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\pagestyle{fancy} \lhead{{\sf Homework 9 $|$ Due October 31
}} \rhead{\thepage} \cfoot{{\sf Math 4120/6120 $|$ Modern Algebra $|$
    Fall 2017 $|$ M.~Macauley}}



\begin{document}

$\;$

\noindent Read the following, which can all be found either in the
textbook or on the course website.
\begin{itemize}
\item Chapter 9.1 of \emph{Visual Group Theory} (VGT). \vspace{-2mm}
\item VGT Exercises 8.15--8.18, 9.17.
\end{itemize} 



\noindent Write up solutions to the following exercises.

\begin{enumerate}


\item The commutator subgroup of a group $G$ is the subgroup
  \[
  G'=\<aba^{-1}b^{-1}\mid a,b\in G\>\,.
  \]
  \begin{enumerate}
  \item Prove that $G$ is abelian if and only if $G'=\{e\}$.
  \item Prove that $G'\normal G$. [\emph{Hint}: Take a ``commutator''
    $c=aba^{-1}b^{-1}$ and prove that $gcg^{-1}\in G'$.]
  \item Prove that $G'$ is the intersection of all normal subgroups of $G$
    that contain the set $C:=\{aba^{-1}b^{-1}\mid a,b\in G\}$:
    \[
    G'=\bigcap_{C\subseteq N\normal\,G} N
    \]
  \item If we quotient $G$ by $G'$, then we are in essence,
    ``killing'' all non-abelian parts of the Cayley diagram, as shown
    below:
    
    \colorlet{lightgrey}{black!15}
    \tikzstyle{v} = [circle, draw, fill=lightgrey,inner sep=0pt,
      minimum size=2.5mm] \tikzstyle{r} = [draw,very thick, Red,
      -stealth] \tikzstyle{b} = [draw,very thick, Blue, -stealth]
    \begin{center}
      \begin{tikzpicture}[scale=.8,auto]
        \node at (-1.75,0) {$ab=ba$};
        \node (e) at (0,0) [v] {{\tiny $*$}};
        \node (a) at (1,-1) [v] {};
        \node (b) at (1,1) [v] {};
        \node (ab) at (2,0) [v] {};
        \draw [b] (e) to (a);
        \draw [r] (a) to (ab);
        \draw [r] (e) to (b);
        \draw [b] (b) to (ab);
        \begin{scope}[shift={(8,0)}]
          \node at (-1.75,0) {$ab\neq ba$};
          \node (e) at (0,0) [v] {{\tiny $*$}};
          \node (a) at (1,1) [v] {};
          \node (b) at (2.3,1) [v] {};
          \node (c) at (3.3,0) [v] {};
          \node (d) at (2.3,-1) [v] {};
          \draw [r] (e) to (a);
          \draw [b] (a) to (b);
          \draw [r] (c) to (b);
          \draw [b] (d) to (c);
        \end{scope}
      \end{tikzpicture}
    \end{center}
    Prove algebraically that $G/G'$ is indeed abelian. \bigskip
  \end{enumerate}



\item For each of the following groups $G$, compute its commutator
  subgroup $G'$ and its abelianization $G/G'$. Finally, draw the
  subgroup lattice of $G$ and circle every normal subgroup, and circle
  twice the one that is $G'$.
  \begin{enumerate}
    \item $V_4$,
    \item $D_3$,
    \item $Q_8$. \bigskip
  \end{enumerate}



\item Find the commutator subgroup of each of the following groups and
  compute its abelianization.
  \begin{enumerate}
    \item An abelian group $A$.
    \item The alternating group $A_n$, for $n\geq 5$. [\emph{Hint}: $A_n$ is a
      \emph{simple group}, which means its only normal subgroups are
      $\<e\>$ and $A_n$.]
    \item The dihedral group $D_n$ for $n$ even.
    \item The dihedral group $D_n$ for $n$ odd. \bigskip
  \end{enumerate}


