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solutions.tex
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solutions.tex
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\chapter*{Some solutions}
\pagestyle{plain}
\fancyhf{}
\fancyhead[LE,RO]{Representation theory of algebras}
\fancyhead[RE,LO]{Some solutions}
\fancyfoot[CE,CO]{\leftmark}
\fancyfoot[LE,RO]{\thepage}
\addcontentsline{toc}{chapter}{Some solutions}
\begin{sol}{xca:deg2}
Assume that $\phi$ is not irreducible. There exists a proper non-zero $G$-invariant
subspace $W$ of $V$. Thus $\dim W=1$. Let $w\in W\setminus\{0\}$.
For each $g\in G$, $\phi_g(w)\in W$. Thus $\phi_g(w)=\lambda w$ for some
$\lambda$. This means that $w$ is a common eigenvector for all the $\phi_g$.
Conversely, if $\phi$ admits a common eigenvector $v\in V$, then
the subspace generated by $v$ is $G$-invariant.
\end{sol}
\begin{sol}{xca:commutators}
Let $C_1,\dots,C_t$ be the conjugacy classes of $G$. For each
$i\in\{1,\dots,t\}$, let $g_i$ be a representative of $C_i$. Assume
that $g_i$ is conjugate to $g$ and
$g_j$ is conjugate to $h$. Let $\gamma\in G$.
Then
\begin{align*}
\sum_{z\in G}\chi(zg_iz^{-1}g_j)
&= \sum_{z\in G}\chi(\gamma zg_iz^{-1}g_j\gamma^{-1})\\
&= \sum_{z\in G}\chi(\gamma zg_iz^{-1}\gamma^{-1}\gamma g_j\gamma^{-1})\\
&=\sum_{y\in G}\chi(yg_iy^{-1}\gamma g_j\gamma^{-1}).
\end{align*}
Hence
\[
\sum_{z\in G}\chi(zg_iz^{-1}g_j)
=\frac{1}{|G|}\sum_{z,\gamma\in G}\chi(zg_iz^{-1}\gamma g_j\gamma^{-1}).
\]
Now $z_1g_iz_1^{-1}=z_2g_iz_2^{-1}$ if and only
if $z_2^{-1}z_1\in C_G(g_i)$. Thus
\begin{align*}
\sum_{z\in G}\chi(zg_iz^{-1}g_j) &= \frac{1}{|G|}|C_G(g_i)||C_G(g_j)|\sum_{\substack{x\in C_i\\y\in C_j}}\chi(xy)\\
&=\frac{|G|}{|C_i||C_j|}\sum_{\substack{x\in C_i\\y\in C_j}}\chi(xy).
\end{align*}
Now
\[
\omega_{\chi}(C_i)\omega_{\chi}(C_j)=\sum_{i=1}^t a_{ijk}\omega_{\chi}(C_k),
\]
where
\[
\omega_{\chi}(C_i)=\frac{|C_i|\chi(C_i)}{\chi(1)}
\]
and
$a_{ijk}$ is the number of solutions of the equation
$xy=z$ with $x\in C_i$, $y\in C_j$ and $z\in C_k$. Therefore
\begin{align*}
\frac{\chi(1)}{|G|}\sum_{z\in G}\chi(zg_iz^{-1}g_j)
&=\frac{\chi(1)}{|C_i||C_j|}\sum_{\substack{x\in C_i\\y\in C_j}}\chi(xy)\\
&=\frac{\chi(1)}{|C_i||C_j|}\sum_{k=1}^t a_{ijk}\chi(g_k)|C_k|\\
&=\frac{\chi(1)^2}{|C_i||C_j|}\sum_{k=1}^t a_{ijk}\omega_{\chi}(C_k)\\
&=\chi(g_i)\chi(g_j).
\end{align*}
To prove the second formula,
set $h=g^{-1}$ in the first formula.
