
$y^{-1}\in\mathcal{O}_{y}=\mathcal{O}_{x}$. Then, Proposition
LABEL:*prop:equivalence of sets (2) tells us that $x(t)=y(t)$.

      1. (a)

If $\frac{n}{t}$ is odd, then $x(t)+t=y(t)+t$ and since this is the only value
used in theorem Theorem 4.3.2 when computing the indicator, we have that the
set of indicators induced from $y$ is the same as the set of indicators
induced from $x$ and that they are induced from the same irreducible group
representations of $F_{x}=F_{y}$.

      2. (b)

If $\frac{n}{t}$ is even, then let $s\leq t$ be such that
$x^{-1}=a^{-x(s)}xa^{s}$ and let $y=y_{l}=a^{-x(l)}xa^{l}\in\mathcal{O}_{x}.$
Then Lemma LABEL:*lem:y in orbit x (1) gives that $y^{-1}=a^{-x((x(l)+s)\mod
t)}xa^{(x(l)+s)\mod t}$. Let $1\leq q\leq\frac{n}{t}$ and $1\leq p\leq t$ be
such that $x(l)+s=qt+p$. Then, $y(p-l)=x(p)-x(l)$ and so

$y^{-1}=a^{-x(p)}xa^{p}=a^{-x(p)}(a^{x(l)}ya^{-l})a^{p}=a^{-(x(p)-x(l))}ya^{p-l}=a^{-y(p-l)}ya^{p-l}$

and so $y^{-1}=a^{-y(p-l)}ya^{p-l}\in\mathcal{O}_{y}$. Finally,
$l=x^{-1}(x(l))=x(x(l)+s)-x(s)$ and so

$\displaystyle y(p-l)+p-l$ $\displaystyle=x(p)-x(l)+p-l$
$\displaystyle=x(x(l)+s-qt)-x(l)+x(l)+s-qt-l$
$\displaystyle=x(x(l)+s)-x(qt)+s-qt-l$ $\displaystyle=l+x(s)+s-q(x(t)+t)-l$
$\displaystyle=x(s)+s-q(x(t)+t)$

Fix $i$ and consider the irreducible group representation
$\rho_{x,i}=\rho_{y,i}=\rho_{i}$ of $F_{x}$ defined by
$\rho_{i}(a^{t})=(\zeta_{\frac{n}{t}})^{i}$. Then,

$(\zeta_{\frac{n}{t}})^{-i\frac{y(p-l)+p-l}{t}}=(\zeta_{\frac{n}{t}})^{-i\frac{x(s)+s-q(x(t)+t)}{t}}=(\zeta_{\frac{n}{t}})^{-i\frac{x(s)+s}{t}}((\zeta_{\frac{n}{t}})^{i\frac{x(t)+t}{t}})^{q}.$

Namely, either the indicator of the associated character
$\nu(\hat{\chi}_{i})=0$ or, if $(\zeta_{\frac{n}{t}})^{i\frac{x(t)+t}{t}}=1$,
then

$(\zeta_{\frac{n}{t}})^{-i\frac{y(p-l)+p-l}{t}}=(\zeta_{\frac{n}{t}})^{-i\frac{x(s)+s}{t}}$

and so in either case, the indicator of
$\nu(\hat{\chi}_{x,i})=\nu(\hat{\chi}_{y,i}).$

  2. (2)

If $y\in\mathcal{O}_{x}$ has order $2$, then $s=t$ and so if
$(\zeta_{\frac{n}{t}})^{i\frac{y(t)+t}{t}}=1$ then clearly
$(\zeta_{\frac{n}{t}})^{-i\frac{y(s)+s}{t}}=((\zeta_{\frac{n}{t}})^{i\frac{y(t)+t}{t}})^{-1}=1^{-1}=1$
as well. By (1), the indicator of the representation induced by $y$ is the
same as that of $x$ and so the indicators are all nonnegative.

  3. (3)

If $t$ is odd, then
$|\mathcal{O}_{x}|=\frac{|F|}{|F_{x}|}=\frac{n}{\frac{n}{t}}=t$ and
$x^{-1}\in\mathcal{O}_{x}$, then every $y\in\mathcal{O}_{x}$ also has its
inverse in $\mathcal{O}_{x}$ and so $\mathcal{O}_{x}$ contains an involution.
By (2), all indicators are nonnegative.

  4. (4)

If $n=2m$ where $m$ is odd, then either $t$ is odd, in which case (3) tells us
$\nu(\hat{\chi})\geq 0$, or $\frac{n}{t}$ is odd, in which case Theorem 4.3.2
tells us $\nu(\hat{\chi})\geq 0.$∎

######  Proposition 4.3.6.

The only $n$ for which $J_{n}$ is totally orthogonal is $n=2.$

Proof.

  1. (1)

$J_{n}$ can only be defined for $n\geq 2$ since its structure is derived from
$S_{n-1}$ and $S_{m}$ is not defined for $m<1$. Letting $n=2$, gives that
$S_{1}=\\{(1)\\}$ so the only possible permutation is the identity $x=(1)$ and
so $F_{x}=F$. In this case, we have that $t=1$. Now, we just plug into Theorem
4.3.2. $u_{1}=\frac{x(1)+1}{1}=2$ so if $\zeta_{2}$ is a primitive square root
of unity, then $\zeta_{2}^{2i}=1$ for all $i=0,1$, we have that
$\nu(\hat{\chi}_{(1),i})=+1$.

  2. (2)

Now, let $n>2$. Again, consider the identity permutation $(1)\in S_{n-1}$.
Then of course, $F_{x}=F$ again, and $u_{1}=2$, but now $\zeta_{n}^{2i}=1$ if
and only if $2i=0\mod n$. If $n$ is odd, $i=0$ is the only solution, and if
$n$ is even then $i=0,\frac{n}{2}$ are both solutions. Since $n>2$, there
exists $0\leq i\leq n-1$ such that $2i\not=0$ and so for these choices,
$\nu(\hat{\chi}_{(1),i})=0$. Therefore, not all the indicators are $+1$ and so
$J_{n}$ is not totally orthogonal for any $n>2.$ ∎

##  Explicitly Counting

Of course, the goal for the project is to try and count explicitly the number
of irreps of $J_{n}$ which have indicator $+1,-1$, and $0$ for a given
dimension. This result is very difficult to obtain directly since there are
$(n-1)!$ permutations and we must check the stabilizer of each permutation in
order to determine how many associated representations are induced from $x$
(see Theorem 3.2). However, using our new understanding of the group actions
which describe the structure of $J_{n}$ and our new reduced formulas from the
previous section, we are able to remove a lot of extraneous computation.

In this section, we show explicitly how to describe all irreps of $J_{n}$ of
dimension $t$ if $t$ is odd. Not only do we present formulas for counting the
total number of such irreps, but we describe explicitly how to construct the
set of permutations $x$ which we will later show produce nonzero indicators
and what these irreps look like.

Throughout this section, all arithmetic will be done modulo $n$ unless
otherwise specified. For example, when we write $x(2t)=2x(t)$ we mean
$x(2t\mod n)=2x(t)\mod n.$

For us, an involution is any permutation $x\in S_{n-1}$ which is its own
inverse (including the identity).

Additionally, we will frequently invoke a result from elementary number
theory: The equation $ax=b\mod n$ has either no solutions, or gcd$(a,n)$
solutions. Specifically, the equation $ax=0\mod n$ always has gcd$(a,n)$
solutions.

###  Structure of Permutations with Given Stabilizer

As we have seen from the previous sections, if $t$ is odd, then all
$t$-dimensional irreps of $J_{n}$ have nonnegative indicators (Proposition
4.3.5). Furthermore, the only way an (odd) $t$-dimensional irrep of $J_{n}$
can have an indicator of $+1$ is if the permutation $x$ which induces that
irrep shares its orbit with an involution.

Therefore, describing the odd dimensional irreps as having indicator either
$+1$ or $0$ is equivalent to understanding and counting the orbits under the
matched pair action which contain involutions.

######  Proposition 5.1.1.

Let $M_{n/t}$ be the set of $x\in S_{n-1}$ such that $F_{x}=\langle
a^{t}\rangle$. Let $\varphi(n)$ be the Euler Totient Function which is the
number of integers coprime to $n$. Then

$\displaystyle|M_{n/1}|$ $\displaystyle=\varphi(n)$ (4)
$\displaystyle|M_{n/t}|$
$\displaystyle=\varphi\left(\frac{n}{t}\right)\left(\frac{n}{t}\right)^{t-1}(t-1)!-\sum_{\begin{subarray}{c}s|t\\\
s\not=t\end{subarray}}|M_{n/s}|$ (5)

###### Proof.

Let $\varphi(n)=\\{1\leq j\leq n\,|\,j\text{ coprime to }n\\}$ be the Euler
Totient function.

The proof is simple counting. Corollary 4.2.4 tells us that $F_{x}=F$ if and
only if $x(u)=ux(1)$, Lemma 4.2.1 says that $x(1)=j$ for some $j$ coprime to
$n$, and Corollary 4.2.3 states that $x$ is determined by where it sends $1$.
Thus, the number of permutations $x\in S_{n-1}$ with full stabilizer is given
by the number of choices of $x(1)$, which is the number of integers coprime to
$n.$ Therefore, $|M_{n/1}|=\varphi(n)$.

Now, let $t>1$ be a divisor of $n.$ Again, Corollary 4.2.3 tell us that if we
want to count all permutations $x\in S_{n-1}$ with stabilizer $F_{x}=\langle
a^{t}\rangle$, then to fully determine $x$, we need only understand where it
sends $1,2,3,...,t$. Since $x(t)=jt$ where $j$ is coprime to $\frac{n}{t}$ by
Lemma 4.2.1, we know there are $\varphi(\frac{n}{t})$ choices for where $x$
can send $t.$

Now, where can $x$ send $1$? Well it certainly cannot send it to any multiple
of $t$ since then $x$ would no longer be a bijection. Thus, it can send $1$ to
any number $u$ where $1\leq u\leq n-1$ and $u$ is not a multiple of $t$. There
are $n-\frac{n}{t}=(t-1)\frac{n}{t}$ such options.

Recall that once we know $x(1)$, we also know $x(1+t)$,
$x(1+2t)$,…,$x(1+(\frac{n}{t}-1)t)$. Namely, once we choose $x(1)$ we have
fixed $\frac{n}{t}$ outputs.

Similarly, $x$ can send $2$ to any available value. Since all multiples of $t$
and now all values of the form $qt+x(1)$ for $1\leq q\leq\frac{n}{t}$ are
taken. So there are $n-2\frac{n}{t}=(t-2)\frac{n}{t}$ possible options for
$x(2)$.

Inductively we obtain:

$\underbrace{\varphi\left(\frac{n}{t}\right)}_{\text{choices for
}x(t)}\cdot\underbrace{(t-1)\frac{n}{t}}_{\text{choices for
}x(1)}\cdot\underbrace{(t-2)\frac{n}{t}}_{\text{choices for
}x(2)}\cdots\underbrace{\frac{n}{t}}_{\text{choices for }x(t-1)}.$

Now, define $x(qt+u)=qx(t)+x(u)$ and because this is a necessary condition we
have that $a^{t}\in F_{x}$ by Corollary 4.2.3.

However, we will have over counted here. Because this now counts all $x\in
S_{n-1}$ for which $a^{t}\in F_{x}$, and it is not necessarily true that
$F_{x}=\langle a^{t}\rangle$ for each of these permutations. Namely, we must
now remove all $x$ for which $F_{x}=\langle a^{s}\rangle$ where $s$ divides
$t$. That is, we have over counted by exactly $|M_{n/s}|$ for each $s$ which
is a divisor of $t$.

Finally, this gives the desired result,

$\displaystyle|M_{n/t}|$
$\displaystyle=\varphi\left(\frac{n}{t}\right)(t-1)\frac{n}{t}(t-2)\frac{n}{t}\cdots\frac{n}{t}-\sum_{\begin{subarray}{c}s|t\\\
s\not=t\end{subarray}}M_{n/s}$
$\displaystyle=\varphi\left(\frac{n}{t}\right)\left(\frac{n}{t}\right)^{t-1}(t-1)!-\sum_{\begin{subarray}{c}s|t\\\
s\not=t\end{subarray}}|M_{n/s}|\qed$

######  Remark 5.1.2.

For example, when $n=12$

$\displaystyle|M_{12/1}|$ $\displaystyle=4$ $\displaystyle|M_{12/2}|$
$\displaystyle=2(6)1!-4=8$ $\displaystyle|M_{12/3}|$
$\displaystyle=2(4)^{2}2!-4=60$ $\displaystyle|M_{12/6}|$
$\displaystyle=1(2)^{5}5!-4-8-60=3,768$ $\displaystyle|M_{12/12}|$
$\displaystyle=1(1)^{9}9!-4-8-60-3768=359,040$

In fact, Proposition 5.1.1 gives a recursive formula for explicitly
constructing all permutaitons $x$ which have stabilizer $F_{x}=\langle
a^{t}\rangle$ for a given $t$.

