\documentclass[t,xcolor=dvipsnames]{beamer}
\usepackage{graphics,multimedia,tikz,amssymb,amsmath,stmaryrd,graphicx,mathrsfs,color}
\usepackage{xcolor,colortbl}% http://ctan.org/pkg/xcolor

\setbeamertemplate{navigation symbols}{}

\theoremstyle{theorem-KL}
\newtheorem{theorem-KL}[theorem]{Theorem (Kearnes, Li)}

%\theoremstyle{Problem}
%\newtheorem{problem}[theorem]{Problem}

\theoremstyle{Conjecture}
\newtheorem{conjecture}[theorem]{Conjecture}

\theoremstyle{Question}
\newtheorem{question}[theorem]{Question}

\theoremstyle{Example}
\newtheorem{exmp}[theorem]{Example}

\theoremstyle{Conjecture2}
\newtheorem{conjecture2}[theorem]{Conjecture 2}

\theoremstyle{Theorems}
\newtheorem{theorems}[theorem]{Theorems}

\theoremstyle{Lemmas}
\newtheorem{lemmas}[theorem]{Lemmas}

\theoremstyle{Proof (with character theory!)}
\newtheorem{myproof}[theorem]{Proof (with character theory!)}

\theoremstyle{Proof (without character theory!)}
\newtheorem{myproof2}[theorem]{Proof (without character theory!)}

%\newenvironment<>{myproof}[1][\proofname]{%
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%  \def\insertproofname{#1\@addpunct{.}}%
%  \pushQED{\qed}
%  \alert{\textbf{\insertproofname}} \hspace*{\fill} \\}
%T{\popQED}


% This file is a solution template for:

% - Giving a talk on some subject.
% - The talk is between 15min and 45min long.
% - Style is ornate.

% Copyright 2004 by Till Tantau <tantau@users.sourceforge.net>.
%
% In principle, this file can be redistributed and/or modified under
% the terms of the GNU Public License, version 2.
%
% However, this file is supposed to be a template to be modified
% for your own needs. For this reason, if you use this file as a
% template and not specifically distribute it as part of a another
% package/program, I grant the extra permission to freely copy and
% modify this file as you see fit and even to delete this copyright
% notice. 

\mode<presentation>
{


\useoutertheme{infolines} 
%\setbeamertemplate{blocks}[rounded][shadow=true] 

  \usetheme{Warsaw}
%  \usetheme{Berkeley}
%%%%%%  \usetheme{Madrid}
%  \usetheme{default}
%  \usetheme{Boadilla}
%  \usetheme{Pittsburgh}
%  \usetheme{Rochester} %[ works best as \usetheme[height=7mm]{Rochester} ]
%  \usetheme{Copenhagen}
%  \usetheme{Singapore}
%  \usetheme{Malmoe}
%  \usetheme{Szeged}

%\usecolortheme[named=PineGreen]{structure} 

%\usecolortheme[named=Apricot]{structure} 
%\usecolortheme[named=Gray]{structure} 
%\usecolortheme[named=Tan]{structure} 
%\usecolortheme[named=ForestGreen]{structure} 
%\usecolortheme[named=Green]{structure} 
%\usecolortheme[named=SeaGreen]{structure} 
%\usecolortheme[named=LimeGreen]{structure} 
%\usecolortheme[named=OliveGreen]{structure} 
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\usecolortheme[named=NavyBlue]{structure} 
%\usecolortheme[named=DarkOrchid]{structure} 
%\usecolortheme[named=Periwinkle]{structure} 
%% more colors in /usr/share/texmf/tex/latex/graphics/dvipsnam.def 

\usefonttheme{serif}

%  \setbeamercovered{transparent}
  % or whatever (possibly just delete it)
}

\usepackage[english]{babel}
% or whatever

\usepackage[latin1]{inputenc}
% or whatever

\usepackage{times}
\usepackage[T1]{fontenc}

\usepackage{circuitikz}
\usetikzlibrary{arrows,shapes.gates.logic.US,shapes.gates.logic.IEC,calc}
\tikzstyle{branch}=[fill,shape=circle,minimum size=3pt,inner sep=0pt]

% Or whatever.  Note that the encoding and the font should match. If T1
% does not look nice, try deleting the line with the fontenc.


