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\pagestyle{fancy} \lhead{{\sf Homework 6 $|$ Due October 7 (Friday)
}} \rhead{\thepage} \cfoot{{\sf Math 8510 $|$ Abstract Algebra $|$
    Fall 2022 $|$ M.~Macauley}}

\begin{document}

$\;$ \vspace{-6mm}

\begin{enumerate}


\item The \emph{short five lemma} says that given the commutative diagram of exact sequences shown at left, if $\nu$ and $\kappa$ are isomorphisms, then $\gamma$ is as well.
\[
\begin{tikzpicture}[scale=1.75]
\begin{scope}[shift={(0,0)}]
\node (1NW) at (0,1) {$1$};
\node (N1) at (1,1) {$N_1$};
\node (G1) at (2,1) {$G_1$};
\node (Q1) at (3,1) {$Q_1$};
\node (1NE) at (4,1) {$1$};
\node (1SW) at (0,0) {$1$};
\node (N2) at (1,0) {$N_2$};
\node (G2) at (2,0) {$G_2$};
\node (Q2) at (3,0) {$Q_2$};
\node (1SE) at (4,0) {$1$};
\draw[->] (1NW) -- (N1) node[midway,above]{};
\draw[->] (N1) -- (G1) node[midway,above]{\footnotesize $\iota_1$};
\draw[->] (G1) -- (Q1) node[midway,above]{\footnotesize $\pi_1$};
\draw[->] (Q1) -- (1NE) node[midway,above]{};
%%
\draw[->] (1SW) -- (N2) node[midway,above]{};
\draw[->] (N2) -- (G2) node[midway,above]{\footnotesize $\iota_2$};
\draw[->] (G2) -- (Q2) node[midway,above]{\footnotesize $\pi_2$};
\draw[->] (Q2) -- (1SE) node[midway,above]{};
%%
\draw[->] (N1) -- (N2) node[midway,right]{\footnotesize $\nu$};
\draw[->] (G1) -- (G2) node[midway,right]{\footnotesize $\gamma$};
\draw[->] (Q1) -- (Q2) node[midway,right]{\footnotesize $\kappa$};
\end{scope}
    \begin{scope}[shift={(4.85,.5)}]
      \node (1L) at (0,0) {$1$};
      \node (N) at (1,0) {$N$};
      \node (G1) at (2,.707) {$G$};
      \node (G2) at (2,-.707) {$G$};
      \node (Q) at (3,0) {$Q$};
      \node (1R) at (4,0) {$1$};
      \draw[->] (1L) -- (N) node[midway,above]{};
      \draw[right hook->] (N) -- (G1) node[midway,above left]{\footnotesize $\iota$};
      \draw[right hook->] (N) -- (G2) node[midway,below left]{\footnotesize $\iota'$};
      \draw[->>] (G1) -- (Q) node[midway,above right]{\footnotesize $\pi$};
      \draw[->>] (G2) -- (Q) node[midway,below right]{\footnotesize $\pi'$};
      \draw[->] (Q) -- (1R);
      \draw[->,dashed] (G1) -- (G2) node[midway, right]{\footnotesize $\gamma$};
    \end{scope}
\end{tikzpicture}
\]
\begin{enumerate}
\item Prove the short five lemma.
\item Two extentions
  $N_i\stackrel{\iota_i}{\into}G_i\stackrel{\pi_i}{\to} Q_i$ are said
  to be \emph{equivalent} if they are related via a commutative
  diagram, like the one above (left). Two extensions of $Q$ by $N$ are
  equivalent if they are related by a commutative diagram like the one
  on the right. Up to \emph{isomorphism}, there are four extensions of
  $\Alert{Q=V_4}$ by $\Balert{N=C_2}$, as shown below.