Using this formula, we will show in Section 7 that for a fixed $t$, the ratio
$|M_{n/t}|/(n-1)!\to 0$ as $n\to\infty.$ This shows that it is rare for a
permutation to have stabilizer $\langle a^{t}\rangle$ for large $n.$

Now, let us describe explicitly how to write $x$ in terms of a permutation in
a smaller symmetric group.

######  Lemma 5.1.3.

Let $x\in S_{n-1}$ with $F_{x}=\langle a^{t}\rangle$. Then there exists an
integer $j$ which is coprime to $\frac{n}{t}$ (and so therefore invertible
modulo $\frac{n}{t}$), a permutation $\sigma_{x}\in S_{t-1}$ and integers
$1\leq u_{i}\leq\frac{n}{t}$ such that

$\displaystyle x(i)$ $\displaystyle=u_{i}t+\sigma_{x}(i)\mod n\qquad\text{ for
all }1\leq i\leq t-1$ $\displaystyle x(qt)$ $\displaystyle=qjt\mod
n\qquad\text{ for all }1\leq q\leq\frac{n}{t}$ $\displaystyle x(qt+i)$
$\displaystyle=qx(t)+x(i)\mod n$

Furthermore,

$\displaystyle x^{-1}(i)$
$\displaystyle=-ju_{\sigma_{x}^{-1}(i)}t+\sigma_{x}^{-1}(i)\mod n\qquad\text{
for all }1\leq i\leq t-1$ $\displaystyle x^{-1}(t)$ $\displaystyle=j^{-1}t\mod
n$ $\displaystyle x^{-1}(qt+i)$ $\displaystyle=qx^{-1}(t)+x^{-1}(i)\mod n$

We call $\sigma_{x}$ a remainder permutation for $x$.

Proof. Let $x\in S_{n-1}$ with stabilizer $F_{x}=\langle a^{t}\rangle$. We
have already established in Corollary 4.2.3 that $x$ is determined by where it
sends $1,...,t$. We now claim that $x$ defines a permutation modulo $t$. From,
Lemma 4.2.1, there exists a $j$ coprime to $\frac{n}{t}$ (and so invertible
modulo $\frac{n}{t}$) with $x(t)=jt.$ Now, if $i,i^{\prime}<t$ and
$x(i)=x(i^{\prime})\mod t$, then $x(i)-x(i^{\prime})$ is a multiple of $t$,
say $x(i)-x(i^{\prime})=qt$ for some value $q$. Then

$x(j^{-1}qt+i^{\prime})=x(j^{-1}qt)+x(i^{\prime})=jj^{-1}qt+x(i^{\prime})=qt+x(i^{\prime})=x(i)-x(i^{\prime})+x(i^{\prime})=x(i).$

Therefore, because $x$ is a bijection, $j^{-1}qt+i^{\prime}=i\mod n$ and so
$i-i^{\prime}$ is a multiple of $t$. However, we assumed that both $i$ and
$i^{\prime}$ were positive integers smaller than $t$ so their difference is
only a multiple of $t$ if their difference is $0$. Thus, $x\mod t$ actually
defines a permutation in $S_{t-1}$.

Specifically, for each $x$, there exists an associated remainder permutation
$\sigma_{x}\in S_{t-1}$ defined by $\sigma_{x}(i):=x(i)\mod t$ for all $i<t$
and a value $j$ which is coprime to $\frac{n}{t}$ such that $x(t)=jt.$
Furthermore, $x(i)=tu_{i}+\sigma_{x}(i)$ for all $1\leq i\leq t-1$.

Now, if we let $x^{-1}(i)=tl_{i}+\sigma_{x^{-1}}(i)$, then we obtain

$\displaystyle i$ $\displaystyle=x(x^{-1}(i))=x(tl_{i}+\sigma_{x^{-1}}(i))$
$\displaystyle=l_{i}x(t)+x(\sigma_{x^{-1}}(i))$
$\displaystyle=l_{i}jt+tu_{\sigma_{x^{-1}}(i)}+\sigma_{x}(\sigma_{x^{-1}}(i))$

Now, note that $t$ divides $l_{i}jt+tu_{\sigma_{x^{-1}}(i)}$, however,
$\sigma_{x},\sigma_{x^{-1}}\in S_{t-1}$ and so they can only output values
strictly smaller than $t$. This gives us two relations, first,
$\sigma_{x}(\sigma_{x^{-1}}(i))=i$ for all $i$, and so
$\sigma_{x}^{-1}=\sigma_{x^{-1}}$ which is very desirable, but also
$l_{i}jt=-tu_{\sigma_{x}^{-1}(i)}$ for all $i$, which we can rewrite to obtain

$l_{i}=-j^{-1}u_{\sigma_{x}^{-1}(i)}\mod\frac{n}{t}\qquad\text{ for all
}i=1,...,t-1.$ ∎

######  Corollary 5.1.4.

Let $x\in S_{n-1}$ have stabilizer $F_{x}=\langle a^{t}\rangle$. Then there
exists integers $j,u_{i}$ for $1\leq i\leq t-1$ and remainder permutation
$\sigma_{x}\in S_{t-1}$ as from Lemma 5.1.3 such that

$\displaystyle x(i)$ $\displaystyle=u_{i}t+\sigma_{x}(i)\mod n\qquad\text{ for
all }1\leq i\leq t-1$ $\displaystyle x(qt)$ $\displaystyle=qjt\mod
n\qquad\text{ for all }1\leq q\leq\frac{n}{t}$ $\displaystyle x(qt+i)$
$\displaystyle=qx(t)+x(i)\mod n$

Then, if $x$ is an involution, $\sigma_{x}$ is as well and
$u_{i}=-ju_{\sigma_{x}(i)}$ for all $1\leq i\leq t-1$.

Conversely, every choice of $j$ coprime to $\frac{n}{t}$, involution
$\sigma\in S_{t-1}$, and integers $u_{i}$ satisfying $u_{i}=-ju_{\sigma(i)}$
will generate an involution $x\in S_{n-1}$ which is stabilized by $a^{t}$.

Proof. Lemma 5.1.3 gives that $\sigma_{x^{-1}}=\sigma_{x}^{-1}$ and so if
$x^{2}=1$ then $\sigma_{x}^{2}=1$.

Note because $x^{-1}=x$, $j^{-1}=j$, and $l_{i}=u_{i}$ for all $i$. Therefore,
$u_{i}=-j^{-1}u_{\sigma_{x}^{-1}(i)}=-ju_{\sigma_{x}(i)}\mod\frac{n}{t}$ for
all $i$. This says that for every fixed point of $\sigma_{x}$, we require that
$u_{i}(j+1)=0\mod\frac{n}{t}$.

The converse follows by direct computation. ∎

Note that knowing $\sigma_{x}$ is an involution does not guarantee that $x$ is
as well. For example, if $t=3$, then every $\sigma\in S_{2}$ has order $2$,
but certainly $x$ need not be.

It must be stressed here that the remainder permutation $\sigma_{x}$ plays a
vital role in understanding (and thus counting) involutions $x$ with certain
properties.

While we do not explore it in this work, we believe that irreps in $J_{n}$ are
in some way connected to and possible even extended from irreps in $J_{t}$
(for $t$ dividing $n$) and that remainder permutations are the key to
understanding this connection.

###  Counting Involutions with Certain Properties

Here we focus on counting formulas for involutions in $x\in S_{n-1}$ with
stabilizer $F_{x}=\langle a^{t}\rangle$ and certain additional properties.
Specifically, we will show how to count involutions with a given number of
fixed points and involutions whose orbits contain a given number of
involutions.

We will see that the ideas presented in this section and the next will be
crucial to constructing a counting formula for the number of irreps of odd
dimension with indicator $+1$.

Although they may seem very different, the ideas used to construct these
counting formulas build upon each other. That is, we must understand what
involutions with stabilizer $F_{x}=\langle a^{t}\rangle$ in order to
understand involutions with stabilizer $F_{x}=\langle a^{t}\rangle$ and $r$
fixed points, and we must understand involutions with stabilizer
$F_{x}=\langle a^{t}\rangle$ and $r$ fixed points in order to understand
involutions with stabilizer $F_{x}=\langle a^{t}\rangle$ and $r$ fixed points
smaller than $t.$

######  Table 1 (Big Table of Commonly Cited Values and Sets).

Number | Set or Value
---|---
(1) | $E_{n/t}=\displaystyle\left\\{1\leq j\leq\frac{n}{t}\,\big{|}\,j^{2}=1\mod\frac{n}{t}\right\\}$
(2) | $K_{j,n/t}=\displaystyle\left\\{1\leq u\leq\frac{n}{t}\,\big{|}\,u(j+1)=0\mod\frac{n}{t}\right\\}$
(3) | $\alpha_{j,n/t}=\displaystyle\text{gcd}\left(j+1,\frac{n}{t}\right)=|K_{j,n/t}|$
(4) | $P_{j,n/t}=\displaystyle\left\\{q(j-1)\mod\frac{n}{t}\,\big{|}\,1\leq q\leq\frac{n}{t}\right\\}$
(5) | $P_{j,n/t}^{(c)}=\displaystyle\left\\{1\leq u\leq\frac{n}{t}\,\big{|}\,u\not\in P_{j,n/t}\right\\}\bigcap K_{j,n/t}$
(6) | $\delta^{P^{c}}=\displaystyle\begin{cases}|P_{j,n/t}^{(c)}|&\text{ if }P_{j,n/t}^{(c)}\not=\varnothing\\\ 1&\text{ otherwise }\end{cases}$
(7) |  $\overline{K}_{j^{\prime},j_{\sigma_{x}},n,t,s}=$$\displaystyle\left\\{1\leq u\leq\frac{n}{t}\,\big{|}\,\left(u\frac{t}{s}+m_{i}\right)(1+j^{\prime})=0\mod\frac{n}{s}\text{ for some }m_{i}\in P_{j_{\sigma_{x}},t/s}\right\\}$
(8) |  $\overline{K}_{j^{\prime},j_{\sigma_{x}},n,t,s}^{(c)}=$$\displaystyle\left\\{1\leq u\leq\frac{n}{t}\,|\,\left(u\frac{t}{s}+m_{i}\right)(1+j^{\prime})=0\mod\frac{n}{s}\text{ for some }m_{i}\in P_{j_{\sigma_{x}},t/s}^{(c)}\right\\}$
(9) | $\delta^{K^{c}}=\displaystyle\begin{cases}|\overline{K}_{j^{\prime},j_{\sigma_{x}},n,t,s}^{(c)}|&\text{ if }\overline{K}_{j^{\prime},j_{\sigma_{x}},n,t,s}^{(c)}\not=\varnothing\\\ 1&\text{otherwise}\end{cases}$
(10) | $\overline{E}_{j_{\sigma_{x}},n,t,s}=\displaystyle\left\\{j_{\sigma_{x}}+\overline{m}\frac{t}{s}\,\big{|}\,1\leq\overline{m}\leq\frac{n}{t}\right\\}\bigcap E_{n/s}$
(11) | $\beta_{j,n/t}=\displaystyle\text{gcd}\left(j-1,\frac{n}{t}\right)$
(12) | $\delta_{j,n/t,r}=\displaystyle\begin{cases}1&\text{ if }\beta_{j,n/t}|r\text{ and }r\leq t\beta_{j,n/t}\\\ 0&\text{ otherwise}\end{cases}$
(13) | $m_{r,j,n/t}=\displaystyle\frac{r}{\beta_{j,n/t}}$
(14) | $K_{j,n/2}^{\prime}=\left\\{u\,|(u+1)(j+1)=0\mod\frac{n}{2}\,\right\\}$
(15) | $\overline{\delta}_{u,j,n/2}=\begin{cases}1&\text{ if }j\not=2u+1\mod\frac{n}{2}\\\ 0&\text{ otherwise}\end{cases}$
(16) | $\delta_{i,u,n/2}^{0}=\begin{cases}1&\text{ if }-i(u+1)=0\mod\frac{n}{2}\\\ 0&\text{ otherwise}\end{cases}$
(17) | $\delta_{i,u,n/2}^{\not=0}=\begin{cases}1&\text{ if }-i(u+1)\not=0\mod\frac{n}{2}\\\ 0&\text{ otherwise}\end{cases}$

Note: A table listing the majority of sets and values which are used heavily
in the statements in this section.