%%%\title[Short Paper Title] % (optional, use only with long paper titles)
%%%{Presentation Title}

\title[Examples, Properties, Applications]{Examples, Properties, Applications}

%%%\subtitle
%%%{Presentation Subtitle} % (optional)

%%%\author[Author, Another] % (optional, use only with lots of authors)
%%%{F.~Author\inst{1} \and S.~Another\inst{2}}
%%%% - Use the \inst{?} command only if the authors have different
%%%%   affiliation.
\author{Math 6270}
%\author{Kearnes\inst{U Colorado}, Szendrei\inst{U Colorado}, Willard\inst{U Waterloo}}
%\author{Kearnes}


%%%\institute[Universities of Somewhere and Elsewhere] % (optional, but mostly needed)
%%%{
%%%  \inst{1}%
%%%  Department of Computer Science\\
%%%  University of Somewhere
%%%  \and
%%%  \inst{2}%
%%%  Department of Theoretical Philosophy\\
%%%  University of Elsewhere}
%%%% - Use the \inst command only if there are several affiliations.
%%%% - Keep it simple, no one is interested in your street address.
%\institute{University of Colorado}

\institute{}


%%%\date[Short Occasion] % (optional)
%%%{Date / Occasion}
%\date{January 20, 2014\\ University of Waterloo}

\date[]{}%Math Club, November 8, 2017}

%76. Arbeitstagung Allgemeine Algebra\\May 22-25, 2008\\Linz, Austria}

%%%\subject{Talks}
%%%% This is only inserted into the PDF information catalog. Can be left
%%%% out. 


%%%% If you have a file called "university-logo-filename.xxx", where xxx
%%%% is a graphic format that can be processed by latex or pdflatex,
%%%% resp., then you can add a logo as follows:

%%\pgfdeclareimage[height=2cm]{algalg}{algalg.pdf}
%%\logo{\pgfuseimage{algalg}}


%%%% Delete this, if you do not want the table of contents to pop up at
%%%% the beginning of each subsection:
%%%\AtBeginSubsection[]
%%%{
%%%  \begin{frame}<beamer>
%%%    \frametitle{Outline}
%%%    \tableofcontents[currentsection,currentsubsection]
%%%  \end{frame}
%%%}


% If you wish to uncover everything in a step-wise fashion, uncomment
% the following command: 


\usepackage{amsmath,amsthm,amsfonts,latexsym,amssymb,amscd,mathrsfs}

\def\magen#1{{\color{magenta}#1\color{black}}}
\def\purple#1{{\color{purple}#1\color{black}}}
\def\blue#1{{\color{blue}#1\color{black}}}
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\def\orange#1{{\color{orange}#1\color{orange}}}
\def\green#1{{\color{green}#1\color{green}}}
\def\gray#1{{\color{gray}#1\color{gray}}}


\begin{document}


%%%\begin{frame}
%%%  \frametitle{Outline}
%%%  \tableofcontents
%%%  % You might wish to add the option [pausesections]
%%%\end{frame}


% Since this a solution template for a generic talk, very little can
% be said about how it should be structured. However, the talk length
% of between 15min and 45min and the theme suggest that you stick to
% the following rules:  

% - Exactly two or three sections (other than the summary).
% - At *most* three subsections per section.
% - Talk about 30s to 2min per frame. So there should be between about
%   15 and 30 frames, all told.



\beamerdefaultoverlayspecification{<+->}
%%%\subsection[Short First Subsection Name]{First Subsection Name}
%%\subsection{Apr 3--7, 2006}


\def\m#1{\mathbf{\uppercase{#1}}}
\def\df{{\mathrm{Def}}}
\def\thm{{\mathrm{Thm}}}
\def\sd{{\mathrm{sd}}}
\def\EE{{\sf E}}
\def\UU{{\sf U}}
\def\MM{{\sf M}}
\def\PP{{\sf P}}
\def\Clo{{\mathrm{Clo}}}
\def\Pol{{\mathrm{Pol}}}
\def\Sub{{\mathrm{Sub}}}
\def\Con{{\mathrm{Con}}}
\def\Aut{{\mathrm{Aut}}}
\def\pr{{\mathrm{pr}}}
\def\var#1{{\mathcal #1}}
\def\wec#1{{\mathbf #1}}

