chap6.tex

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\lhead{\small\emph{Baitian Li}}
\chead{}
\rhead{\textsc{Representation Theory of Finite Groups}}
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\begin{document}

\setcounter{chapter}{5}
\chapter{Burnside's Theorem}

\setcounter{exercise}{1}
\begin{exercise}
  Let $G$ be a non-abelian group of order $39$.
  \begin{enumerate}
    \item Determine the degrees of the irreducible representations of $G$ and how many
    irreducible representations $G$ has of each degree (up to equivalence).
    \begin{answer}
      The degree of an irreducible representation must be $1$ or $3$.
      Let $n$ be the number of degree one representations, $m$ be the number
      of degree $3$ representations, then we have
      $n + 9m = 39$, thus $n \equiv 3 \pmod 9$. Lemma 6.2.7 asserts that
      $n \mid 39$, we must have $n=3$, $m=4$.
    \end{answer}
    \item Determine the number of conjugacy classes of $G$.
    \begin{answer}
      $n + m = 7$.
    \end{answer}
  \end{enumerate}
\end{exercise}

\begin{exercise}
  Let $G$ be a non-abelian group of order $21$.
  \begin{enumerate}
    \item Determine the degrees of the irreducible representations of $G$ and how many
    irreducible representations $G$ has of each degree (up to equivalence).
    \begin{answer}
      The degree of an irreducible representation must be $1$ or $3$.
      Let $n$ be the number of degree one representations, $m$ be the number
      of degree $3$ representations, then we have
      $n + 9m = 21$, thus $n \equiv 3 \pmod 9$. Lemma 6.2.7 asserts that
      $n \mid 21$, we must have $n=3, m = 2$.
    \end{answer}
    \item Determine the number of conjugacy classes of $G$.
    \begin{answer}
      $n + m = 5$.
    \end{answer}
  \end{enumerate}
\end{exercise}

\setcounter{exercise}{4}
\begin{exercise}
  Show that if $\varphi\colon G \to \GL_d(\bbC)$ is a representation with character
  $\chi$, then $g \in \ker \varphi$ if and only if $\chi(g) = d$.
  \begin{proof}
    Since $\varphi_g$ has $d$ eigenvalues that are roots of unity $\lambda_1^n = \dots = \lambda_d^n = 1$,
    we have $\chi(g) = \lambda_1 + \dots + \lambda_d = d$ iff $\lambda_1 = \dots = \lambda_d = 1$,
    i.e., $\varphi_g = I$.
  \end{proof}
\end{exercise}

\begin{exercise}
  For $\alpha\in \bbC$, denote by $\bbZ[\alpha]$ the smallest
  subring of $\bbC$ containing $\alpha$.
  \begin{enumerate}
    \item Prove that $\bbZ[\alpha]=\{a_0 +a_1 \alpha+\dots +a_n \alpha^n
    \mid n\in \bbN, a_0,\dots,a_n \in \bbZ \}$.
    \begin{proof}
      Clearly these elements are closed under addition
      and multiplication, so it is a ring. By the axiom of ring, every $a_0 +a_1 \alpha+\dots +a_n \alpha^n$
      must be in $\bbZ[\alpha]$, so we have the minimality.
    \end{proof}
    \item Prove that the following are equivalent:
    \begin{enumerate}
      \item $\alpha$ is an algebraic integer;
      \item The additive group of $\bbZ[\alpha]$ is finitely generated;
      \item $\alpha$ is contained in a finitely generated subgroup of the additive group of $\bbC$,
      which is closed under multiplication by $\alpha$.
    \end{enumerate}
    \begin{proof}
      (a) implies (b): Suppose $\alpha$ is an algebraic integer, we have
      $a_0 + a_1\alpha + \dots + a_{n-1}\alpha^{n-1} + \alpha^n = 0$.
      Therefore $1, \alpha, \dots, \alpha^{n-1}$ generates $\bbZ[\alpha]$.
      
      (b) clearly implies (c).

      (c) implies (a): Let $y_1, \dots, y_n$ be a set of generators, $\alpha y_i$
      then can be represented by $y_1, \dots, y_n$, by Lemma 6.1.5, $\alpha$ is an
      algebraic integer.
    \end{proof}
  \end{enumerate}
\end{exercise}

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