######  Proposition 5.2.1.

Let $T_{n/t}$ be the set of involutions $x\in S_{n-1}$ such that
$F_{x}=\langle a^{t}\rangle$. Let $E_{n/t}$ be as from (1) and
$\alpha_{j,n/t}$ be as from (3) in 1 ( (Big Table of Commonly Cited Values and
Sets).). Then

$\displaystyle|T_{n/1}|$ $\displaystyle=|E_{n/1}|$ (6)
$\displaystyle|T_{n/t}|$ $\displaystyle=\sum_{j\in
E_{n/t}}\sum_{l=0}^{\left\lfloor\frac{t-1}{2}\right\rfloor}(\alpha_{j,n/t})^{t-1-2l}\left(\frac{n}{t}\right)^{l}\frac{(t-1)!}{(t-1-2l)!2^{l}l!}-\sum_{\begin{subarray}{c}s|t\\\
s\not=t\end{subarray}}|T_{n/s}|$ (7) $\displaystyle|T_{n/n}|$
$\displaystyle=\sum_{l=0}^{\frac{n-1}{2}}\frac{(n-1)!}{(n-1-2l)!2^{l}l!}-\sum_{\begin{subarray}{c}s|n\\\
s\not=n\end{subarray}}|T_{n/s}|$ (8)

Proof. Let $x\in S_{n-1}$ be an involution with stabilizer $F_{x}=\langle
a^{t}\rangle$ and let

$E_{n/t}:=\left\\{1\leq j\leq n\,|\,j^{2}=1\mod\frac{n}{t}\right\\}.$

From Corollary 5.1.4, let $\sigma_{x}\in S_{t-1}$, write
$x(i)=tu_{i}+\sigma_{x}(i)$ for all $1\leq i\leq t-1$ and $x(t)=jt$ for some
$j\in E_{n/t}$ which determines $x$ completely.

Therefore, the number of $x$ satisfying $x^{2}=1$ can be counted by fixing a
value $j\in E_{n/t}$, then for each involution $\sigma\in S_{t-1}$, we count
the number of fixed points of $\sigma$ which requires counting the number of
transpositions present in the disjoint cycle decomposition of $\sigma.$

Note, there are $\frac{(t-1)!}{(t-1-2l)!2^{l}l!}$ permutations $\sigma\in
S_{t-1}$ with $\sigma^{2}=1$ and $\sigma$ being a product of $l$ disjoint
cycles. This is because we choose any two integers between $1$ and $t-1$ to
put in a transposition (so $\binom{t-1}{2}=\frac{(t-1)(t-2)}{2}$, then we
choose any two of the remaining $t-3$ integers to pair up in a second
transposition so $\frac{(t-3)(t-4)}{2}$), and we repeat this $l$ times which
gives

$\frac{(t-1)(t-2)(t-3)\cdots(t-1-2l)}{2^{l}l!}$

where we must divide by $l!$ because the order in which we obtain each
transposition does not matter, that is,
$(1\,\,\,2)(3\,\,\,4)=(3\,\,\,4)(1\,\,\,2)$ and there are $l!$ ways to re-
order the product of transpositions.

Now, if $x$ has remainder permutation $\sigma$, then from Lemma 5.1.3, we know
that $u_{i}=-ju_{\sigma(i)}\mod\frac{n}{t}$ and so each fixed point of
$\sigma$ (of which there are $t-1-2l$) forces $u_{i}(j+1)=0\mod\frac{n}{t}$
which contributes gcd$(j+1,\frac{n}{t})$ possible choices for $u_{i}$ where
$\sigma(i)=i$ (see last paragraph of the proof of Lemma 5.1.3). Furthermore,
each transposition $(i\,\,\,i^{\prime})$ in $\sigma$ which pairs $u_{i}$ with
$u_{i^{\prime}}$ and so contributes $\frac{n}{t}$ choices for $u_{i}$ (which
then determines $u_{i^{\prime}}$).

Let $\alpha_{j,n/t}:=\text{gcd}\left(j+1,\frac{n}{t}\right)$. Therefore, for a
fixed $j$, and fixed $\sigma$ comprised of $l$ disjoint transpositions, there
are $(\alpha_{j,n/t})^{t-1-2l}\left(\frac{n}{t}\right)^{l}$ choices for the
$u_{i}$ and so if we multiply this by the number of such $\sigma$, which is
$\frac{(t-1)!}{(t-1-2l)!2^{l}l!}$ and then sum over all $j\in E_{n/t}t$, and
subtract those we have already counted, we obtain our result

$|T_{n/t}|=\underbrace{\sum_{j\in E_{n/t}}}_{\text{choices for
}x(t)}\underbrace{\sum_{l=0}^{\left\lfloor\frac{t-1}{2}\right\rfloor}}_{\begin{subarray}{c}\\#\text{
of fixed}\\\ \text{points of
}\sigma_{x}\end{subarray}}\underbrace{(\alpha_{j,n/t})^{t-1-2l}}_{\begin{subarray}{c}\text{choices
for }x(i)\\\ \text{ when
}\sigma_{x}(i)=i\end{subarray}}\underbrace{\left(\frac{n}{t}\right)^{l}}_{\begin{subarray}{c}\text{choices
for }x(i)\\\ \text{ when
}\sigma_{x}(i)\not=i\end{subarray}}\underbrace{\frac{(t-1)!}{(t-1-2l)!2^{l}l!}}_{\text{
choice for }\sigma_{x}}-\underbrace{\sum_{\begin{subarray}{c}s|t\\\
s\not=t\end{subarray}}|T_{n/s}|}_{\text{remove over count}}$

In the case where $t=1$, then $l=0$ and all powers are $0$ so we are simply
summing $1$ for every $j\in E_{n/1}$.

Of course, if $t=n$, all sets trivialize and so we simply count all
involutions in $S_{n-1}$ and then remove those we have already counted. ∎

Again, as with Proposition 5.1.1, explicitly writing out the set $T_{n/t}$ is
done recursively by using the formula given.

Note that we can now verify the statements made at the end of the discussion
under Theorem 4.3.2 that if $n=p^{r}$ for $p$ and odd prime and
$r\in\mathbb{N}$, then there exists only one nontrivial involution $x\in
S_{n-1}$ with $F_{x}=F$. Since $E_{p^{r}/1}=\\{1,-1\\}$, we have only two
permutations in

$T_{p^{r}/1}=\\{(1),(1\,\,\,p^{r}-1)(2\,\,\,p^{r}-2)\cdots(\frac{p^{r}-1}{2}\,\,\,\frac{p^{r}+1}{2})\\}.$

######  Lemma 5.2.2.

Let $x\in S_{n-1}$ be an involution with $F_{x}=\langle a^{t}\rangle$. Let
$j\in E_{n/t}$ as from 1 ( (Big Table of Commonly Cited Values and Sets).) and
write $x(t)=jt$ and for all $1\leq i<t$, and $x(i)=u_{i}t+\sigma_{x}(i)$ for
some involution $\sigma_{x}\in S_{t-1}$ and constants $1\leq
u_{i}\leq\frac{n}{t}$. Let

$\beta_{j,n/t}=\text{gcd}\left(j-1,\frac{n}{t}\right)=\text{gcd}\left(1-j,\frac{n}{t}\right).$

Then, $x(qt+i)=qt+i$ for $\beta_{j,n/t}$ different values $q$ if and only if
$\sigma_{x}(i)=i$ and $u_{i}=q(1-j)\mod\frac{n}{t}$ for some $q$.

Proof.

  1. $\implies$

Let $x(qt+i)=qt+i$ be a fixed point of $x$. Then this says that

$x(qt+i)=qx(t)+x(i)=qjt+u_{i}t+\sigma_{x}(i)=qt+i.$

Because $\sigma_{x}(i)<t$ and $i<t$, this forces $\sigma_{x}(i)=i$. Thus,
subtracting we obtain that

$u_{i}t=qt-qjt\implies u_{i}=q(1-j)\mod\frac{n}{t}.$

  2. $\impliedby$

Let $\sigma_{x}(i)=i$ and $u_{i}=q(1-j)\mod\frac{n}{t}$ for some $q$. Now,
since there exists a $q$ for which the equation $u_{i}=q(1-j)\mod\frac{n}{t}$
has a solution, there exists a total of $\beta_{j,n/t}$ solutions and so there
are $\beta_{j,n/t}$ possible values $q$ for which

$x(qt+i)=qjt+u_{i}t+\sigma_{x}(i)=qjt+q(1-j)t+i=qt+i.$ ∎

######  Proposition 5.2.3.

Let $R_{n/t,r}$ be the set of involutions $x\in S_{n-1}$ such that
$F_{x}=\langle a^{t}\rangle$ and $x$ has exactly $r$ fixed points (including
$x(n)=n$ as one of the fixed points). Let $E_{n/t}$, $\beta_{j,n/t}$,
$m_{r,j,n/t}$, $\delta_{j,n/t,r}$, $P_{j,n/t}$, $\delta^{P^{c}}$ as from 1 (
(Big Table of Commonly Cited Values and Sets).). Let $\delta_{r}$ be the
Kronecker-Delta step function. Then

$\displaystyle|R_{n/1,r}|$ $\displaystyle=\sum_{j\in
E_{n/1}}\delta_{r}(\beta_{j,n/1})$ (9) $\displaystyle|R_{n/t,r}|$
$\displaystyle=\sum_{j\in
E_{n/t}}\sum_{l=0}^{\lfloor\frac{t-m_{r,j,n/t}}{2}\rfloor}\delta_{j,n/t,r}\delta_{j,n/t,r}^{l}\binom{t-1-2l}{m_{r,j,n/t}-1}|P_{j,n/t}|^{m_{r,j,n/t}-1}(\delta^{P^{c}})^{t-2l-m_{r,j,n/t}}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\left(\frac{n}{t}\right)^{l}\frac{(t-1)!}{(t-1-2l)!2^{l}l!}-\sum_{\begin{subarray}{c}s|t\\\
s\not=t\end{subarray}}|R_{n/s,r}|$ (10) $\displaystyle|R_{n/n,r}|$
$\displaystyle=\frac{(n-1)!}{(r-1)!2^{\frac{n-r}{2}}\left(\frac{n-r}{2}\right)!}-\sum_{\begin{subarray}{c}s|n\\\
s\not=n\end{subarray}}|R_{n/s,r}|$ (11) $\displaystyle|R_{n/t,r}|$
$\displaystyle=0\text{ if }n-r\text{ is not even}.$

Proof. Let $x\in S_{n-1}$ be an involution with stabilizer $F_{x}=\langle
a^{t}\rangle$. For some $\sigma_{x}\in S_{t-1}$, write
$x(i)=tu_{i}+\sigma_{x}(i)$ for all $1\leq i\leq t-1$ and $x(t)=jt$ for some
$j\in E_{n/t}$ which determines $x$ completely as from Corollary 5.1.4.

Then, assume $x$ has $r$ fixed points (including $x(n)=n$ as a fixed point).

Since $x(qt)=jqt=qt$ if and only if $q(j-1)=0\mod\frac{n}{t}$, we immediately
get $\beta_{j,n/t}:=$gcd$(j-1,\frac{n}{t})$ fixed points for $x$ from the
given choice of $j.$

Additionally, from Lemma 5.2.2, we know that for a fixed $1\leq i<t$,
$x(qt+i)=qt+i$ for $\beta_{j,n/t}$ different values of $q$ if and only if
$\sigma_{x}(i)=i$ and $u_{i}=q(1-j)\mod\frac{n}{t}$.