\def\FF{{\mathfrak F}}


\begin{frame}
  \titlepage
\vspace{-2cm}
  \begin{center}
  $$
\begin{array}{|c||c|c|c|c|}
\hline
 & 1 & k_2 & \cdots & k_r \\
G & 1 & g_2 & \cdots & g_r \\
\hline
\hline
\chi_1 & 1 & 1 & \cdots & 1 \\
\hline
\chi_2 & d_2 & \chi_2(g_2) & \cdots & \chi_2(g_r)\\
\hline
\vdots & \vdots &\vdots & \ddots &\vdots \\
\hline
\chi_r & d_r & \chi_r(g_2) & \cdots & \chi_r(g_r) \\
\hline
\end{array}
$$

{\begin{center}
Each $d_i$ divides $|G| = 1^2+d_2^2+\cdots+d_r^2$

Each $k_j$ divides $|G| = 1+k_2+\cdots+k_r$
\end{center}}
%\includegraphics[height=2cm]{algalg.pdf}
\end{center}
\end{frame}

\begin{frame}[t]
  \frametitle{Character tables of abelian groups}

  \pause

$$
  \begin{array}{|c||c|c|}
    \hline
         {}      & 1 & 1  \\
    \mathbb Z_2& 0 & 1 \\
    \hline    \hline
    \xi_1& 1 & \hphantom{-}1 \\
    \hline
    \xi_2& 1 & -1\\
    \hline
  \end{array}
  \quad\pause
  \begin{array}{|c||c|c|c|}
    \hline
         {}      & 1 & 1 & 1 \\
    \mathbb Z_3& 0 & 1 & 2\\
    \hline    \hline
    \chi_1& 1 & 1 & 1\\
    \hline
    \chi_2& 1 & \omega & \omega^2\\
    \hline
    \chi_2^2& 1 & \omega^2 & \omega\\
    \hline
  \end{array}
  \quad\pause
  \begin{array}{|c||c|c|c|c|}
    \hline
        {}       & 1 & 1 & 1 & 1\\
    \mathbb Z_4& 0 & 1 & 2 & 3\\
    \hline    \hline
    \chi_1& 1 & \hphantom{-}1 & \hphantom{-}1 & \hphantom{-}1\\
    \hline
    \chi_2& 1 & \hphantom{-}i & -1 & -i\\
    \hline
    \chi_2^2& 1 & -1 & \hphantom{-}1 & -1\\
    \hline
    \chi_2^3& 1 & -i & -1 & \hphantom{-}i\\
    \hline
  \end{array}
$$

  \pause
\begin{center}
$$  \begin{array}{|c||c|c|c|c|}
    \hline
               & 1 & 1 & 1 & 1\\
    \mathbb Z_2\times \mathbb Z_2& (0,0) & (0,1) & (1,0) & (1,1)\\
    \hline    \hline
    \xi_1(x)\xi_1(y)& 1 & \hphantom{-}1 & \hphantom{-}1 & \hphantom{-}1\\
    \hline
    \xi_2(x)\xi_1(y)& 1 & \hphantom{-}1 & -1 & -1\\
    \hline
    \xi_1(x)\xi_2(y)& 1 & -1 & \hphantom{-}1 & -1\\
    \hline
    \xi_2(x)\xi_2(y)& 1 & -1 & -1 & \hphantom{-}1\\
    \hline
  \end{array}$$
 \end{center}
\end{frame}

\begin{frame}[t]
  \frametitle{Completing a table from partial information}
  
\pause
Let $G$ be a nonabelian group of order $8$. Necessarily
$G/Z(G)\cong \mathbb Z_2\times \mathbb Z_2$, so
inflation gives us partial information about the character
table.

\pause
\begin{center}
$$  \begin{array}{|c||c|c||c|c|c|}
    \hline
               & 1 & 1 & 2 & 2 & 2\\
    G& 1 & g_2 & g_3 & g_4 & g_5\\
    \hline    \hline
    \chi_1& 1 & 1 & \hphantom{-}1 & \hphantom{-}1 & \hphantom{-}1\\
    \hline
    \chi_2& 1 & 1 & \hphantom{-}1 & -1 & -1\\
    \hline
    \chi_3& 1 & 1 & -1 & \hphantom{-}1 & -1\\
    \hline
    \chi_4& 1 & 1 & -1 & -1 & \hphantom{-}    1\\
    \hline
    \chi_5& ? & ? & \hphantom{-}? & \hphantom{-}? & \hphantom{-}?\\
    \hline
  \end{array}
  $$
\end{center}
  \pause
  The entries in the last row can be determined
  by: \hfill\linebreak \pause
  (i) finding the missing irrep, \pause
  (ii) column orthogonality, \pause
  (iii) row orthogonality, \pause
  (iv) $\sum_i d_i\chi_i(h) = \chi_{\textrm{reg}}(h)$, ETC.