\newpage

$\;$\vspace{-2mm}


  \colorlet{lred}{red!70!white}
  \colorlet{lblue}{blue!65!white}
  \colorlet{lime}{green!80!white}
  \colorlet{lyellow}{yellow!70!white}
  \definecolor{rouge}{rgb}{0.79 0.0 0.1}
  \definecolor{lgreen}{rgb}{0.25 0.7 0.15}  
  \definecolor{lpurple}{rgb}{0.7 0.45 0.9}
  \colorlet{lorange}{orange!70!white}
  

  \tikzstyle{n} = [inner sep=1pt]
  \tikzstyle{o} = [draw,bend left=45,orange,-stealth]
  \tikzstyle{p} = [draw,bend left=45,lpurple,-stealth]
  
\item Recall that the automorphism group of $V_4=\<h,v\>=\{e,h,v,r\}$,
  where $r=hv$, has presentation 
  \[
  \Aut(V_4)=\big\<\alpha,\beta\mid
  \alpha^3=\beta^2=(\alpha\beta)^2=id\big\>,\hspace{5mm}\text{where}\hspace{5mm} 
  \begin{array}{ll} h\stackrel{\alpha}{\longmapsto}v \\ v\longmapsto r\end{array}\hspace{5mm}\text{and}\hspace{5mm} 
  \begin{array}{ll} h\stackrel{\beta}{\longmapsto}v \\ v\longmapsto h.\end{array}
\]
  The generating automorphisms are the following permuations of $V_4$:
  \begin{center}
    $\alpha$ :\hspace{5mm} \begin{tikzpicture}[scale=.8,auto,baseline=-.6ex]
    \node [n] (0) at (0,0) {{\small $e$}};
    \node [n] (1) at (1,.05) {{\small $h$}};
    \node [n] (2) at (2,0) {{\small $v$}};
    \node [n] (3) at (3,0) {{\small $r$}};
    \path [o] (1) to (2);
    \path [o] (2) to (3);
    \path [o] (3) to (1);
  \end{tikzpicture} \hspace{10mm}
  %%
  and \hspace{10mm} $\beta$ :\hspace{5mm}
  %%
  \begin{tikzpicture}[scale=.8,auto,baseline=-.6ex]
    \node [n] (0) at (0,0) {{\small $e$}};
    \node [n] (1) at (1,.05) {{\small $h$}};
    \node [n] (2) at (2,0) {{\small $v$}};
    \node [n] (3) at (3,0) {{\small $r$}};
    \path [p] (1) to (2);
    \path [p] (2) to (1);
  \end{tikzpicture}
  \end{center}  
  The multiplication table and Cayley diagram of
  $\Aut(V_4)=\<\alpha,\beta\>$, which highlights how automorphisms are
  ``re-wirings'', are shown below:

  \newcommand*{\n}{7}%
  %%
  \tikzstyle{v} = [circle, draw, fill=grey,inner sep=0pt, minimum size=1.2mm] 
  \tikzstyle{r} = [draw, thick,Red]
  \tikzstyle{g} = [draw, thick,darkgreen]
  \tikzstyle{b} = [draw, thick,Blue]
  \tikzstyle{o} = [draw, very thick,lorange, -stealth]
  \tikzstyle{p} = [draw, very thick,lpurple]
  %%