Thus, the total number of fixed points of $x$ is a multiple of $\beta_{j,n/t}$
and so $\beta_{j,n/t}$ must divide $r$. To ensure the existence of an $x$ with
the given choice of $j$ and $r$, the value $\delta_{j,n/t,r}$ must be included
in our sum where

$\delta_{j,n/t,r}:=\begin{cases}1&\text{ if }\beta_{j,n/t}|r\text{ and }r\leq
t\beta_{j,n/t}\\\ 0&\text{ otherwise }\end{cases}.$

Furthermore, letting $m_{r,j,n/t}:=\frac{r}{\beta_{j,n/t}}$, the ratio minus
one, $m_{r,j,n/t}-1$ tells us how many $u_{i}$ for $1\leq i<t$ must be of the
form $q(1-j)$ for some $q$. Let

$P_{j,n/t}:=\left\\{q(1-j)\mod\frac{n}{t}\,|\,1\leq
q\leq\frac{n}{t}\right\\}.$

Then $m_{r,j,n/t}-1$ of the $u_{i}$ come from $P_{j,n/t}$, and
$t-1-(m_{r,j,n/t}-1)=t-m_{r,j,n/t}$ of the remaining $u_{i}$ come from its
compliment

$P_{j,n/t}^{c}:=\left\\{u\,|\,u\not=q(1-j)\mod\frac{n}{t}\text{ for all
}q\right\\}.$

However, recall that if $\sigma_{x}(i)=i$, then $u_{i}(j+1)=0\mod\frac{n}{t}$
is required by Corollary 5.1.4. Of course, if $u_{i}\in P_{j,n/t}$, then
$u_{i}=q(j-1)$ for some $q$ and so certainly
$u_{i}(j+1)=q(j-1)(j+1)=q(j^{2}-1)=0$ since $j^{2}=1$. However, to guarantee
this for $u_{i}\in P_{j,n/t}^{c}$, we must actually require that $u_{i}\in
P_{j,n/t}^{(c)}:=P_{j,n/t}^{c}\cap K_{j,n/t}$ where

$K_{j,n/t}:=\left\\{1\leq
u\leq\frac{n}{t}\,|\,u(j+1)=0\mod\frac{n}{t}\right\\}.$

In summary:

  1. (1)

If $\sigma_{x}(i)=i$, then either:

    1. (i)

$u_{i}\in P_{j,n/t}$ and so contributes $\beta_{j,n/t}$ fixed points to $x$

    2. (ii)

or $u_{i}\in P_{j,n/t}^{(c)}$ and so contributes $0$ fixed points to $x$

  2. (2)

If $\sigma_{x}(i)\not=i$ then $x(qt+i)\not=qt+i$ for any $q$ and so we can
allow $u_{i}$ to take any value between $1$ and $\frac{n}{t}$ and this will
not add any fixed points to $x$.

Let $t-1-2l$ be the number of fixed points of $\sigma_{x}$ for some $l$. Then,
since $m_{r,j,n/t}-1$ of the $u_{i}$ come from $P_{j,n/t}$, $\sigma_{x}$ must
have at least $m_{r,j,n/t}-1$ fixed points which means $t-1-2l\geq
m_{r,j,n/t}-1$ and so $t-m_{r,j,n/t}\geq 2l$ so finally
$l\leq\frac{t-m_{r,j,n/t}}{2}$.

Additionally, for the remaining $t-1-2l-(m_{r,j,n/t}-1)=t-2l-m_{r,j,n/t}$
fixed points of $\sigma_{x}$, $u_{i}$ must be taken from $P_{j,n/t}^{(c)}$,
however, there is an edge case we must address here.

Assume $P_{j,n/t}^{(c)}=\varnothing$. Then either $P_{j,n/t}^{c}=\varnothing$
(which is possible) or $K_{j,n/t}\subset P_{j,n/t}$. In the first case, we get
$P_{j,n/t}=\\{1,2,...,n/t\\}$ and since $P_{j,n/t}\subset K_{j,n/t}$, we get
$K_{j,n/t}=P_{j,n/t}$. In the second case, $K_{j,n/t}\subset P_{j,n/t}$ and so
again $K_{j,n/t}=P_{j,n/t}$. In either case, if $i$ is a fixed point of
$\sigma_{x}$, and so $\sigma_{x}(i)=i$, then because $u_{i}\in
K_{j,n/t}=P_{j,n/t}$ by Corollary 5.1.4, $x$ has $\beta_{j,n/t}$ fixed points
of the form $qt+i$ for some $q$.

Namely, if $P_{j,n/t}^{(c)}=\varnothing$, every fixed point of $\sigma_{x}$
contributes exactly $\beta_{j,n/t}$ fixed points to $x$ and so the number of
fixed points of $\sigma_{x}$ is equal to $m_{j,n/t}-1$. That is
$t-1-2l=m_{j,n/t}-1$ and so $t-m_{j,n/t}=2l$.

All this is to say that if $P_{j,n/t}^{(c)}=\varnothing$, then we must have
$t-m_{j,n/t}=2l$ so we introduce

$\delta_{j,n/t,r}^{l}=\begin{cases}0&\text{ if
}|P_{j,n/t}^{(c)}|=\varnothing\text{ and }2l\not=t-m_{j,n/t}\\\
1&\text{otherwise}\end{cases}$

However, this condition is messy to write down as we do not want to be stuck
with a $0^{0}$ indeterminate in our sum so to avoid this, the set
$P_{j,n/t}^{(c)}$ is never used. Instead, we include the messy conditional

$\delta^{P^{c}}:=\begin{cases}|P_{j,n/t}^{(c)}|&\text{ if
}P_{j,n/t}^{(c)}\not=\varnothing\\\ 1&\text{ otherwise}\end{cases}.$

Now, we are at last ready to fully describe all involutions with stabilizer
$F_{x}=\langle a^{t}\rangle$ with and $r$ fixed points.

In the base case, if $t=1$, then $x(i)=ij=i$ if and only if $i(j-1)=0\mod n$.
So $x$ has $r$ fixed points if and only if the number of solutions to this
equation is $r$. Since $i=n$ is always a solutions, the number of solutions to
this equation is $\beta_{j,n/1}=$gcd$(j-1,n)$. Let $\delta_{r}$ be the step
function so

$\delta_{r}(u)=\begin{cases}1&\text{ if }r=u\\\ 0&\text{
otherwise}\end{cases}$

Then

$|R_{n/1,r}|=\sum_{j\in E_{n/1}}\delta_{r}(\beta_{j,n/1}).$

Finally,

$|R_{n/t,r}|=\underbrace{\sum_{j\in E_{n/t}}}_{\text{choices for
}x(t)}\underbrace{\sum_{l=0}^{\lfloor\frac{t-m_{r,j,n/t}}{2}\rfloor}}_{\begin{subarray}{c}\\#\text{
of fixed}\\\ \text{points of }\sigma_{x}\\\
\end{subarray}}\underbrace{\delta_{j,n/t,r}}_{x\text{
exists}}\underbrace{\delta_{j,n/t,r}^{l}}_{x\text{ matches
}\sigma_{x}}\underbrace{\binom{t-1-2l}{m_{r,j,n/t}-1}}_{\begin{subarray}{c}\text{
choose }u_{i}\\\ \text{ in
}P_{j,n/t}\end{subarray}}\,\,\,\underbrace{|P_{j,n/t}|^{m_{r,j,n/t}-1}}_{\begin{subarray}{c}\sigma_{x}(i)=i\text{
and }\\\
u_{i}=q(1-j)\end{subarray}}\underbrace{(\delta^{P^{c}})^{t-2l-m_{r,j,n/t}}}_{\begin{subarray}{c}\sigma_{x}(i)=i\text{
and }\\\ u_{i}\not=q(1-j)\end{subarray}}$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\underbrace{\left(\frac{n}{t}\right)^{l}}_{\sigma_{x}(i)\not=i}\underbrace{\frac{(t-1)!}{(t-1-2l)!2^{l}l!}}_{\text{
choices for }\sigma_{x}}-\underbrace{\sum_{\begin{subarray}{c}s|t\\\
s\not=t\end{subarray}}R_{n/s,r}}_{\text{ remove over count}}$

Note that this agrees with our base case since if $t=1$ we know that $1\geq
m_{r,j,n/t}$ and that $m_{r,j,n/t}$ is an integer so $m_{r,j,n/t}=1$. Thus,
necessarily $r=\beta_{j,n/t}$ (or of course no such $x$ exists) and so $l=0$.
So the sum becomes

$\sum_{j\in E_{n/1}}\delta_{j,n/t,r}=\sum_{j\in
E_{n/1}}\delta_{r}(\beta_{j,n/t})$

as desired.

Furthermore, If $t=n$, all sets trivialize and so we simply count all
involutions with $r$ fixed points in $S_{n-1}$ and then remove those we have
already counted.

Note also that if $n-r$ is not even, then $x$ cannot be an involution in
$S_{n-1}$ with $r$ fixed points, and so we say that $|R_{n/t,r}|=0$ if $n-r$
is not even. ∎

######  Lemma 5.2.4.

Either $P_{j,n/t}=K_{j,n/t}$ and $\alpha_{j,n/t}\beta_{j,n/t}=\frac{n}{t}$ or
$|P_{j,n/t}|=\frac{1}{2}|K_{j,n/t}|=|P_{j,n/t}^{(c)}|$ and
$\alpha_{j,n/t}\beta_{j,n/t}=2\frac{n}{t}$

Proof. Recall that $\beta_{j,n/t}:=\text{gcd}\left(1-j,\frac{n}{t}\right)$ and
$P_{j,n/t}:=\\{q(1-j)\mod\frac{n}{t}\\}$. We claim
$|P_{j,n/t}|=\frac{n}{t\beta_{j,n/t}}$. This is because
$q(1-j)=q^{\prime}(1-j)$ if and only if $(q-q^{\prime})(1-j)=0\mod\frac{n}{t}$
if and only if $q=q^{\prime}\mod\frac{n}{t\beta_{j,n/t}}$ which proves the
claim.

Recall now that $P_{j,n/t}\subset K_{J,n/t}\subset\\{1,2,...,\frac{n}{t}\\}$
is always true and so the results follow by checking the cases where
$P_{j,n/t}^{(c)}=\varnothing$ and $P_{j,n/t}^{(c)}\not=\varnothing$ separately
and using basic properties of the gcd. ∎

######  Remark 5.2.5.

By definition $\sum_{r=1}^{n}|R_{n/t,r}|=|T_{n/t}|$ must be true, however,
using some careful computation it can be shown that both equations from
Proposition 5.2.1 and Proposition 5.2.3 satisfy this relation.

###  Prerequisites for Counting Indicators

Now, lets say we want to count the number of $x\in S_{n-1}$ which have
indicators of a certain value. Note that if $x^{2}=1$, then
$x=x^{-1}\in\mathcal{O}_{x}$ and so by Proposition 4.3.5, all indicators are
nonnegative. Now, what determines whether the indicator of $x$ is zero or not?
Well, $\nu(\hat{V})=1$ if $\zeta_{n}^{i\frac{x(t)+t}{t}}=1$ and $0$ otherwise.

Therefore, knowing the value of $x(t)$ tells us exactly which (and so how
many) $i$ exist for which $\nu(\hat{V})\not=0$. Now, assume that $t$ is odd.
Then $\mathcal{O}_{x}$ is closed under inversion if and only if it contains an
element $y$ of order $2$. Since $\mathcal{O}_{y}=\mathcal{O}_{x}$, we may
relabel so as to only consider orbits $\mathcal{O}_{x}$ where $x$ is an
involution.

And so, in the case where $t$ is odd, the strategy is this:

  1. (1)

$x^{-1}\in\mathcal{O}_{x}$ if and only if $\mathcal{O}_{x}$ contains an
element of order $2$.

  2. (2)

The number of involutions (excluding $x$ itself) in $\mathcal{O}_{x}$ is given
by the number of fixed points of $\sigma_{x}\in S_{t-1}$ which generates $x$
(by Lemma LABEL:*lem:y in orbit x (2) ).

  3. (3)

The number of representations of $J_{n}$ induced form $x$ which have positive
indicator is given by the number of values $i$ for which
$\zeta_{n}^{i\frac{x(t)+t}{t}}=1$ (by Theorem 4.3.2).

Namely, if we want to count the nonzero indictors of irreps of $J_{n}$ of a
fixed odd dimension $t$, we must be able to count the number of orbits which
contain exactly $r$ involutions.