\end{frame}


\begin{frame}[t]
  \frametitle{Completing a table from partial information, $Q_8, D_4$}
  
\pause
The table for either of $D_4(D_8)$ or $Q_8$ is:

\pause
\begin{center}
$$  \begin{array}{|c||c|c||c|c|c|}
    \hline
               & 1 & 1 & 2 & 2 & 2\\
    G& 1 & g_2 & g_3 & g_4 & g_5\\
    \hline    \hline
    \chi_1& 1 & 1 & \hphantom{-}1 & \hphantom{-}1 & \hphantom{-}1\\
    \hline
    \chi_2& 1 & 1 & \hphantom{-}1 & -1 & -1\\
    \hline
    \chi_3& 1 & 1 & -1 & \hphantom{-}1 & -1\\
    \hline
    \chi_4& 1 & 1 & -1 & -1 & \hphantom{-}    1\\
    \hline
    \hline
    \chi_5& 2 & -2 & \hphantom{-}0 & \hphantom{-}0 & \hphantom{-}0\\
    \hline
  \end{array}
  $$
  \end{center}
    
  \pause
 The groups $D_4$ and $Q_8$ can be distinguished by the
 fact that $\det(\chi_5^{Q_8})=\chi_1\neq \det(\chi_5^{D_4})$.
  
\end{frame}


\begin{frame}[t]
  \frametitle{$A_4$}
  
\pause
$A_4/K\cong \mathbb Z_3$, so $3$ linear characters
arise from inflation. \pause
Also, $A_4$ acts $2$-transitively on $S=\{1,2,3,4\}$,
so $\chi_4 = \chi_S-\chi_1$ is an irrep of degree $3$.
\pause
This must be all.

\pause
\begin{center}
$$  \begin{array}{|c||c|c||c|c|}
    \hline
               & 1 & 3 & 4 & 4 \\
    A_4& 1 & (1\;2)(3\;4) & (1\;2\;3) &(1\;3\;2) \\
    \hline    \hline
    \chi_1& 1 & \hphantom{-}1 & 1 & 1 \\
    \hline
    \chi_2& 1 & \hphantom{-}1 & \omega & \omega^2 \\
    \hline
    \chi_3& 1 & \hphantom{-}1 & \omega^2 & \omega \\
    \hline
    \hline
    \chi_4& 3 & -1 & 0 & 0\\
    \hline
  \end{array}
  $$
  \end{center}
    
  \pause
  Another way to produce $\chi_4$ is to realize
  $A_4$ as the rotation group of the tetrahedron. \pause
  Or use orthogonality. \pause
  Or use the regular representation.
  
\end{frame}


\begin{frame}[t]
  \frametitle{$S_4$}
  
\pause
$S_4/K\cong S_3$, so get $3$ irreps by inflation.
\pause
$S_4$ acts $2$-transitively on $\{1, 2, 3, 4\}$,
so get another irrep from that.
\pause
Or, realize $S_4$ as the rotation group of the cube.

\pause
\begin{center}
$$  \begin{array}{|c||c|c||c||c|c|}
    \hline
               & 1 & 3 & 8 & 4 & 4 \\
    S_4& 1 & (1\;2)(3\;4) & (1\;2\;3) & (1\;2) & (1\;2\;3\;4)\\
    \hline    \hline
    \chi_1& 1 & \hphantom{-}1 & \hphantom{-}1 & \hphantom{-}1 & \hphantom{-}1\\
    \hline
    \chi_2& 1 & \hphantom{-}1 & \hphantom{-}1 & -1 & -1 \\
    \hline
    \chi_3& 2 & \hphantom{-}2 & -1 & \hphantom{-}0 & \hphantom{-}0 \\
    \hline 
    \hline
    \chi_4& 3 & -1 & \hphantom{-}0 & \hphantom{-}1 & -1\\
    \hline 
    \chi_5& 3 & -1 & \hphantom{-}0 & -1 & \hphantom{-}1\\
    \hline
  \end{array}
  $$
  \end{center}
    