  \begin{center}
  \begin{tikzpicture}[scale=.8]
    %%
    \begin{scope}[scale=0.9,shift={(-13,-3)}]
      \path[fill=lyellow] (-0.1,6) rectangle ++(1,1);
      \path[fill=lorange] (-0.1,5) rectangle ++(1,1);
      \path[fill=lime] (-0.1,4) rectangle ++(1,1);
      \path[fill=lpurple] (-0.1,3) rectangle ++(1,1);
      \path[fill=lgreen] (-0.1,2) rectangle ++(1,1);
      \path[fill=lred] (-0.1,1) rectangle ++(1,1);
      %%
      \path[fill=lyellow] (1,7.1) rectangle ++(1,.9);
      \path[fill=lorange] (2,7.1) rectangle ++(1,.9);
      \path[fill=lime] (3,7.1) rectangle ++(1,.9);
      \path[fill=lpurple] (4,7.1) rectangle ++(1,.9);
      \path[fill=lgreen] (5,7.1) rectangle ++(1,.9);
      \path[fill=lred] (6,7.1) rectangle ++(1,.9);
      %%
      \path[fill=lyellow] (1,6) rectangle ++(1,1);
      \path[fill=lorange] (1,5) rectangle ++(1,1);
      \path[fill=lime] (1,4) rectangle ++(1,1);
      \path[fill=lpurple] (1,3) rectangle ++(1,1);
      \path[fill=lgreen] (1,2) rectangle ++(1,1);
      \path[fill=lred] (1,1) rectangle ++(1,1);
      %%
      \path[fill=lorange] (2,6) rectangle ++(1,1);
      \path[fill=lime] (2,5) rectangle ++(1,1);
      \path[fill=lyellow] (2,4) rectangle ++(1,1);
      \path[fill=lred] (2,3) rectangle ++(1,1);
      \path[fill=lpurple] (2,2) rectangle ++(1,1);
      \path[fill=lgreen] (2,1) rectangle ++(1,1);
      %%
      \path[fill=lime] (3,6) rectangle ++(1,1);
      \path[fill=lyellow] (3,5) rectangle ++(1,1);
      \path[fill=lorange] (3,4) rectangle ++(1,1);
      \path[fill=lgreen] (3,3) rectangle ++(1,1);
      \path[fill=lred] (3,2) rectangle ++(1,1);
      \path[fill=lpurple] (3,1) rectangle ++(1,1);
      %%
      \path[fill=lpurple] (4,6) rectangle ++(1,1);
      \path[fill=lgreen] (4,5) rectangle ++(1,1);
      \path[fill=lred] (4,4) rectangle ++(1,1);
      \path[fill=lyellow] (4,3) rectangle ++(1,1);
      \path[fill=lorange] (4,2) rectangle ++(1,1);
      \path[fill=lime] (4,1) rectangle ++(1,1);
      %%
      \path[fill=lgreen] (5,6) rectangle ++(1,1);
      \path[fill=lred] (5,5) rectangle ++(1,1);
      \path[fill=lpurple] (5,4) rectangle ++(1,1);
      \path[fill=lime] (5,3) rectangle ++(1,1);
      \path[fill=lyellow] (5,2) rectangle ++(1,1);
      \path[fill=lorange] (5,1) rectangle ++(1,1);
      %%
      \path[fill=lred] (6,6) rectangle ++(1,1);
      \path[fill=lpurple] (6,5) rectangle ++(1,1);
      \path[fill=lgreen] (6,4) rectangle ++(1,1);
      \path[fill=lorange] (6,3) rectangle ++(1,1);
      \path[fill=lime] (6,2) rectangle ++(1,1);
      \path[fill=lyellow] (6,1) rectangle ++(1,1);
      %%
      