######  Proposition 5.3.1.

Let $X_{n/t,r}$ be the set of involutions $x\in S_{n-1}$ with $F_{x}=\langle
a^{t}\rangle$, and such that $\mathcal{O}_{x}$ has $r\not=0$ involutions. Let
$T_{n/1}$ be as from Proposition 5.2.1, $R_{n/t,r}$ as from Proposition 5.2.3,
and we pull the entries (1), (2), (4), (6), (7), (9), (10), (11), (12), (13)
from 1 ( (Big Table of Commonly Cited Values and Sets).). Then

$\displaystyle|X_{n/1,r}|$ $\displaystyle=|T_{n/1}|=|E_{n/1}|$ (12)
$\displaystyle|X_{n/t,r}|$ $\displaystyle=\sum_{j\in
E_{n/t}}(\alpha_{j,n/t})^{r-1}\left(\frac{n}{t}\right)^{\frac{t-r}{2}}\frac{(t-1)!}{(r-1)!2^{\frac{t-r}{2}}\left(\frac{t-r}{2}\right)!}-\sum_{\begin{subarray}{c}s|t\\\
s\not=t\end{subarray}}|C_{n,t,s,r}|$ (13) $\displaystyle|X_{n/n,r}|$
$\displaystyle=\frac{(n-1)!}{(r-1)!2^{\frac{n-r}{2}}\left(\frac{n-r}{2}\right)!}-\sum_{\begin{subarray}{c}s|n\\\
s\not=n\end{subarray}}|C_{n,n,s,r}|$ (14) $\displaystyle|X_{n/t,r}|$
$\displaystyle=0\text{ if }t-r\text{ is negative or odd}$

where

$\displaystyle|C_{n,t,1,r}|$ $\displaystyle=\sum_{j_{\sigma_{x}}\in
E_{t/1}}\delta_{r}(\beta_{j_{\sigma_{x}},t/1})|\overline{E}_{j_{\sigma_{x}},n,t,1}|$
(15) $\displaystyle|C_{n,t,s,r}|$ $\displaystyle=\sum_{j_{\sigma_{x}}\in
E_{t/s}}\sum_{j^{\prime}\in\overline{E}_{j_{\sigma_{x}},n,t,s}}\sum_{l=0}^{\lfloor\frac{s-m_{r,j_{\sigma_{x}},t/s}}{2}\rfloor}\delta_{j_{\sigma_{x}},t/s,r}\delta_{j_{\sigma_{x}},t/s,r}^{l}\binom{s-1-2l}{m_{r,j_{\sigma_{x}},t/s}-1}|P_{j_{\sigma_{x}},t/s}|^{m_{r,j_{\sigma_{x}},t/s}-1}$
$\displaystyle\qquad(\delta^{P^{c}})^{s-2l-m_{r,j_{\sigma_{x}},t/s}}\frac{(s-1)!}{(s-1-2l)!2^{l}l!}|\overline{K}_{j^{\prime},j_{\sigma_{x}},n,t,s}|^{m_{r,j_{\sigma_{x}},t/s}-1}(\delta^{K^{c}})^{s-2l-m_{r,j_{\sigma_{x}},t/s}}\left(\frac{n}{s}\right)^{l}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad-\sum_{\begin{subarray}{c}p|s\\\
p\not=d\end{subarray}}|C_{n,t,p,r}|$ (16) $\displaystyle|C_{n,n,s,r}|$
$\displaystyle=|R_{n/s,r}|$ (17)

Proof. Once again, we count. Assume $x\in S_{n-1}$ is an involution with
$F_{x}=\langle a^{t}\rangle$ such that $\mathcal{O}_{x}$ contains exactly $r$
involutions. Our goal is to fully describe the conditions necessary to
construct $x$ and then count the possible ways to achieve those conditions.

We know that there exists integers $u_{i}$ and permutation $\sigma_{x}\in
S_{t-1}$ such that for all $1\leq i\leq t-1$, $x(i)=tu_{i}+\sigma_{x}(i)$.
Furthermore, we know that $\sigma_{x}$ has exactly $r$ fixed points (including
$\sigma_{x}(t)=t$). This is because Lemma LABEL:*lem:y in orbit x (2) tells us
that if we write $\mathcal{O}_{x}=\\{x=y_{t},y_{1},y_{2},...,y_{t-1}\\}$, then
$y_{i}\in\mathcal{O}_{x}$ has order $2$ if and only if $i=x(i)\mod
t=\sigma_{x}(i)$. Since $\mathcal{O}_{x}$ is assume to have $r$ involutions
(including $x$), $\sigma_{x}$ must then have $r$ fixed points.

Now, we can adapt the formulas from Proposition 5.2.1 here, however, there
will be a few key differences. Let $j\in E_{n/t}$ be fixed. Then recall that
the number of involutions $\sigma\in S_{t-1}$ which are a product of $l$
disjoint cycles is given by $\frac{(t-1)!}{(t-1-2l)!2^{l}l!}$. In this case,
$\sigma$ has $t-1-2l$ fixed points. In our case we want $t-1-2l=r-1$ (we now
exclude $\sigma_{x}(t)$ as a fixed point since we are viewing $\sigma_{x}$ as
strictly a permutation in $S_{t-1}$) and so we obtain that $l=\frac{t-r}{2}$.
Note that $t-r$ must be even (since this is the number of permutations
$y\in\mathcal{O}_{x}$ which are not involutions) this value is an integer.
Finally, we obtain that there are
$\frac{(t-1)!}{(r-1)!2^{\frac{t-r}{2}}\left(\frac{t-r}{2}\right)!}$
involutions $\sigma\in S_{t-1}$ with $r-1$ fixed points.

Thus, there should be

$\sum_{j\in
E_{n/t}}(\alpha_{j,n/t})^{r-1}\left(\frac{n}{t}\right)^{\frac{t-r}{2}}\frac{(t-1)!}{(r-1)!2^{\frac{t-r}{2}}\left(\frac{t-r}{2}\right)!}$

involutions $x\in S_{n-1}$ with stabilizer $\langle a^{t}\rangle$ which have
$r$ fixed points mod $t$. This should seem familiar as it is the same formula
as from Proposition 5.2.1 with $l$ fixed.

However, we have–unfortunately–over counted. What this value really gives us,
is the number of involutions $x\in S_{n-1}$ which have $r$ fixed points mod
$t$ and which are stabilized by $a^{t}$. However, their actual stabilizer set
could be larger than $\langle a^{t}\rangle.$

To compensate for this, we must remove those permutations which have smaller
stabilizer. However, we cannot just do this recursively. Note that we are
counting the number of fixed points of $x$ modulo $t$, which we will not do if
we simply replace every $t$ with an $s$. Rather, we must work harder here.
Certainly, if $x$ is actually stabilized by $a^{s}$ for some $s|t$, then there
exists a $\tau_{x}\in S_{s-1}$ such that $x(i)=l_{i}s+\tau_{x}(i)$ for all
$i<s$. This is to say that we have two equivalent ways of writing $x(i)$.
Namely, for all $i<s$ we have

$\displaystyle x(i)$ $\displaystyle=tu_{i}+\sigma_{x}(i)$ $\displaystyle x(i)$
$\displaystyle=sl_{i}+\tau_{x}(i).$

Namely, $\sigma_{x}(i)=sl_{i}-tu_{i}+\tau_{x}(i)\mod n$ and so
$\sigma_{x}(i)=\tau_{x}(i)\mod s$ for all $i<s$ and $\sigma_{x}(s)=s\mod s$.
Note that this actually tells us that $\sigma_{x}$ is generated from
$\tau_{x}$ in the same way that $x$ is generated from $\sigma_{x}$. That is,
$\sigma_{x}$ is stabilized by $a_{t}^{s}$ where
$a_{t}=(1\,\,\,2\,\,\,\cdots\,\,\,t)$ is the standard $t$-cycle and so there
exists integers $1\leq m_{i}\leq\frac{t}{s}$ such that
$\sigma_{x}(i)=m_{i}s+\tau_{x}(i)$. We can now write (for all $i<s)$

$\displaystyle x(i)$ $\displaystyle=tu_{i}+\sigma_{x}(i)\qquad 1\leq
u_{i}\leq\frac{n}{t},\text{ and }\sigma_{x}\in S_{t-1}$ $\displaystyle x(i)$
$\displaystyle=sl_{i}+\tau_{x}(i)\qquad 1\leq l_{i}\leq\frac{n}{s},\text{ and
}\tau_{x}\in S_{s-1}$ $\displaystyle\sigma_{x}(i)$
$\displaystyle=sm_{i}+\tau_{x}(i)\qquad 1\leq m_{i}\leq\frac{t}{s}.$

Note, our goal is to write all possible $l_{i}$. Furthermore, there exists
$j_{\sigma_{x}}\in E_{t/s}$ such that $\sigma_{x}(s)=j_{\sigma_{x}}s$. Now,
using again that $\sigma_{x}(i)=l_{i}s-u_{i}t+\tau_{x}(i)$, we have that

$l_{i}s=m_{i}s+u_{i}t\implies l_{i}=u_{i}\frac{t}{s}+m_{i}\mod\frac{n}{s}.$

Note that
$x(s)=tu_{s}+\sigma_{x}(s)=tu_{s}+j_{\sigma_{x}}s=s(\frac{t}{s}u_{s}+j_{\sigma_{x}})$
and so if we hope to have $x$ be an involution stabilized by $a^{s}$ we must
have $\frac{t}{s}u_{i}+j_{\sigma_{x}}\mod\frac{n}{s}\in E_{n/s}$. Let

$\overline{E}_{j_{\sigma_{x}},n,t,s}:=\left\\{j_{\sigma_{x}}+\overline{m}\frac{t}{s}\,|\,1\leq\overline{m}\leq\frac{n}{t}\right\\}\bigcap
E_{n/s}.$

Note here that if $x(s)=j^{\prime}s$ then
$x(t)=\frac{t}{s}x(s)=\frac{t}{s}j^{\prime}s=tj^{\prime}=tj\mod n$.

Now, after choosing a $j^{\prime}\in\overline{E}_{j_{\sigma_{x}},n,t,s}$ such
that $x(s)=j^{\prime}s$, our goal is to construct all such $x$ are stabilzied
by $a^{s}$. This requires determining $\tau_{x}$, $m_{i}$ and $u_{i}$ such
that $\sigma_{x}(i)=sm_{i}+\tau_{x}(i)$ for all $i<s$ and
$x(i)=tu_{i}+\sigma_{x}(i)$. If we determine all such values, then we can let
$l_{i}=u_{i}\frac{t}{s}+m_{i}\mod\frac{n}{s}$ and this will give us exactly a
permutation $x$ stabilized by both $a^{t}$ and $a^{s}$ with remainder
permutations $\sigma_{x}$ and $\tau_{x}$. Then, for all $i<s$ we have

$x(i)=tu_{i}+\sigma_{x}(i)=tu_{i}+sm_{i}+\tau_{x}(i)=s(\frac{t}{s}u_{i}+m_{i})+\tau_{x}(i)=sl_{i}+\tau_{x}(i).$

Note now that $x(\sigma_{x}(i))=u_{\sigma_{x}(i)}t+i$ and
$x(\sigma_{x}(i))=x(m_{i}s+\tau_{x}(i))=m_{i}j^{\prime}s+l_{\tau_{x}(i)}s+i$
where we are assuming $x$ is stabilized by $a^{s}$ and that
$x(s)=j^{\prime}s$. Therefore, we also have that

$l_{\tau_{x}(i)}s=u_{\sigma_{x}(i)}t-m_{i}j^{\prime}s$

Finally, recall from Corollary 5.1.4 that $u_{\sigma_{x}(i)}t=-ju_{i}t\mod n$
for all $1\leq i\leq t-1$ and that
$jt=x(t)=\frac{t}{s}x(s)=\frac{t}{s}j^{\prime}s=tj^{\prime}$ so substituting
we get

$l_{\tau_{x}(i)}s=-ju_{i}t-m_{i}j^{\prime}s\implies
l_{\tau_{x}(i)}=-j^{\prime}u_{i}\frac{t}{s}-m_{i}j^{\prime}s\mod\frac{n}{s}.$

If $\tau_{x}(i)=i$, then these two must be equal so
$m_{i}s+u_{i}t=-ju_{i}t-m_{i}j^{\prime}s=-j^{\prime}tu_{i}-m_{i}j^{\prime}s$
which gives that

$\displaystyle u_{i}\frac{t}{s}+m_{i}$
$\displaystyle=-j^{\prime}u_{i}\frac{t}{s}-m_{i}j^{\prime}s\mod\frac{n}{s}$
$\displaystyle\implies u_{i}(1+j^{\prime})\frac{t}{s}+m_{i}(1+j^{\prime})$
$\displaystyle=0\mod\frac{n}{s}$
$\displaystyle\implies\left(\frac{t}{s}u_{i}+m_{i}\right)(1+j^{\prime})$
$\displaystyle=0\mod\frac{n}{s}.$

Note that if $\sigma_{x}(i)=i$ for $i<s$, then $u_{i}(j+1)=0\mod\frac{n}{t}$.
However, this will always be the case since if $\sigma_{x}(i)=i$, then
necessarily $\tau_{x}(i)=i$ and there is no $m_{i}$ so $m_{i}=0$. Namely,

$\displaystyle m_{i}s+u_{i}t$
$\displaystyle=-j^{\prime}tu_{i}-m_{i}j^{\prime}s$ $\displaystyle\implies
u_{i}t$ $\displaystyle=-jtu_{i}\qquad j^{\prime}t=jt\mod n$
$\displaystyle\implies u_{i}(1+j)$ $\displaystyle=0\mod\frac{n}{t}$

as desired.