\end{frame}




\begin{frame}[t]
  \frametitle{$A_5$}
  
  \pause
  $A_5$ is the smallest nonabelian simple group,
  which makes it an interesting example. \pause
  $\chi_1$ is the only irrep of degree $1$.
  \pause
  $A_5$ may be realized as the rotation group
  of the dodecahedron, yielding an irrep of degree $3$.
  \pause
  $A_5$ acts $2$-transitively on $\{1,2,3,4,5\}$, yielding
  an irrep of degree $4$. 
  \pause
  $A_5$ acts $2$-transitively by conjugation
  on its six $5$-Sylow subgroups, yielding
  an irrep of degree $5$. 

\pause
\begin{center}
$$  \begin{array}{|c||c|c|c|c|c|}
    \hline
               & 1 & 15 & 20 & 12 & 12 \\
    A_5& 1 & (1\;2)(3\;4) & (1\;2\;3) & (1\;2\;3\;4\;5) & (1\;2\;3\;5\;4)\\
    \hline    \hline
    \chi_1& 1 & \hphantom{-}1 & \hphantom{-}1 & \hphantom{-}1\hphantom{-1} & \hphantom{-}1\hphantom{-1}\\
    \hline
    \chi_2& 3 & -1 & \hphantom{-}0 & \hphantom{-}\phi\hphantom{-1} & \hphantom{-}\phi^{-1} \\
    \hline
    \chi_3& 3 & -1 & \hphantom{-}0 & \hphantom{-}\phi^{-1} & \hphantom{-}\phi\hphantom{-1}  \\
    \hline 
    \chi_4& 4 & \hphantom{-}0 & \hphantom{-}1 & -1\hphantom{-1} & -1\hphantom{-1}\\
    \hline 
    \chi_5& 5 & \hphantom{-}1 & -1 & \hphantom{-}0\hphantom{-1} & \hphantom{-}0\hphantom{-1}\\
    \hline
  \end{array}
  $$
  \end{center}
    
\end{frame}




\begin{frame}[t]
  \frametitle{Silly applications}

  \begin{theorem}
If $|G|=p$, $p$ prime, then $G$ is abelian.
  \end{theorem}

  \begin{proof}
    \pause
    The first column of the character table for $G$ consists
    of at most $p$ integers $d_1,\ldots,d_r$ such that (i) $d_1=1$, (ii)
    $d_j|p$ for all $j$, and (iii) $d_1^2+\cdots +d_r^2 = p$. \pause
    It must be that $d_1=\cdots=d_r=1$, so $G$ is abelian.
  \end{proof}
  \pause
  
  \begin{theorem}
Any group of order $<6$ is abelian.
  \end{theorem}

  \begin{proof}
    Same idea. \pause
    The smallest number $n$ that is a sum of squares
    $d_1^2+\cdots +d_r^2$ where $d_1=1$, some $d_j>1$,
    and all $d_j$ divide $n$ is $6$.
  \end{proof}
  
\end{frame}


\begin{frame}[t]
  \frametitle{More silly applications}

  \begin{theorem}
If $|G|=p^2$, $p$ prime, then $G$ is abelian.
  \end{theorem}

  \begin{proof}
    \pause
    Each degree $d_j$ must be $1, p$ or $p^2$. \pause
    The choices $d_j=p$ or $p^2$ are too big
    if $1^2+d_2^2+\cdots+d_r^2=p^2$.
  \end{proof}
  \pause
  
  \begin{theorem}
    If $|G|=pq$, $q<p$ primes, and $G$ is nonabelian,
    then $p\equiv 1\pmod{q}$.
  \end{theorem}

  \begin{proof}
    \pause
    Possible degrees are $d_j\in \{1,q\}$, and some $d_j=q$.
    \pause
    Assume that there are $m$ degrees satisfying $d_j=q$.
    \pause
    The number of
    $d_j$ equal to $1$ is $n:=[G:G']\in \{1, q, p\}$.
    \pause
    We have $pq=d_1^2+\cdots+d_r^2=mq^2+n$. \pause
    From this,
    $q|n$, so $q=n$. \pause
    Dividing $pq=mq^2+q$ by $q$ we get
    $p=mq+1$.
  \end{proof}
  