  \[
  \hspace*{-14mm}
 \begin{tikzpicture}[shorten >= -3pt, shorten <= -3pt,scale=.95]
    \tikzstyle{every node}=[font=\footnotesize]
    \tikzstyle{E} = [draw, gray];
    \tikzstyle{V} = [gray];
    \tikzstyle{N} = [draw, very thick, Blue];
    \tikzstyle{H} = [draw, very thick, Red];
    %%
\begin{scope}[shift={(0,0)}]
      \node[Red] (Q8) at (0,3) {$Q_8$};
      \node[Red] (i) at (-1,2) {$C_4$};
      \node[Red] (j) at (0,2) {$C_4$};
      \node[Red] (k) at (1,2) {$C_4$};
      \node[Blue] (-1) at (0,1) {$C_2$};
      \node[Blue] (1) at (0,0) {$C_1$};
      \draw[H] (Q8)--(i); 
      \draw[H] (Q8)--(j); 
      \draw[H] (Q8)--(k);
      \draw[H] (-1)--(i); 
      \draw[H] (-1)--(j); 
      \draw[H] (-1)--(k);
      \draw[N] (1)--(-1); 
    \end{scope}
    %%
    \begin{scope}[shift={(3.25,0)}]
      \node[Red] (D4) at (0,3) {$D_4$};
      \node[Red] (r2-f) at (-1,2) {$V_4$};
      \node[Red] (r) at (0,2) {$C_4$};
      \node[Red] (r2-rf) at (1,2) {$V_4$};
      %%
      \node[V] (f) at (-2,1) {$C_2$};
      \node[V] (r2f) at (-1,1) {$C_2$};
      \node[Blue] (r2) at (0,1) {$C_2$};
      \node[V] (r3f) at (1,1) {$C_2$};
      \node[V] (rf) at (2,1) {$C_2$};
      %%
      \node[Blue] (1) at (0,0) {$C_1$};
      %%
      \draw[H] (D4) to (r2-f); 
      \draw[H] (D4) to (r); 
      \draw[H] (D4) to (r2-rf);
      \draw[E] (r2-f) to (f); 
      \draw[E] (r2-f) to (r2f); 
      \draw[H] (r2-f) to (r2);
      \draw[E] (r2-rf) to (rf); 
      \draw[E] (r2-rf) to (r3f); 
      \draw[H] (r2-rf) to (r2);
      \draw[H] (r2) to (r); 
      \draw[N] (r2) to (1); 
      \draw[E] (f) to (1);
      \draw[E] (r2f) to (1); 
      \draw[E] (r3f) to (1); 
      \draw[E] (rf) to (1);
    \end{scope}
       %%
    \begin{scope}[shift={(9,0)}]
    \node[Red] (G) at (0,3) {$C_2^3$};
    %%
    \node[V] (100-001) at (-2,2) {$V_4$}; % 100,001,101    
    \node[V] (100-010) at (-3,2) {$V_4$}; % 100,010,110
    \node[Red] (010-001) at (-1,2) {$V_4$}; % 010,001,011
    \node[Red] (100-011) at (0,2) {$V_4$}; % 100,011,111
    \node[Red] (110-011) at (1,2) {$V_4$};  % 110,011,101
    \node[V] (010-101) at (3,2) {$V_4$};  % 010,101,111
    \node[V] (110-001) at (2,2) {$V_4$};  % 110,001,111

    %%
    \node[V] (100) at (-3,1) {$C_2$};
    \node[V] (010) at (-2,1) {$C_2$};
    \node[V] (001) at (-1,1) {$C_2$};
    \node[Blue] (011) at (0,1) {$C_2$};
    \node[V] (101) at (1,1) {$C_2$};
    \node[V] (110) at (2,1) {$C_2$};
    \node[V] (111) at (3,1) {$C_2$};
    %%
    \node[Blue] (000) at (0,0) {$C_1$};
    %%
    \draw[E] (000) to (100); 
    \draw[E] (000) to (010); 
    \draw[E] (000) to (110);
    \draw[E] (000) to (101); 
    \draw[E] (000) to (001); 
    \draw[N] (000) to (011);
    \draw[E] (000) to (111);
    %%
    \draw[E] (100)to(100-010); 
    \draw[E] (100)to(100-001); 
    \draw[E] (100)to(100-011);
    \draw[E] (010)to(100-010); 
    \draw[E] (010)to(010-001); 
    \draw[E] (010)to(010-101);
    \draw[E] (110)to(100-010); 
    \draw[E] (110)to(110-001); 
    \draw[E] (110)to(110-011);
    \draw[E] (101)to(100-001); 