Namely, if we let

$\overline{K}_{j^{\prime},j_{\sigma_{x}},n,t,s}:=\left\\{u\,|\,(u\frac{t}{s}+m_{i})(1+j^{\prime})=0\mod\frac{n}{s}\text{
for any }m_{i}\in P_{j_{\sigma_{x}},t/s}\right\\}$

then $\\{0\\}\subset\overline{K}_{j^{\prime},j_{\sigma_{x}},n,t,s}\subset
K_{j,n/t}$ where $j=j^{\prime}\mod\frac{n}{t}$. Let

$\overline{K}_{j^{\prime},j_{\sigma_{x}},n,t,s}^{(c)}:=\left\\{u\,|\,(u\frac{t}{s}+m_{i})(1+j^{\prime})=0\mod\frac{n}{s},m_{i}\in
P_{j_{\sigma_{x}},t/s}^{(c)}\right\\}.$

Note that it is possible for
$\overline{K}_{j^{\prime},j_{\sigma_{x}},n,t,s}^{(c)}$ to be empty which is
why must introduce

$\delta^{K^{c}}:=\begin{cases}|\overline{K}_{j^{\prime},j_{\sigma_{x}},n,t,s}^{(c)}|&\text{
if }\overline{K}_{j^{\prime},j_{\sigma_{x}},n,t,s}^{(c)}\not=\varnothing\\\
1&\text{otherwise}\end{cases}$

This gives the necessary requirements for $x.$ Namely, we must have that if
$\tau_{x}(i)=i$, there is a relationship between the possible $u_{i}$ for
which $x(i)=tu_{i}+\sigma_{x}(i)$ and possible $m_{i}$ for which
$\sigma_{x}(i)=sm_{i}+\tau_{x}(i)$. However, if $\tau_{x}(i)\not=i$, then no
such relationship is required and so the values $l_{i}$ such that
$x(i)=sl_{i}+\tau_{x}(i)$ may be anything smaller than $\frac{n}{s}$ and their
only dependence is that $l_{\tau_{x}(i)}=-j^{\prime}l_{i}\mod\frac{n}{s}$.

Then we sum.

$C_{n,t,s,r}=\underbrace{\sum_{j_{\sigma_{x}}\in E_{t/s}}}_{\text{choices for
}\sigma_{x}(s)}\underbrace{\sum_{j^{\prime}\in\overline{E}_{j_{\sigma_{x}},n,t,s}}}_{\text{choices
for
}x(s)}\underbrace{\sum_{l=0}^{\lfloor\frac{s-m_{r,j_{\sigma_{x}},t/s}}{2}\rfloor}}_{\begin{subarray}{c}\\#\text{
of fixed}\\\ \text{ points of
}\tau_{x}\end{subarray}}\,\,\,\underbrace{\delta_{j_{\sigma_{x}},t/s,r}}_{\sigma_{x}\text{
exists}}\,\,\,\underbrace{\delta_{j_{\sigma_{x}},t/s,r}^{l}}_{x\text{ and
}\sigma_{x}\text{
agree}}\,\,\underbrace{\binom{s-1-2l}{m_{r,j_{\sigma_{x}},t/s}-1}}_{\begin{subarray}{c}\text{
choose }m_{i}\\\ \text{ in
}P_{j_{\sigma_{x}},t/s}\end{subarray}}\qquad\,\qquad\,$
$\qquad\underbrace{|P_{j_{\sigma_{x}},t/s}|^{m_{r,j_{\sigma_{x}},t/s}-1}}_{\begin{subarray}{c}\tau_{x}(i)=i\text{
and }\\\
m_{i}=q(1-j_{\sigma_{x}})\end{subarray}}\underbrace{(\delta^{P^{c}})^{s-2l-m_{r,j_{\sigma_{x}},t/s}}}_{\begin{subarray}{c}\tau_{x}(i)=i\text{
and }\\\
m_{i}\not=q(1-j_{\sigma_{x}})\end{subarray}}\underbrace{\frac{(s-1)!}{(s-1-2l)!2^{l}l!}}_{\text{
choices for }\tau_{x}}$
$\qquad\qquad\qquad\underbrace{|\overline{K}_{j^{\prime},j_{\sigma_{x}},n,t,s}|^{m_{r,j_{\sigma_{x}},t/s}-1}}_{\begin{subarray}{c}u_{i}\text{
satisfying req.}\\\ \text{ when }\tau_{x}(i)=i\text{ and }m_{i}\in
P_{j_{\sigma_{x}},t/s}\end{subarray}}\underbrace{(\delta^{K^{c}})^{s-2l-m_{r,j_{\sigma_{x}},t/s}}}_{\begin{subarray}{c}u_{i}\text{
satisfying req.}\\\ \text{ when }\tau_{x}(i)=i\text{ and }m_{i}\in
P_{j_{\sigma_{x}},t/s}^{c}\end{subarray}}\underbrace{\left(\frac{n}{s}\right)^{l}}_{\tau_{x}(i)\not=i}-\underbrace{\sum_{\begin{subarray}{c}p|s\\\
p\not=d\end{subarray}}C_{n,t,p,r}}_{\text{ remove over count}}$

Specifically, if $s=1$, then $\sigma_{x}$ has full stabilizer and so
$\sigma_{x}(i)=ij_{\sigma_{x}}$ for all $i.$ Thus, since $\sigma_{x}$ has $r$
fixed points, $\beta_{j_{\sigma_{x}},n/t}$ which is the number of $\sigma_{x}$
must equal $r$. Therefore,

$C_{n,t,1,r}=\sum_{j_{\sigma_{x}}\in
E_{t/1}}\sum_{j^{\prime}\in\overline{E}_{j_{\sigma_{x}},n,t,1}}\delta_{r}(\beta_{j_{\sigma_{x}},t/1})=\sum_{j_{\sigma_{x}}\in
E_{t/1}}\delta_{r}(\beta_{j_{\sigma_{x}},t/1})|\overline{E}_{j_{\sigma_{x}},n,t,1}|.$

In the case where $t=n$, then we are just counting inovlutions $x$ with $r$
fixed points and we then remove all overcounts.

This is the number of involutions $x$ which can be generated by a permutation
$\sigma_{x}\in S_{t-1}$ with $r$ fixed points, and which are stabilized by
$a^{s}$ for $s|t$. ∎

######  Proposition 5.3.2.

Let $O_{n/t,r}$ be the set of orbits of length $t$ containing $r$ involutions.
Then

$|O_{n/t,r}|=\begin{cases}\frac{1}{r}|X_{n/t,r}|&\text{ if }r\not=0\\\
\frac{1}{t}|M_{n/t}|-\sum_{r=1}^{t}\frac{1}{r}|X_{n/t,r}|&\text{ if
}r=0\end{cases}.$ (18)

Where $X_{n/t,r}$ is from Proposition 5.3.1 and $M_{n/t}$ is from Proposition
5.1.1.

Proof. If $r\not=0$, then we have seen an involution $x$ with stabiliser
$\langle a^{t}\rangle$ is in an orbit containing $r$ total involutions if and
only if it has $r$ fixed points smaller than $t$. Namely, the number of orbits
containing $r$ involutions is $\frac{1}{r}|X_{n/t,r}|.$

If $r=0$, then $O_{n/t,0}$ is the number of orbits containing no involutions.
This value can be computed by taking the total number of orbtis containing $t$
permutations which given by $\frac{1}{t}|M_{n/t}|$ and subtracting those
orbits which contain at least one involution. Thus, we subtract off
$\frac{1}{r}|X_{n/t,r}|$ for all $1\leq r\leq t$. ∎

Lets take the time now to show that our formula here, unwieldy as it is,
aligns perfectly with that of [5] in Theorem 3.9 in the case where $n=p$.

######  Proposition 5.3.3.

Our formula agrees with those of [5] in Theorem 3.9. That is,

  1. (1)

$m_{p,1}=\sum_{r=1}^{p}|O_{p/p,r}|$

  2. (2)

$m_{p,1}+m_{p,0}=\frac{1}{p}|M_{p/p}|$

  3. (3)

$p+1$ of the $p(p-1)$ $1$-dimensional irreps of $J_{p}$ have indicator $+1$

Proof.

  1. (1)

Recall, that $m_{p,1}:=\frac{i_{p}-1}{p}-1$ is the number of orbits
$\mathcal{O}_{x}$ of length $t=p$ containing at least one involution where
$i_{p}$ is the number of involutions (including the identity) in $S_{p}.$ Note
also that in the proof of Proposition 5.2.1, we show that

$i_{p}=\sum_{l=0}^{\lfloor\frac{p}{2}\rfloor}\frac{p!}{(p-2l)!2^{l}l!}.$

We would like to show that

$m_{p,1}=\sum_{r=1}^{p}O_{p/p,r}=\sum_{r=1}^{p}\frac{1}{r}|X_{p/p,r}|.$

Now, we simplify,

$\displaystyle|X_{p/p,r}|$
$\displaystyle=\frac{(n-1)!}{(r-1)!2^{\frac{n-r}{2}}\left(\frac{n-r}{2}\right)!}-|C_{p,p,1,r}|$
$\displaystyle=\frac{(p-1)!}{(r-1)!2^{\frac{p-r}{2}}\left(\frac{p-r}{2}\right)!}-\sum_{j_{\sigma_{x}}\in
E_{p/1}}\delta_{r}(\beta_{j_{\sigma_{x}},p/1})|\overline{E}_{j_{\sigma_{x}},p,p,1}|$
$\displaystyle=\frac{(p-1)!}{(r-1)!2^{\frac{p-r}{2}}\left(\frac{p-r}{2}\right)!}-\delta_{r}(\beta_{1,p/1})|\overline{E}_{1,p,p,1}|-\delta_{r}(\beta_{p-1,p/1})|\overline{E}_{p-1,p,p,1}|$
$\displaystyle=\frac{(p-1)!}{(r-1)!2^{\frac{p-r}{2}}\left(\frac{p-r}{2}\right)!}-\delta_{r}(p)|\overline{E}_{1,p,p,1}|-\delta_{r}(1)|\overline{E}_{p-1,p,p,1}|\qquad\beta_{j,p/1}=\text{
gcd}(j-1,p)$
$\displaystyle=\frac{(p-1)!}{(r-1)!2^{\frac{p-r}{2}}\left(\frac{p-r}{2}\right)!}-\delta_{r}(p)-\delta_{r}(1)\qquad|\overline{E}_{j,p,p,1}|=1\text{
for }j=1,p-1$

Therefore,

$\displaystyle\sum_{r=1}^{p}\frac{1}{r}|X_{p/p,r}|$
$\displaystyle=\sum_{r=1}^{p}\frac{1}{r}\left[\frac{(p-1)!}{(r-1)!2^{\frac{p-r}{2}}\left(\frac{p-r}{2}\right)!}-\delta_{r}(p)-\delta_{r}(1)\right]$
$\displaystyle=\sum_{r=1}^{p}\left[\frac{(p-1)!}{r!2^{\frac{p-r}{2}}\left(\frac{p-r}{2}\right)!}-\frac{\delta_{r}(p)}{r}-\frac{\delta_{r}(1)}{r}\right]$
$\displaystyle=\left[\sum_{r=1}^{p}\frac{(p-1)!}{r!2^{\frac{p-r}{2}}\left(\frac{p-r}{2}\right)!}\right]-\frac{1}{1}-\frac{1}{p}$
$\displaystyle=\left[\sum_{l=0}^{\frac{p-1}{2}}\frac{(p-1)!}{(p-2l)!2^{l}l!}\right]-1-\frac{1}{p}\qquad\text{substituting
}l=\frac{p-r}{2}$
$\displaystyle=\frac{1}{p}\left[\sum_{l=0}^{\lfloor}\frac{p}{2}\rfloor\frac{p!}{(p-2l)!2^{l}l!}\right]-1-\frac{1}{p}\qquad\text{since
}p\text{ is odd }\left\lfloor\frac{p}{2}\right\rfloor=\frac{p-1}{2}$
$\displaystyle=\frac{1}{p}i_{p}-1-\frac{1}{p}$
$\displaystyle=\frac{i_{p}-1}{p}-1$

as desired. Of course, Theorem 3.5 tells us that each permutation $x$
satisfying $F_{x}=F$ and $x^{-1}\in\mathcal{O}_{x}$ induces a single irrep
with indicator $+1$. Thus, there are $pm_{p,1}$ total $p$-dimensional irreps
with indicator $+1$.

  2. (2)

The number of orbits of length $p$ is given by

$\frac{1}{p}|M_{p/p}|==\frac{1}{p}((p-1)!-\varphi(p))=(p-1)\frac{(p-2)!-1}{p}$

which also aligns with Theorem 3.9.