\end{frame}



\begin{frame}[t]
\frametitle{Unsilly applications}

\begin{theorem}
A finite nonabelian simple group $G$ has no
conjugacy class $h^G$ of prime power size other than
the class $\{1\}$.
\end{theorem}

\begin{proof}
  \pause
  If  $\chi\in\textrm{Irr}(G)$ is not the principal character, then
    $K_{\chi}=Z_{\chi}=\{1\}$, since these are proper normal
  subgroups. \pause
  If $p$ is the prime of the theorem, then
  $p\!\not{\mid}~\chi(1)$ implies $\chi(h)=0$ for any such $\chi$.
  \pause
  Column orthogonality, applied to the columns of $1$ and $h$, yields
  $
1 + \chi_2(1)\chi_2(h) + \cdots + \chi_r(1)\chi_r(h) = 0,
$
\pause
hence
$$(\chi_2(1)\chi_2(h))/p + \cdots + (\chi_r(1)\chi_r(h))/p=-1/p.$$
\pause
Each summand on the left is an algebraic integer,
but $-1/p$ is not.
\end{proof}
  
  
\end{frame}


\begin{frame}[t]
\frametitle{Unsilly applications}

\begin{theorem}
If $G$ is a finite nonabelian simple group
and $H$ is a subgroup of prime power index,
then $H$ is centerless.
\end{theorem}

\begin{proof}
%This is a restatement of the previous theorem
%in the language of subgroups instead of
%conjugacy classes.
\pause
If $H$ is a subgroup of $G$ of $p$-power index
with a nontrivial central element $h$, then $h^G$
is a conjugacy class of $G$ of $p$-power size.
\pause
If $h^G$ is a conjugacy class of $G$ of $p$-power size, then
$C_G(h)$ is a subgroup of $G$ of $p$-power index
with a nontrivial central element $h$.
\end{proof}
  

  
\end{frame}


\begin{frame}[t]
\frametitle{Burnside's $p^aq^b$ Theorem}

\begin{theorem}
  There is no nonabelian simple group $G$ such that 
  $|G|=p^aq^b$, $p, q$ primes.
\end{theorem}

\begin{proof}
  \pause
  Apply the preceding theorem to $H\in\textrm{Syl}_p(G)$
  to derive a contradiction.
\end{proof}
  
\pause
  
\begin{corollary}
If $|G|=p^aq^b$, $p, q$ primes, then $G$ is solvable.
\end{corollary}

\begin{proof}
  \pause
  If $|G|=p^aq^b$ and $G$ is not solvable, then it must have a
  nonabelian simple section $G'$
  whose order divides $p^aq^b$. \pause
  There is no such group.
\end{proof}
\end{frame}


\begin{frame}[t]
\frametitle{A theorem about simple groups}

\pause

\begin{theorem}
  If $G$ is a nonabelian simple group
  and some irreducible character degree is a prime $p$,
  then the Sylow $p$-subgroups of $G$ have order $p$.  
\end{theorem}

\pause
Recall, as an example:

\pause
\begin{center}
$$  \begin{array}{|c||c|c|c|c|c|}
    \hline
               & 1 & 15 & 20 & 12 & 12 \\
    A_5& 1 & (1\;2)(3\;4) & (1\;2\;3) & (1\;2\;3\;4\;5) & (1\;2\;3\;5\;4)\\
    \hline    \hline
    \chi_1& 1 & \hphantom{-}1 & \hphantom{-}1 & \hphantom{-}1\hphantom{-1} & \hphantom{-}1\hphantom{-1}\\
    \hline
\chi_2& 3 & {-1} & \hphantom{-}0 & \hphantom{-}\phi\hphantom{-1} & \hphantom{-}\phi^{-1} \\
    \hline
    \chi_3& 3 & -1 & \hphantom{-}0 & \hphantom{-}\phi^{-1} & \hphantom{-}\phi\hphantom{-1}  \\
    \hline 
    \chi_4& 4 & \hphantom{-}0 & \hphantom{-}1 & -1\hphantom{-1} & -1\hphantom{-1}\\
    \hline 
    \chi_5& 5 & \hphantom{-}1 & -1 & \hphantom{-}0\hphantom{-1} & \hphantom{-}0\hphantom{-1}\\
    \hline
  \end{array}
  $$
  \end{center}
   