    \draw[E] (101)to(010-101); 
    \draw[E] (101)to(110-011);
    \draw[E] (111)to(100-011); 
    \draw[E] (111)to(010-101); 
    \draw[E] (111)to(110-001);
    \draw[E] (001)to(100-001); 
    \draw[E] (001)to(010-001); 
    \draw[E] (001)to(110-001);
    %%
    \draw[E] (100-010) to (G); 
    \draw[E] (100-001) to (G); 
    \draw[E] (010-101) to (G); 
    \draw[E] (110-001) to (G);
    \draw[H] (110-011) to (G);
    \draw[H] (010-001) to (G);
    \draw[H] (100-011) to (G); 
    \draw[H] (011)to(010-001); 
    \draw[H] (011)to(100-011); 
    \draw[H] (011)to(110-011);
  \end{scope}
  %%
   \begin{scope}[shift={(13.75,0)}]
      \node[Red] (G) at (1,3) {$C_4\!\times\!C_2$};
      \node[Red] (01-20) at (0,2) {$V_4$};
      \node[Red] (11) at (1,2) {$C_4$};
      \node[Red] (10) at (2,2) {$C_4$};
      \node[V] (01) at (-1,1) {$C_2$};
      \node[V] (21) at (0,1) {$C_2$};
      \node[Blue] (20) at (1,1) {$C_2$};
      \node[Blue] (00) at (0,0) {$C_1$};
      \draw[H] (G)--(01-20); 
      \draw[H] (G)--(11); 
      \draw[H] (G)--(10);
      \draw[E] (01-20)--(01); 
      \draw[E] (01-20)--(21); 
      \draw[H] (01-20)--(20); 
      \draw[H] (11)--(20); 
      \draw[H] (10)--(20);
      \draw[E] (01)--(00); 
      \draw[E] (21)--(00); 
      \draw[N] (20)--(00);
    \end{scope}
  \end{tikzpicture}
 \]
 However, up to equivalence, there are more than four.
 For example, finding all extesnions $C_2\stackrel{\iota}{\into}D_4\stackrel{\pi}{\to}V_4$ amounts to finding all $\gamma\in\Aut(D_4)$ that makes the diagram commute. Since each $\gamma$ fixes the cosets $\{1,r^2\}$ and $\{r,r^3\}$, the following diagram illustrates two quotients that cannot be equivalent. 
 \[
 \hspace*{-15mm}
\begin{tikzpicture}
\begin{scope}[shift={(-4.5,0)},scale=2]
\tikzstyle{v-big} = [circle,fill=lightgrey,draw,inner sep=0pt, minimum size=8mm]
   \node [v,fill=lightestpurple] (e) at (0,1) {$r^2$};
    \node [v,fill=lyellow] (h) at (0,0) {$1$};
    \draw [rr] (e) to (h);
    \draw [right hook->, thick] (.5,.7) to [bend left=0] node[above left] {\normalsize $\iota$} (1.9,1.5);
    \draw [right hook->, thick] (.5,.3) to [bend left=0] node[below left] {\normalsize $\iota'$} (1.9,-.6);
    \node at (0,-.75) {$\large\bm{N=C_2}$};
    \end{scope}
\begin{scope}[shift={(0,2.25)},scale=.8]
\tikzstyle{every node}=[font=\scriptsize]
%%
\node at (1,3.2) {\scriptsize Every $\gamma\in\Aut(D_4)$};
\node at (1,2.8) {\scriptsize stabilizes these cosets};
\draw[decorate,decoration={brace,amplitude=6pt}] 
      (0,2.15) --  (2,2.15); 
\draw[lightestgreen,fill=lightestgreen] (3,0) rectangle (4,2);
\draw[lightestblue,fill=lightestblue] (2,0) rectangle (3,2);
\draw[lightestred,fill=lightestred] (1,0) rectangle (2,2);
\draw[lightestpurple,fill=lightestpurple] (0,1) rectangle (1,2);
\draw[lyellow,fill=lyellow] (0,0) rectangle (1,1);
\draw[thick] (0,0) rectangle (4,2);
\draw[thick] (1,0) to (1,2);
\draw[thick] (2,0) to (2,2);
\draw[thick] (3,0) to (3,2);
%%
\node at (.5,1.5) {$r^2$}; \node at (1.5,1.5) {$r^3$}; \node at (2.5,1.5) {$r^2f$}; \node at (3.5,1.5) {$r^3f$};
%%