      \foreach \i in {1,...,\n} {
        \draw [very thin] (\i,1) -- (\i,\n); 
        \draw [very thin] (\i,\n+.1) -- (\i,\n+1); 
        \draw [very thin] (1,\i) -- (\n,\i); 
        \draw [very thin] (-0.1,\i) -- (.9,\i); 
      } 
      \draw [very thin] (1,\n+.1) rectangle (\n,\n+1);
      \draw [very thin] (-0.1,1) rectangle (.9,\n);
      \node at (0.4,6.5) {$id$};
      \node at (0.4,5.5) {$\alpha$};
      \node at (0.4,4.5) {$\alpha^2$};
      \node at (0.4,3.5) {$\beta$};
      \node at (0.4,2.5) {$\alpha\beta$};
      \node at (0.4,1.5) {$\alpha^2\!\beta$};
      %%
      \node at (1.5,7.5) {$id$};
      \node at (2.5,7.5) {$\alpha$};
      \node at (3.5,7.5) {$\alpha^2$};
      \node at (4.5,7.5) {$\beta$};
      \node at (5.5,7.5) {$\alpha\beta$};
      \node at (6.5,7.5) {$\alpha^2\!\beta$};
      %%
      \node at (1.5,6.5) {$id$};
      \node at (1.5,5.5) {$\alpha$};
      \node at (1.5,4.5) {$\alpha^2$};
      \node at (1.5,3.5) {$\beta$};
      \node at (1.5,2.5) {$\alpha\beta$};
      \node at (1.5,1.5) {$\alpha^2\!\beta$};
      %%
      \node at (2.5,6.5) {$\alpha$};
      \node at (2.5,5.5) {$\alpha^2$};
      \node at (2.5,4.5) {$id$};
      \node at (2.5,3.5) {$\alpha^2\!\beta$};
      \node at (2.5,2.5) {$\beta$};
      \node at (2.5,1.5) {$\alpha\beta$};
      %%
      \node at (3.5,6.5) {$\alpha^2$};
      \node at (3.5,5.5) {$id$};
      \node at (3.5,4.5) {$\alpha$};
      \node at (3.5,3.5) {$\alpha\beta$};
      \node at (3.5,2.5) {$\alpha^2\!\beta$};
      \node at (3.5,1.5) {$\beta$};
      %%
      \node at (4.5,6.5) {$\beta$};
      \node at (4.5,5.5) {$\alpha\beta$};
      \node at (4.5,4.5) {$\alpha^2\!\beta$};
      \node at (4.5,3.5) {$id$};
      \node at (4.5,2.5) {$\alpha$};
      \node at (4.5,1.5) {$\alpha^2$};
      %%
      \node at (5.5,6.5) {$\alpha\beta$};
      \node at (5.5,5.5) {$\alpha^2\!\beta$};
      \node at (5.5,4.5) {$\beta$};
      \node at (5.5,3.5) {$\alpha^2$};
      \node at (5.5,2.5) {$id$};
      \node at (5.5,1.5) {$\alpha$};
      %%
      \node at (6.5,6.5) {$\alpha^2\!\beta$};
      \node at (6.5,5.5) {$\beta$};
      \node at (6.5,4.5) {$\alpha\beta$};
      \node at (6.5,3.5) {$\alpha$};
      \node at (6.5,2.5) {$\alpha^2$};
      \node at (6.5,1.5) {$id$};
      %%
    \end{scope}
    %%%%
    %%%%%=============================================================
    %%%%
     \begin{scope}[shift={(0,0)},scale=.12]
      \path[o] (3,15) to [bend right] (-2,5);
      \path[p] (-5,-3) to (-11,-8);
      \draw[grey, fill=lightgrey] (0,0) circle (48mm);
      \node (e) at (-3,3) [v] {};
      \node (v) at (-3,-3) [v] {};
      \node (h) at (3,3) [v] {};
      \node (hv) at (3,-3) [v] {};
      \draw [r] (e) to (v);
      \draw [r] (h) to (hv);
      \draw [b] (e) to (hv);
      \draw [b] (v) to (h);
    \end{scope}
    %%
    \begin{scope}[shift={(2,0)},scale=.12]
      \draw[grey, fill=lightgrey] (0,0) circle (48mm);
      \node (e) at (-3,3) [v] {};
      \node (v) at (-3,-3) [v] {};
      \node (h) at (3,3) [v] {};
      \node (hv) at (3,-3) [v] {};