\end{frame}



\begin{frame}[t]
\frametitle{A theorem about simple groups}

\begin{theorem}
  If $G$ is a nonabelian simple group
  and some character degree is a prime $p$,
  then the Sylow $p$-subgroups of $G$ have order $p$.
\end{theorem}

Recall, as an example:

\begin{center}
$$  \begin{array}{|c||c|c|c|c|c|}
    \hline
               & 1 & 15 & \cellcolor{yellow}20 & 12 & 12 \\
    A_5& 1 & (1\;2)(3\;4) & \cellcolor{yellow}(1\;2\;3) & (1\;2\;3\;4\;5) & (1\;2\;3\;5\;4)\\
    \hline    \hline
    \chi_1& 1 & \hphantom{-}1 & \cellcolor{yellow}\hphantom{-}1 & \hphantom{-}1\hphantom{-1} & \hphantom{-}1\hphantom{-1}\\
    \hline
    \cellcolor{yellow}\chi_2& \cellcolor{yellow}3 & \cellcolor{yellow}{-1} & \cellcolor{yellow}\hphantom{-}0 & \cellcolor{yellow}\hphantom{-}\phi\hphantom{-1} & \cellcolor{yellow}\hphantom{-}\phi^{-1} \\
    \hline
    \cellcolor{yellow}\chi_3& \cellcolor{yellow}3 & \cellcolor{yellow}{-1} & \cellcolor{yellow}\hphantom{-}0 & \cellcolor{yellow}\hphantom{-}\phi^{-1} & \cellcolor{yellow}\hphantom{-}\phi\hphantom{-1} \\
    \hline 
    \chi_4& 4 & \hphantom{-}0 & \cellcolor{yellow}\hphantom{-}1 & -1\hphantom{-1} & -1\hphantom{-1}\\
    \hline 
    \chi_5& 5 & \hphantom{-}1 & \cellcolor{yellow}-1 & \hphantom{-}0\hphantom{-1} & \hphantom{-}0\hphantom{-1}\\
    \hline
  \end{array}
  $$
  \end{center}
   
\end{frame}


\begin{frame}[t]
\frametitle{A theorem about simple groups}

\begin{theorem}
  If $G$ is a nonabelian simple group
  and some character degree is a prime $p$,
  then the Sylow $p$-subgroups of $G$ have order $p$.
\end{theorem}

Recall, as an example:

\begin{center}
$$  \begin{array}{|c||c|c|c|c|c|}
    \hline
               & 1 & 15 & \cellcolor{yellow}20 & \cellcolor{green}12 & \cellcolor{green}12 \\
    A_5& 1 & (1\;2)(3\;4) & \cellcolor{yellow}(1\;2\;3) & \cellcolor{green}(1\;2\;3\;4\;5) & \cellcolor{green}(1\;2\;3\;5\;4)\\
    \hline    \hline
    \chi_1& 1 & \hphantom{-}1 & \cellcolor{yellow}\hphantom{-}1 & \cellcolor{green}\hphantom{-}1\hphantom{-1} & \cellcolor{green}\hphantom{-}1\hphantom{-1}\\
    \hline
    \cellcolor{yellow}\chi_2& \cellcolor{yellow}3 & \cellcolor{yellow}{-1} & \cellcolor{yellow}\hphantom{-}0 & \cellcolor{green}\cellcolor{yellow}\hphantom{-}\phi\hphantom{-1} & \cellcolor{green}\cellcolor{yellow}\hphantom{-}\phi^{-1} \\
    \hline
    \cellcolor{yellow}\chi_3& \cellcolor{yellow}3 & \cellcolor{yellow}{-1} & \cellcolor{yellow}\hphantom{-}0 & \cellcolor{green}\cellcolor{yellow}\hphantom{-}\phi^{-1} & \cellcolor{green}\cellcolor{yellow}\hphantom{-}\phi\hphantom{-1} \\
    \hline 
    \chi_4& 4 & \hphantom{-}0 & \cellcolor{yellow}\hphantom{-}1 & \cellcolor{green}-1\hphantom{-1} & \cellcolor{green}-1\hphantom{-1}\\
    \hline 
    \cellcolor{green}\chi_5& \cellcolor{green}5 & \cellcolor{green}\hphantom{-}1 & \cellcolor{green}\cellcolor{yellow}-1 & \cellcolor{green}\hphantom{-}0\hphantom{-1} & \cellcolor{green}\hphantom{-}0\hphantom{-1}\\
    \hline
  \end{array}
  $$
  \end{center}
   