\node at (.5,.5) {$1$}; \node at (1.5,.5) {$r$}; \node at (2.5,.5) {$f$}; \node at (3.5,.5) {$rf$};
%%
\node[faded] at (.5,-.5) {\large $\mathbf{e}$};
\node[faded] at (1.5,-.42) {\large $\mathbf{h}$};
\node[faded] at (2.5,-.5) {\large $\mathbf{v}$};
\node[faded] at (3.5,-.42) {\large $\mathbf{hv}$};
%%
\draw [->>, thick] (4.5,1) to [bend left=0] node[above right] {\normalsize $\pi$} (9,-1.5);
\end{scope}
%%
\begin{scope}[shift={(0,-2.25)},scale=.8]
\tikzstyle{every node}=[font=\scriptsize]
\draw[lightestgreen,fill=lightestgreen] (3,0) rectangle (4,2);
\draw[lightestred,fill=lightestred] (2,0) rectangle (3,2);
\draw[lightestblue,fill=lightestblue] (1,0) rectangle (2,2);
\draw[lightestpurple,fill=lightestpurple] (0,1) rectangle (1,2);
\draw[lyellow,fill=lyellow] (0,0) rectangle (1,1);
\draw[thick] (0,0) rectangle (4,2);
\draw[thick] (1,0) to (1,2);
\draw[thick] (2,0) to (2,2);
\draw[thick] (3,0) to (3,2);
%%
\node at (.5,1.5) {$r^2$}; \node at (1.5,1.5) {$r^2f$}; \node at (2.5,1.5) {$r^3$}; \node at (3.5,1.5) {$r^3f$};
%%
\node at (.5,.5) {$1$}; \node at (1.5,.5) {$f$}; \node at (2.5,.5) {$r$}; \node at (3.5,.5) {$rf$};
%%
\node at (.5,2.4) {$\bm{N}$};
\node at (1.5,2.4) {$\bm{fN}$};
\node at (2.5,2.4) {$\bm{rN}$};
\node at (3.5,2.4) {$\bm{rfN}$};
%%
\node[faded] at (.5,-.5) {\large $\mathbf{e}$};
\node[faded] at (1.5,-.42) {\large $\mathbf{h}$};
\node[faded] at (2.5,-.5) {\large $\mathbf{v}$};
\node[faded] at (3.5,-.42) {\large $\mathbf{hv}$};
\draw [->>, thick] (4.5,1) to [bend left=0] node[below right] {\normalsize $\pi'$} (9,3.5);
%%
\end{scope}
\begin{scope}[shift={(8,0)},scale=2]
   \node [v] (e) at (0,1) {$e$};
    \node [v] (v) at (0,0) {$v$};
    \node [v] (h) at (1,1) {$h$};
   \node [v] (hv) at (1,0) {$hv$};
    \draw [bb] (e) to (h);
    \draw [rr] (e) to (v);
    \draw [rr] (hv) to (h);
    \draw [bb] (hv) to (v);
    \node at (.5,-.75) {$\large\bm{Q=V_4}$};
    \end{scope}
\draw [->, thick,dashed] (1.6,1.7) to [bend left=0] node[left] {\normalsize $\nexists\,\gamma$} (1.6,0);
\end{tikzpicture}
\]
Each of the three choices of the image $\pi(r)\in\{h,v,hv\}$ characterizes an extension $C_2\stackrel{\iota}{\into}D_4\stackrel{\pi}{\to}V_4$. Examples of these are shown below.
  \[
  \hspace*{-15mm}
  \begin{tikzpicture}
    \tikzstyle{coset}=[rectangle, draw, 
      minimum width=11.5mm, minimum height=6mm,font=\scriptsize]
    \tikzstyle{coset2}=[rectangle, draw, fill=conj-orange,
      minimum width=11.5mm, minimum height=6mm,font=\scriptsize]
    \tikzstyle{coset-f}=[rectangle, draw, fill=lightestblue,
      minimum width=11.5mm, minimum height=6mm,font=\scriptsize]
    \tikzstyle{coset-rf}=[rectangle, draw, fill=lightestgreen,
      minimum width=11.5mm, minimum height=6mm,font=\scriptsize]
    \tikzstyle{coset-r}=[rectangle, draw, fill=lightestred,
      minimum width=11.5mm, minimum height=6mm,font=\scriptsize]
    \begin{scope}[shift={(-1,0)},scale=2]
      \node [v] (e) at (0,1) {$e$};
      \node [v] (v) at (0,0) {$v$};
      \node [v] (h) at (1,1) {$h$};
      \node [v] (hv) at (1,0) {$hv$};