  3. (3)

Finally, $|M_{p/1}|=\varphi(p)=p-1$, so there are $p(p-1)$ $1$-dimensional
irreps of $J_{p}$. Of these, $|T_{p/1}|=|E_{p/1}|=2$.

Thus, there are $p(p-3)$ irreps of $J_{p}$ of dimension $1$ with indicator $0$
(induced from non-involutions). Of the two involutions with full stabilizer,
the identity induces the trivial irrep which has indicator $+1$ and the rest
have indicator $0$, and a non-identity $x$ which satisfies $x(1)=p-1$ and so
plugging into Theorem 4.3.2, gives all $p$ irreps have indicator $+1$. That
is, there are $p+1$ total $1$-dimensional irreps of $J_{p}$ with indicator
$+1$ and the rest have indicator $0$. ∎

Let us summarize what we have found so far. We have seen that any permutation
$x\in S_{n-1}$ with stabilizer $\langle a^{t}\rangle$ has a remainder
permutation $\sigma_{x}\in S_{t-1}$ such that $x(i)\mod t=\sigma_{x}(i)$ for
all $1\leq i<t$. Furthermore, $\sigma_{x}$ shares certain properties with $x$.
That is, if $x^{-1}=a^{-x(s)}xa^{s}\in\mathcal{O}_{x}$, then
$\sigma_{x}^{-1}=\sigma_{x^{-1}}=a_{t}^{-\sigma_{x}(s)}\sigma_{x}a_{t}^{s}\in\mathcal{O}_{\sigma_{x}}$
where $a_{t}$ is the standard $t$ cycle in $S_{t}$.

We have used remainder permutations to help us determine whether
$\mathcal{O}_{x}$ contains an involution. Most helpfully, we showed that if
$x$ is an involution, then the permutation
$y_{l}=a^{-x(l)}xa^{l}\in\mathcal{O}_{x}$ is an involution if and only if
$\sigma_{x}(l)=l$ (Lemma 4.2.5).

This connection between permutations in $S_{n}$ and $S_{t}$ was an unexpected
one and we believe the matched pair action connects $S_{n}$ with $S_{t}$ where
$t|n$ in more ways than we have been able to uncover in this paper.

##  Counting Irreps with a given dimension

Here we state one of our greatest results. Given an $n$ and any odd divisor
$t$, we can explicitly write down and count the number of $t$-dimensional
irreps of $J_{n}$ with indicator $+1$ and indicator $0.$

###  Counting the Number of odd dimensional Irreps with Indicator $+1$

######  Proposition 6.1.1.

Let $O_{n/t,r,j}$ be the set of orbits of length $1<t<n$ containing $r$
involutions where for each permutation $y$ in any of these orbits, $y(t)=jt$.
Recall entries (1), (2), (4), (6), (7), (9), (10), (11), (12), (13) from 1 (
(Big Table of Commonly Cited Values and Sets).). Then,

$r|O_{n/t,r,j}|=(\alpha_{j,n/t})^{r-1}\left(\frac{n}{t}\right)^{\frac{t-r}{2}}\frac{(t-1)!}{(r-1)!2^{\frac{t-r}{2}}\left(\frac{t-r}{2}\right)!}-\sum_{\begin{subarray}{c}s|t\\\
s\not=t\end{subarray}}\tilde{C}_{n,t,s,r,j}$ (19)

where

$\displaystyle\delta_{j,j^{\prime},n/t}^{\prime}$
$\displaystyle=\begin{cases}1&\text{ if }j^{\prime}t=jt\mod n\\\ 0&\text{
otherwise}\end{cases}$ $\displaystyle\tilde{C}_{n,t,1,r,j}$
$\displaystyle=\sum_{j_{\sigma_{x}}\in
E_{t/1}}\sum_{j^{\prime}\in\overline{E}_{j_{\sigma_{x}},n,t,1}}\delta_{j,j^{\prime},n/t}^{\prime}\delta_{r}(\beta_{j_{\sigma_{x}},t/1,r})$
(20) $\displaystyle\tilde{C}_{n,t,s,r,j}$
$\displaystyle=\sum_{j_{\sigma_{x}}\in
E_{t/s}}\sum_{j^{\prime}\in\overline{E}_{j_{\sigma_{x}},n,t,s}}\sum_{l=0}^{\lfloor\frac{s-m_{r,j_{\sigma_{x}},t/s}}{2}\rfloor}\delta_{j,j^{\prime},n/t}^{\prime}\delta_{j_{\sigma_{x}},t/s,r}\binom{s-1-2l}{m_{r,j_{\sigma_{x}},t/s}-1}|P_{j_{\sigma_{x}},t/s}|^{m_{r,j_{\sigma_{x}},t/s}-1}$
$\displaystyle\qquad\qquad(\delta^{P^{c}})^{s-2l-m_{r,j_{\sigma_{x}},t/s}}\frac{(s-1)!}{(s-1-2l)!2^{l}l!}|\overline{K}_{j^{\prime},j_{\sigma_{x}},n,t,s}|^{m_{r,j_{\sigma_{x}},t/s}-1}(\delta^{K^{c}})^{s-2l-m_{r,j_{\sigma_{x}},t/s}}\left(\frac{n}{s}\right)^{l}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad-\sum_{\begin{subarray}{c}p|s\\\
p\not=d\end{subarray}}\tilde{C}_{n,t,p,r,j}$ (21)

Proof. This is a direct adaptation of Proposition 5.3.1. The only change is
that we are assuming a fixed $j$ here and so we omit the first sum over the
set $E_{n/t}$. Additionally, in the overcount function we must ensure that we
only remove those $x$ which have the same $j$ value. Namely, since our choice
of $j^{\prime}$ such that $x(s)=j^{\prime}s$ forces $x(t)=j^{\prime}t$, we
must have that $j^{\prime}t=jt$ for our fixed $j.$

Therefore, we include the value

$\delta_{j,j^{\prime},n/t}^{\prime}=\begin{cases}1&\text{ if
}j^{\prime}t=jt\mod n\\\ 0&\text{ otherwise}\end{cases}$

in our overcount function. ∎

######  Theorem 6.1.2.

Let $t$ be an odd divisor of $n.$ Let $I_{n/t}^{(+1)}$ be the number of
representations of $J_{n}$ of dimension $t$ with indicator $+1$. Recall (3)
from 1 ( (Big Table of Commonly Cited Values and Sets).) and $O_{n/t,r,j}$ as
from Proposition 6.1.1. Then

$\displaystyle I_{n/1}^{(+1)}$ $\displaystyle=\sum_{j\in
E_{n/1}}\alpha_{j,n/1}$ (22) $\displaystyle I_{n/t}^{(+1)}$
$\displaystyle=t\sum_{j\in E_{n/t}}\sum_{r=1}^{t}\alpha_{j,n/t}O_{n/t,r,j}$
(23) $\displaystyle I_{n/n}^{(+1)}$ $\displaystyle=n\sum_{r=1}^{n}O_{n/n,r}$
(24)

Finally, the number of irreps of $J_{n}$ of dimension $t$ with indicator $0$
is given by

$I_{n/t}^{(0)}=\frac{n}{t}|M_{n/t}|-I_{n/t}^{(+1)}$ (25)

where $M_{n/t}$ is from Proposition 5.1.1.

Proof. Let $t$ be an odd divisor of $n$ and $x$ a permutation with stabilizer
$F_{x}=\langle a^{t}\rangle$ such that $x(t)=jt$ and $\mathcal{O}_{x}$
contains at least one involution.

Let $\alpha_{j,n/t}=\text{gcd}(j+1,\frac{n}{t})$, this is the number of $i$
such that
$\zeta_{n}^{i\frac{x(t)+t}{t}}=\zeta_{n}^{i\frac{jt+t}{t}}=\zeta_{n}^{i(j+1)}=1$.
Namely, $\alpha_{j,n/t}$ is the number of (group) irreps $\rho_{x,i}$ such
that the indicator of the induced character $\nu(\hat{\chi}_{x,i})=+1$ from
Theorem 4.3.2.

Since Proposition LABEL:*prop:observations (1) tells us that each
$y\in\mathcal{O}_{x}$ gives the same number of indicators of value $+1$, we
have $t\alpha_{j,n/t}$ total indicators of value $+1$ coming from this orbit.

Then, the number of representations $\hat{V}$ of $J_{n}$ of dimension $t$ with
indicator $+1$ is given by

$t\sum_{j\in E_{t}}\sum_{r=1}^{t}\alpha_{j,n/t}O_{n/t,r,j}.$

Of course, if $t=n$ then each $x$ with stabilizer $F_{x}=\\{1\\}$ either
satisfies that $x^{-1}\in\mathcal{O}_{x}$ and so induces a single irrep of
$J_{n}$ with indicator $+1$, or $x^{-1}\not\in\mathcal{O}_{x}$ and so $x$
induces an irrep of indicator $0$ (see Theorem 3.5). Since $t=n$ is odd by
assumption, any orbit closed under inversion must contain at least one
involution. Thus, the number of irreps of dimension $n$ with indicator $+1$ is
given by counting the number of orbits containing at least one involution and
multiplying by $n.$ This gives

$I_{n/n}^{(+1)}=n\sum_{r=1}^{n}O_{n/n,r}.$

Finally, since there are $|M_{n/t}|$ permutations with stabilizer $\langle
a^{t}\rangle$ (proved in Proposition 5.1.1) and each of these generates
$\frac{n}{t}$ irreps of $J_{n}$, there are $\frac{n}{t}|M_{n/t}|$ total
irreprs of $J_{n}$ of dimension $t.$ Since when $t$ is odd, there are no
irreps with indicator $-1$ (from Proposition 4.3.5), we conclude that
$I_{n/t}^{(0)}=\frac{n}{t}|M_{n/t}|-I_{n/t}^{(+1)}.$ ∎

Theorem 6.1.2 gives an explicit method of constructing the irreps of dimension
$t$ of $J_{n}$ which have indicator $+1$ in the case where $t$ is odd. Recall
that $|M_{n/t}|$ is the number of permutations $S_{n-1}$ which have stabilizer
$F_{x}=\langle a^{t}\rangle$ and that this value is equivalent to the number
of permutations in $S_{n-1}$ which induce an irrep of $J_{n}$ of dimension $t$
(see Proposition 5.1.1 and Theorem 3.2 and Lemma 3.3).

######  Corollary 6.1.3.

Let $t$ be an odd divisor of $n$. Then,

  1. (1)

every involution $x\in S_{n-1}$ with $F_{x}=\langle a^{t}\rangle$ induces at
least one $t$-dimensional irrep of $J_{n}$ with indicator $+1$

  2. (2)

NOT every $t$-dimensional irrep of $J_{n}$ with indicator $+1$ is induced from
an involution

  3. (3)

every $t$-dimensional irrep of $J_{n}$ with indicator $+1$ is induced from a
permutation whose orbit contains at least one involution

###  The Case $t=2$

Now, we turn our attention to the otherwise completely illusive even case.
Using some computation, we are able to derive formulas when $t=2$, although
due to the increased difficulty when considering even $t$, we do not believe
this is a viable strategy for deriving relationships in general.

######  Proposition 6.2.1.

Let $n$ be even and let $x\in S_{n-1}$ and $F_{x}=\langle a^{2}\rangle$ (so
$t=2$). Let $x(2)=2j$ where $j$ satisfies $j^{2}=1\mod\frac{n}{2}$. Then, the
number of $x$ such that $x^{-1}\in\mathcal{O}_{x}$ with
$x^{-1}=a^{-x(s)}xa^{s}$ is the number $u$ satisfying

  1. (1)

$j\not=2u+1\mod\frac{n}{2}$

  2. (2)

and either

    1. (i)

$u(j+1)=0\mod\frac{n}{2}$ in which case $s=t$ (equivalently, $s=0$)

    2. (ii)

or $(u+1)(j-1)=0\mod\frac{n}{2}$ in which case $s=1$

In fact, the or is exclusive and so the number of $x\in S_{n-1}$ with
$F_{x}=\langle a^{2}\rangle$ is the number of solutions to $u$ to
$u(j+1)=0\mod\frac{n}{2}$ (which counts the number of involutions $x$) plus
the number of solutions $(u+1)(j-1)=0\mod\frac{n}{2}$ (which counts the number
of non-involutions $x$). Here we are excluding all $u$ for which
$j\not=2u+1\mod\frac{n}{2}$ in both cases.

Furthermore, all such $x$ are defined by $x(1)=2u+1$ for any $u$ satisfying
the above.