Then
\begin{align*}
\chi(g)\chi(g^{-1})=\frac{\chi(1)}{|G|}\sum_{z\in G}\chi(zgz^{-1}g^{-1}) &\Longleftrightarrow
\chi(g)\overline{\chi(g)}=\frac{\chi(1)}{|G|}\sum_{z\in G}\chi([z,g])\\
& \Longleftrightarrow |\chi(g)|^2=\frac{\chi(1)}{|G|}\sum_{z\in G}\chi([z,g])\\
& \Longleftrightarrow \frac{|G|}{\chi(1)}|\chi(g)|^2=\sum_{z\in G}\chi([z,g]).
\end{align*}
\end{sol}
\begin{sol}{xca:least_p}
Let $g_1,\dots,g_m$ be the representatives of non-trivial conjugacy classes. Then $C_G(g_i)$ is non-tivial for all $i$. Since $p$ is the smallest prime dividing the order of $G$, it follows that
$(G:C_G(g_i))\geq p$. Now use the class equation to get
\[
|G|\geq |Z(G)|+pm,
\]
which is equivalent to $m\leq \frac1p (|G|-|Z(G)|)$. Since $G$ is non-abelian, $G/Z(G)$ is not cyclic. Thus $(G:Z(G))\geq p^2$. Now
\[
\frac{k(G)}{|G|}=\frac{|Z(G)|+m}{|G|}\leq \frac{(p-1)|Z(G)|+|G|}{p|G|}\leq \frac{p^2+p-1}{p^3}.
\]
This bound is reached if and only if $(G:Z(G))=p^2$.
\end{sol}
\begin{sol}{xca:5/8}
If $\cp(G)>5/8$, then $|[G,G]|<2$. Thus $[G,G]$ is the trivial group
and hence $G$ is abelian.
\end{sol}
\begin{sol}{xca:cp_NS}
\begin{enumerate}
\item If $\cp(G)>1/2$, then $|[G,G]|<3$ by Theorem \ref{thm:[GG]}. If $|[G,G]|=1$,
then $G$ is abelian and hence $G$ is nilpotent. If $|[G,G]|=2$, then
$[G,G]\subseteq Z(G)$.
%In fact, a more general fact is true.
%If $N$ is a normal subgroup of $G$ and
%$|N|=2$, then $N\subseteq Z(G)$. Write $N=\{1,x\}$. If $g\in G$, then
%$gxg^{-1}\in N$. Thus either $gxg^{-1}=1$ or $gxg^{-1}=x$. The first
%case implies $x=1$, a contradiction. Thus $x\in Z(G)$.
It follows that
$G/Z(G)$ is abelian (and hence nilpotent), so $G$ is nilpotent.
\item If $\cp(G)<21/80$, then
$|[G,G]|<60$. Thus $[G,G]$ is solvable, as groups of order $<60$ are solvable.
Hence $G$ is solvable.
\end{enumerate}
\end{sol}
\begin{sol}{xca:isoclinism}
\begin{enumerate}
\item
\item Using that $\sigma$ and
$\tau$ are automorphisms and
the commutativity of the diagram~\eqref{eq:isoclinism},
we compute
\begin{align*}
(G:Z(G))^2\cp(G) &= \frac{1}{|Z(G)|^2}|\{(x,y)\in G\times G:xy=yx\}|\\
&=\frac{1}{|Z(G)|^2}|\{(x,y)\in G\times G:[x,y]=1\}|\\
&=\frac{1}{|Z(G)|^2}|\{(x,y)\in G\times G:c_G(x,y)=1\}|\\
&=|\{(u,v)\in (G/Z(G))^2:c_G(u,v)=1\}|\\
&=|\{(u,v)\in (G/Z(G))^2:\tau c_G(u,v)=1\}|\\
&=|\{(u,v)\in (G/Z(G))^2:c_G(\sigma u,\sigma v)=1\}|\\
&=|\{(a,b)\in (H/Z(H))^2:c_H(a,b)=1\}|.
\end{align*}
It follows that $(G:Z(G))^2\cp(G)=(H:Z(G))^2\cp(H)$.
\end{enumerate}
\end{sol}