      \draw [bb] (e) to (h);
      \draw [rr] (e) to (v);
      \draw [rr] (hv) to (h);
      \draw [bb] (hv) to (v);
    \end{scope}
    %%
    \begin{scope}[shift={(3,0)},scale=2,xscale=1.2]
      \node [coset] (e) at (0,1) {};
      \node [coset-f] (v) at (0,0) {$f\;\;r^2\!f$};
      \node [coset-r] (h) at (1,1) {$r\quad r^3$};
      \node [coset-rf] (hv) at (1,0) {$rf\;\,r^3\!f$};
      \draw [bb] (e) to (h);
      \draw [rr] (e) to (v);
      \draw [rr] (hv) to (h);
      \draw [bb] (hv) to (v);
      %%
      \filldraw[lyellow][] 
      (e.north) -- (e.north west) -- (e.south west)
      -- (e.south)--cycle;
      \filldraw[lightestpurple][] 
      (e.north)-- (e.north east)
      -- (e.south east)-- (e.south)--cycle;
      \draw (e.north west) -- (e.north east) -- (e.south east) -- (e.south west) -- cycle;
      \node at (0,1) {\scriptsize $1\quad r^2$};
      \node[anchor=west] at (.2,.67) {\small $r\mapsto h$};
      \node[anchor=west] at (.2,.33) {\small $f\mapsto v$};
      %%
    \end{scope}
    %%
    \begin{scope}[shift={(7.5,0)},scale=2,xscale=1.2]
      \node [coset] (e) at (0,1) {};
      \node [coset-rf] (v) at (0,0) {$rf\;\,r^3\!f$};
      \node [coset-f] (h) at (1,1) {$f\;\;r^2\!f$};
      \node [coset-r] (hv) at (1,0) {$r\quad r^3$}; 
      \draw [bb] (e) to (h);
      \draw [rr] (e) to (v);
      \draw [rr] (hv) to (h);
      \draw [bb] (hv) to (v);
      %%
      \node[anchor=west] at (.2,.67) {\small $r\mapsto hv$};
      \node[anchor=west] at (.2,.33) {\small $f\mapsto h$};
      %%
      \filldraw[lyellow][] 
      (e.north) -- (e.north west) -- (e.south west)
      -- (e.south)--cycle;
      \filldraw[lightestpurple][] 
      (e.north)-- (e.north east)
      -- (e.south east)-- (e.south)--cycle;
      \draw (e.north west) -- (e.north east) -- (e.south east) -- (e.south west) -- cycle;
      \node at (0,1) {\scriptsize $1\quad r^2$};
    \end{scope}
    %%
    \begin{scope}[shift={(12,0)},scale=2,xscale=1.2]
      \node [coset] (e) at (0,1) {};
      \node [coset-r] (v) at (0,0) {$r\quad r^3$};
      \node [coset-f] (h) at (1,1) {$f\;\;r^2\!f$};
      \node [coset-rf] (hv) at (1,0) {$rf\;\,r^3\!f$};
      \draw [bb] (e) to (h);
      \draw [rr] (e) to (v);
      \draw [rr] (hv) to (h);
      \draw [bb] (hv) to (v);
      %%
      \node[anchor=west] at (.2,.67) {\small $r\mapsto v$};
      \node[anchor=west] at (.2,.33) {\small $f\mapsto h$};
      %%
      \filldraw[lyellow][] 
      (e.north) -- (e.north west) -- (e.south west)
      -- (e.south)--cycle;
      \filldraw[lightestpurple][] 
      (e.north)-- (e.north east)
      -- (e.south east)-- (e.south)--cycle;
      \draw (e.north west) -- (e.north east) -- (e.south east) -- (e.south west) -- cycle;
      \node at (0,1) {\scriptsize $1\quad r^2$};
    \end{scope}
  \end{tikzpicture}
  \]
  Carry out the previous steps, including the visuals, to classify the
  extensions $C_2\stackrel{\iota}{\into}C_4\times C_2\stackrel{\pi}{\to}V_4$,
  and then do the same for the extensions
  $C_2\stackrel{\iota}{\into}Q_8\stackrel{\pi}{\to}V_4$. 