Proof. Corollary 4.2.3 tells us that we need only understand $x(1)$ and $x(2)$
to fully determine $x$. Additionally, $x(2)=2j$ where $j^{2}=1\mod\frac{n}{2}$
by Lemma 4.2.1.

Now, because $x$ sends even numbers to even numbers and sends odd numbers to
odd numbers, we can let $x(1)=2u+1$ for some $u$.

  1. (i)

Now, assume that $x^{2}=1$. we have that

$\displaystyle 1$
$\displaystyle=x^{2}(1)=x(x_{1})=x(2u+1)=ux(2)+x(1)=2uj+x_{1}$
$\displaystyle\implies x_{1}$ $\displaystyle=1-2uj=2u+1$
$\displaystyle\implies 2u(j+1)$ $\displaystyle=0\mod n$ $\displaystyle\implies
u(j+1)$ $\displaystyle=0\mod\frac{n}{2}.$

In fact, any $u$ satisfying this formula will guarantee that $x^{2}=1$ since
it ensures that $x^{2}(1)=1$ and we already have that $x^{2}(2)=2$. However,
similarly to the proof of Proposition 5.1.1, we will have over counted.
Namely, we must remove all $x$ such that $x^{2}=1$ and $F_{x}=\langle
a\rangle$. Namely, we cannot have that

$x(2)=2j=2x(1)=2(2u+1)=4u+2\\\ \implies j=2u+1\mod\frac{n}{2}$

so we throw out all $u$ satisfying this formula.

  2. (ii)

We proceed similarly with the second case: $x^{-1}=a^{-x(1)}xa$. This time, we
note that

$\displaystyle 1$ $\displaystyle=x^{-1}(2u+1)=ux^{-1}(2)+x^{-1}(1)$
$\displaystyle=ux(2)+x^{-1}(1)=2uj+x^{-1}(1)$
$\displaystyle=2uj+x(2)-x(1)=2uj+2j-2u-1$

Which implies that

$0=2uj+2j-2u-2=2j(u+1)-2(u+1)=2(u+1)(j-1).$

Thus, $(u+1)(j-1)=0\mod\frac{n}{2}$. Again, all $u$ and $j$ satisfying this
relation will induce an element $x$ for which $x^{-1}(i)=x(i+1)-x(1)$. Of
course, once again, we must remove the case where $x(2)=2j=2x(1)$.

Since these are the only two cases, and since both are satisfied by the same
$u$ exactly when $j=2u+1\mod\frac{n}{2}$, which we have already ruled out, we
obtain all permutations $x$ with stabilizer $F_{x}=\langle a^{2}\rangle$ such
that $x^{-1}\in\mathcal{O}_{x}$ broken down into the two disjoint cases where
$x^{2}=1$ and $x^{2}\not=1.$ ∎

Actually, Proposition 6.2.1 gives us a full description of what permutations
give what indicators.

Note that the way we prove Proposition 6.2.1 is by checking cases for $x$.
Since $x(2)$ must be even, we force $x(1)=2u+1$ for some $u$. Then, setting
$x=x^{-1}$ or assuming $x\not=x^{-1}$ we rule out options for $u$.

The next smallest even value $t$ can take is $4$. But here our cases explode.
$x(4)$ must certainly be a multiple of $4$, but can have any remainder
permutation in $S_{3}$. For example, we could let $x(1)=4u_{1}+2$ and
$x(2)=4u_{2}+1$ and $x(3)=4u_{3}+3$ meaning $x$ would have remainder
permutation $\sigma_{x}=(1\,\,\,2)\in S_{3}$. From each of these six cases we
would need to work through possible choices of $u_{1},u_{2},u_{3}$ which lead
to either $x$ being an involution or not. This would result in twelve total
cases to consider.

Clearly this method cannot scale, since for any given even $t$, we would need
to consider all $(t-1)!$ possible ways to write $x$ then double it to check
the conditions under which $x$ is an involution or not an involution.

However, we can still get some insight by at least understanding $t=2$.

######  Theorem 6.2.2.

Let $n$ be even and let $x\in S_{n-1}$ and $F_{x}=\langle a^{2}\rangle$ (so
$t=2$). Let $I_{n/2}^{(+1)},I_{n/2}^{(-1)},I_{n/2}^{(0)}$ be the total number
of irreps of $J_{n}$ dimension $2$ which have indicator $+1,-1,0$
respectively. Recall (1), (2), (3), (14), (15), (16), (17) in 1 ( (Big Table
of Commonly Cited Values and Sets).).

Then,

$\displaystyle I_{n/2}^{(+1)}$ $\displaystyle=\sum_{j\in E_{n/2}}\sum_{u\in
K_{j,n/2}}2\overline{\delta}_{u,j,n/2}\left(\alpha_{j,n/2}+\sum_{i\in
K_{j,n/2}}\delta_{i,u,n/2}^{0}\right)$ (26) $\displaystyle I_{n/2}^{(-1)}$
$\displaystyle=\sum_{j\in E_{n/2}}\sum_{u\in K_{j,n/2}^{\prime}}\sum_{i\in
K_{j,n/2}}2\overline{\delta}_{u,j,n/2}\delta_{i,u,n/2}^{\not=0}$ (27)
$\displaystyle I_{n/2}^{(0)}$
$\displaystyle=\frac{n}{2}|M_{n/2}|-I_{n/2}^{(+1)}-I_{n/2}^{(-1)}$ (28)

Proof. Let

$\overline{\delta}_{u,j,n/2}=\begin{cases}1&\text{ if
}j\not=2u+1\mod\frac{n}{2}\\\ 0&\text{ otherwise}\end{cases}$

and

$K_{j,n/2}^{\prime}=\left\\{u\,|(u+1)(j-1)=0\mod\frac{n}{2}\,\right\\}.$

We break into two cases.

  1. (1)

If $x^{2}=1$ then, because $u(j+1)=0\mod\frac{n}{2}$ always has a solution
(namely, $u=0$) there are $\alpha_{j,n/2}=\text{gcd}(j+1,\frac{n}{2})$
possible $u$ which satisfy this relation. Let

$K_{j,n/2}:=\left\\{u\,|\,u(j+1)=0\mod\frac{n}{2}\right\\}.$

Again, $|K_{j,n/2}|=\alpha_{j,n/2}$ Thus, there are
$z_{j,n/2}:=\frac{n}{2}-\alpha_{j,n/2}$ possible $u$ which satisfy
$(u+1)(j-1)=0\mod\frac{n}{2}$.

Now, Proposition 4.3.5, tells us that if $\mathcal{O}_{x}$ contains an
involution, $\nu(\hat{\chi})\geq 0$. Thus, if $s^{2}=1$ (so we are in the
first case), irreps induced from $x$ either have indicator $+1$ or $0$. Using
Theorem 4.3.2, the indicator is $+1$ for all values $1\leq i\leq\frac{n}{2}$
such that $i(j+1)=0\mod\frac{n}{2}$. There are exactly $\alpha_{j,n/2}$ such
$i$. Let $\mathcal{O}_{x}=\\{x,y\\}$ (both are involutions in this case), then
$x$ contributes $\alpha_{j,n/2}$ indicators which are $+1$ and so does $y.$
Namely, there are

$\underbrace{\sum_{j\in E_{n/2}}}_{\text{choices for
}x(2)}\underbrace{\sum_{u\in K_{j,n/2}}}_{\text{choices for
}x(1)}\underbrace{\overline{\delta}_{u,j,n/2}}_{x^{-1}\in\mathcal{O}_{x}}\underbrace{2}_{x\text{
and }y}\underbrace{\alpha_{j,n/2}}_{\begin{subarray}{c}\text{ choices for
}i\\\ \text{ where }i(j+1)=0\mod\frac{n}{2}\end{subarray}}$

indicators with value $+1$ which are contributed to by involutions.

  2. (2)

If $x^{2}\not=1$, then because $t=2$ $\mathcal{O}_{x}=\\{x,x^{-1}\\}$ so
$x^{-1}=a^{-x(1)}xa$. Now, $x(1)+1=2u+2=2(u+1)$ and so $u_{2}=u+1$ as in
Theorem 4.3.2. Now, for any $i$, if $i(j+1)=0\mod\frac{n}{2}$, then either
$-i(u+1)=0\mod\frac{n}{2}$ in which case the indicator is $+1$ for that irrep,
or $-i(u+1)=\frac{n}{4}\mod\frac{n}{2}$ in which case the indicator is $-1$.
Let

$\delta_{i,u,n/2}^{0}=\begin{cases}1&\text{ if }-i(u+1)=0\mod\frac{n}{2}\\\
0&\text{ otherwise}\end{cases}$
$\delta_{i,u,n/2}^{\not=0}=\begin{cases}1&\text{ if
}-i(u+1)\not=0\mod\frac{n}{2}\\\ 0&\text{ otherwise}\end{cases}$

Of course, if $i(j+1)\not=0\mod\frac{n}{2}$, then the indicator is $0.$

Finally, the number of $+1$ indicators coming from this case is

$\underbrace{\sum_{j\in E_{n/2}}}_{\text{choices for
}x(2)}\underbrace{\sum_{u\in K_{j,n/2}^{\prime}}}_{\text{choices for
}x(1)}\underbrace{\sum_{i\in K_{j,n/2}}}_{\begin{subarray}{c}\text{choices for
}i\\\ \text{ where
}i(j+1)=0\mod\frac{n}{2}\end{subarray}}\underbrace{2}_{x\text{ and
}x^{-1}}\underbrace{\overline{\delta}_{u,j,n/2}}_{\text{ ensure
}j\not=2u+1}\underbrace{\delta_{i,u,n/2}^{0}}_{\text{ ensure }-i(u+1)=0}$

And the number of $-1$ indicators come from the remaining values $i$ for which
$i(j+1)=0$ but $-i(u+1)\not=0$. That is,

$\underbrace{\sum_{j\in E_{n/2}}}_{\text{choices for
}x(2)}\underbrace{\sum_{u\in K_{j,n/2}^{\prime}}}_{\text{choices for
}x(1)}\underbrace{\sum_{i\in K_{j,n/2}}}_{\begin{subarray}{c}\text{choices for
}i\\\ \text{ where
}i(j+1)=0\mod\frac{n}{2}\end{subarray}}\underbrace{2}_{x\text{ and
}x^{-1}}\underbrace{\overline{\delta}_{u,j,n/2}}_{\text{ ensure
}j\not=2u+1}\underbrace{\delta_{i,u,n/2}^{\not=0}}_{\text{ ensure
}-i(u+1)\not=0}$

Finally, we have that the total number of irreps with indicator $+1$ is

$\displaystyle I_{n/2}^{(+1)}$ $\displaystyle=\sum_{j\in E_{n/2}}\sum_{u\in
K_{j,n/2}}\overline{\delta}_{u,j,n/2}2\alpha_{j,n/2}+\sum_{j\in
E_{n/2}}\sum_{u\in K_{j,n/2}^{\prime}}\sum_{i\in
K_{j,n/2}}2\overline{\delta}_{u,j,n/2}\delta_{i,u,n/2}^{0}$
$\displaystyle=\sum_{j\in E_{n/2}}\sum_{u\in
K_{j,n/2}}2\overline{\delta}_{u,j,n/2}\left(\alpha_{j,n/2}+\sum_{i\in
K_{j,n/2}}\delta_{i,u,n/2}^{0}\right)$

and the minus ones come only from the second case

$I_{n/2}^{(-1)}=\sum_{j\in E_{n/t}}\sum_{u\in K_{j,n/2}^{\prime}}\sum_{i\in
K_{j,n/2}}2\overline{\delta}_{u,j,n/2}\delta_{i,u,n/2}^{\not=0}$

and the rest must be zero

$I_{n/2}^{(0)}=\frac{n}{2}|M_{n/2}|-I_{n/2}^{(+1)}-I_{n/2}^{(-1)}$ ∎

And so we now give a summary of the results we were able to obtain. Fix a
positive integer $n.$

######  Table 2 (Table of Obtainable Results).

Statement | Conditions | Result | Citation
---|---|---|---
List all permutations with a given stabilizer size | Fix $t|n$ |  There are $|M_{n/t}|$ permutations with stabilizer $F_{x}=\langle a^{t}\rangle$ | Proposition 5.1.1
List all involutions with a given stabilizer size | Fix $t|n$ |  There are $|T_{n/t}|$ involutions with stabilizer $F_{x}=\langle a^{t}\rangle$ | Proposition 5.2.1
List all involutions with given stabilizer size and $r$ fixed points | Fix $t|n$ and $r\leq n$ |  There are $|R_{n/t,r}|$ involutions with $r$ fixed points and stabilizer $F_{x}=\langle a^{t}\rangle$ | Proposition 5.2.3