      \draw [r] (e) to (hv);
      \draw [r] (h) to (v);
      \draw [b] (e) to (h);
      \draw [b] (v) to (hv);
      \path[p] (5,-3) to (11,-8);
      \path[o] (-14,-5) to [bend right] (-2,-5);
    \end{scope}
    %%
    %%
    \begin{scope}[shift={(1,1.73)},scale=.12]
      \path[o] (11,-10) to [bend right] (6,0);
      %\path[o] (6,0) to [bend left] (11,-10);
      \draw[grey, fill=lightgrey] (0,0) circle (48mm);
      \node (e) at (-3,3) [v] {};
      \node (v) at (-3,-3) [v] {};
      \node (h) at (3,3) [v] {};
      \node (hv) at (3,-3) [v] {};
      \draw [b] (e) to (v);
      \draw [b] (h) to (hv);
      \draw [r] (e) to (h);
      \draw [r] (v) to (hv);
      \path[p] (0,6) to (0,13);
    \end{scope}
    %%
    %%----------------------------------------------------------------
    %%
    \begin{scope}[shift={(-2,-1.2)},scale=.12]
      \path[o] (-2,5) to [bend left] (20,40);
      \draw[grey, fill=lightgrey] (0,0) circle (48mm);
      \node (e) at (-3,3) [v] {};
      \node (v) at (-3,-3) [v] {};
      \node (h) at (3,3) [v] {};
      \node (hv) at (3,-3) [v] {};
      \draw [r] (e) to (h);
      \draw [r] (v) to (hv);
      \draw [b] (e) to (hv);
      \draw [b] (h) to (v);
    \end{scope}
    %%
    \begin{scope}[shift={(1,4)},scale=.12]
      \draw[grey, fill=lightgrey] (0,0) circle (48mm);
      \node at (0,0) {$id$};
      \node (e) at (-3,3) [v] {};
      \node (v) at (-3,-3) [v] {};
      \node (h) at (3,3) [v] {};
      \node (hv) at (3,-3) [v] {};
      \draw [r] (e) to (v);
      \draw [r] (h) to (hv);
      \draw [b] (e) to (h);
      \draw [b] (v) to (hv);
    \end{scope}
    \begin{scope}[shift={(4,-1.2)},scale=.12]
      \path[o] (-20,40) to [bend left] (2,5);
      \draw[grey, fill=lightgrey] (0,0) circle (48mm);
      \node (e) at (-3,3) [v] {};
      \node (v) at (-3,-3) [v] {};
      \node (h) at (3,3) [v] {};
      \node (hv) at (3,-3) [v] {};
      \draw [r] (e) to (hv);
      \draw [r] (h) to (v);
      \draw [b] (e) to (v);
      \draw [b] (h) to (hv);
    \end{scope}
    %%
    % stupid hack beacuse of ``Dimension too large'' TikZ error.
    \begin{scope}[shift={(3.6,-1.7)},scale=1]
      \path[o] (0,0) to [bend left] (-5.15,0);
    \end{scope}
    %%
    \end{tikzpicture}
  \end{center}
  Repeat the above steps for each of the following groups. Use the
  Cayley diagram defined by the generating set given. Recall that
  $\Aut(\Z_n)\cong U_n$.
  \begin{enumerate}
    \item $\Z_5=\<1\>$,
    \item $\Z_6=\<1\>$,
    \item $\Z_2\times\Z_3=\<(1,0),(0,1)\>$,\, [Recall that
      $\Z_2\times\Z_3\cong\Z_6$]
    \item $\Z_8=\<1\>$. \bigskip
  \end{enumerate}




\item Let $G$ act on a set $S$. Prove that $\Stab(s)$ is a subgroup of
  $G$ for every $s\in S$. \bigskip



\item Suppose the cyclic group $C_5$ acts on a set $S=\{A,B,C,D\}$.
  \begin{enumerate}
    \item What are the possible sizes of the orbits?
    \item What are the possible stabilizer subgroups of each element?
    \item Draw the action diagram. 
  \end{enumerate}
  Fully explain your reasoning for each part.


\end{enumerate}

\end{document}