\end{frame}

\begin{frame}[t]
  \frametitle{Preliminaries}
  \pause
  Let $G$ be a nonabelian simple group with
  irreducible character $\psi$ of degree $p$.
  \pause
  If $\rho\colon G\to \textrm{GL}_p(\mathbb C)$
  affords $\psi$, then $\rho$ is faithful.
  \pause
  In fact, $K_{\psi}=Z_{\psi}=\{1\}$.

\bigskip
  
  \pause
  Let $P$ be a Sylow $p$-subgroup of $G$.
  \pause
  Choose any $z\in Z(P)\setminus \{1\}$.
  \pause
  $[G:C_G(z)]$ divides $[G:P]$,
  hence $z^G$ has size prime to $p$.
  \pause
  Thus $\psi(z)=0$ 
  for any $z\in Z(P)\setminus\{1\}$.

  \bigskip

  \pause
  \color{red}
  What we are seeing here is important: \pause
  a character is vanishing on the nonidentity elements
  of a subgroup, $Z(P)$.
  \color{black}
\end{frame}

\begin{frame}[t]
  \frametitle{Recognizing multiples of the regular representation}
  \pause
  \begin{lemma}
    Let $\rho\colon H\to \textrm{GL}_n(\mathbb C)$
    be a (not necessarily irreducible) representation
    of a finite group, and let $\psi$ be the character
    afforded by $\rho$.
    If $\psi$ vanishes 
    on $H\setminus\{1\}$, then $|H|$ divides $\psi(1)$. \linebreak \pause
    \color{red} Thus, $\psi$
    is a multiple of the regular character, \pause and so
$\rho$
    is a multiple of the regular representation.
    \color{black}
  \end{lemma}

  \pause

  \begin{proof}
    \pause
    Compare $\chi_1$ and $\psi$:
    $$  \begin{array}{|c||c|c|c|c|}
    \hline
    & 1 & k_2 & \cdots    & k_r \\
    H& 1 & a_2 & \cdots & a_r\\
    \hline    \hline
\chi_1& 1 & 1 & \cdots & 1 \\
    \hline
    \psi& \psi(1) & 0 & \cdots & 0 \\
    \hline
  \end{array}.
    $$
    \pause
    $\langle \chi_1,\psi\rangle = \frac{\psi(1)}{|H|}$ must be an integer.
    \pause
    Now use that representations are determined by their characters.
  \end{proof}
\end{frame}

\begin{frame}[t]
  \frametitle{Proof of theorem}
  \pause
  To be proved:
  If there is an irreducible
  representation
  $\rho\colon G\to \textrm{GL}_p(\mathbb C)$
  (affording $\psi$)
  then a Sylow $p$-subgroup $P\leq G$ has order $p$.

  \pause
  \begin{proof}
    We have shown that $\psi$ vanishes on the nonidentity
    elements of $Z(P)$, so $|Z(P)|=p$. Moreover,
    $\rho_{Z(P)}$ is the regular representation
    of $Z(P)$, \pause
    so if $z\in Z(P)$ is a generator, then $\rho(z)$
    has distinct e-values. \pause
    It follows that $C_{\textrm{GL}_p(\mathbb C)}(z)$
    consists of diagonal matrices, hence is abelian. \pause
A fortiori, $C_{G}(z)$ is abelian. \pause
Thus, $P=Z(P)$, which has order $p$.
    \end{proof}

\bigskip

\pause
The converse is false. \pause
$\textrm{PSL}_2(q)$ has Sylow $3$-subgroups of size $3$
for infinitely many prime powers $q$, but only
$\textrm{PSL}_2(5)$ and $\textrm{PSL}_2(7)$
have irreps of degree $3$.
\end{frame}

\end{document}