\end{enumerate}
    
  
\item A short exact sequence of groups is \emph{left split} if there is a ``backwards map'' $\alpha\colon G\to N$ for which $\alpha\circ \iota=\Id_N$, like the following shows.
\[
\begin{tikzpicture}[scale=1.6,baseline=(current bounding  box.center)]
%%
\begin{scope}[shift={(6,0)}]
\node (1L) at (0,0) {$1$};
\node (N) at (1,0) {$N$}; 
\node (G) at (2.3,0) {$G$};
\node (Q) at (3.6,0) {$H$};
\node (1R) at (4.6,0) {$1$};
\draw[->] (1L) -- (N) node[midway,above]{};
\draw[->] (N) -- (G) node[midway,above]{\footnotesize $\iota$};
\draw[->] (G) -- (Q) node[midway,above]{\footnotesize $\pi$};
\draw[->,dashed] (G) to [bend left=30]  node[midway,below]{\footnotesize $\alpha$} (N);
\draw[->] (Q) -- (1R);
\end{scope}
\end{tikzpicture}
\]
\vspace{-2mm} A picture of a left split exact sequence $1\to V_4\to C_6\times C_2\to C_3\to 1$ is shown below.
%%
%% Left split short exact sequence
%%
\[
 \newcommand\Hf{6.495} % height of C_12 
  \newcommand\He{5.196} % height of C_6  
  \newcommand\Hd{3.897} % height of C_4 
  \newcommand\Hc{2.598} % height of C_3  
  \newcommand\Hb{1.299} % height of C_2 
  \newcommand\Ha{0} % height of C_1   
 \begin{tikzpicture}[scale=.75,shorten >= -2.5pt, shorten <= -2.5pt]
    \tikzstyle{every node}=[font=\footnotesize]
    \tikzstyle{E} = [draw, gray];
    \tikzstyle{V} = [gray];
    \tikzstyle{N} = [draw, very thick, Blue];
    \tikzstyle{H} = [draw, very thick, Red];
    \tikzstyle{P} = [draw, very thick, amethyst];
    \tikzstyle{H2} = [draw, ultra thick, amethyst];
    %% 1
    \begin{scope}[shift={(-.5,0)}]
      \node[Red] (1) at (0,0) {$C_1$};
      \draw [->, thick] (.75,0) to [bend left=0] node[above] {} (2.75,0);
    \end{scope}
    %% N=C_2
    \begin{scope}[shift={(3,0)}]
      \node[Red] (G) at (0,\Hd) {$V_4$};
      \node[Red] (h) at (-.75,\Hb) {$C_2$};
      \node[Red] (hv) at (0,\Hb) {$C_2$};
      \node[Red] (v) at (.75,\Hb) {$C_2$};
      \node[Blue] (1) at (0,0) {$C_1$};
      \draw[H] (G) to (h); \draw[H] (G) to (hv); 
      \draw[H] (G) to (v);  \draw[H] (1) to (v); 
      \draw[H] (1) to (h); \draw[H] (1) to (hv); 
      \draw [right hook->, thick] (1,\Hd) to [bend left=0] node[above] {$\iota$} (4,\Hd);
    \end{scope}
    %%  G=Q_8
    \begin{scope}[shift={(9,0)}]
          \draw[->,dashed,very thick] (.25,3.45) to [bend left=10]  node[pos=.7,below]{\footnotesize $\alpha$} (-5,2);
        \node [Red] (G) at (-0,\Hf) {$C_6\times C_2$};
        \node [amethyst] (12) at (.25,\He) {$C_6$};
        \node [amethyst] (11) at (1,\He) {$C_6$};
        \node [amethyst] (01) at (1.75,\He) {$C_6$};
        \node [Blue] (10-03) at (-1,\Hd) {$V_4$};
        \node [amethyst] (02) at (1,\Hc) {$C_3$};
        \node [Blue] (10) at (-1.75,\Hb) {$C_2$};
        \node [Blue] (13) at (-1,\Hb) {$C_2$};
        \node [Blue] (03) at (-.25,\Hb) {$C_2$};
        \node [Blue] (00) at (-0,0) {$C_1$};
        \draw [P] (G) to (12); 
        \draw [P] (G) to (11); 
        \draw [P] (G) to (01);
        \draw [P] (02) to (12); 
        \draw [P] (02) to (11); 
        \draw [P] (02) to (01);
        \draw [H] (10-03) to (G);
        \draw [E] (12) to (10); 
        \draw [E] (11) to (13); 
        \draw [E] (01) to (03);
        \draw [E] (00) to (02);
        \draw [N] (10-03) to (10); 
        \draw [N] (10-03) to (13); 
        \draw [N] (10-03) to (03); 
        \draw [N] (00) to (10); 
        \draw [N] (00) to (13); 
        \draw [N] (00) to (03); 
        \draw [->>, thick] (1.5,\Hf) to [bend left=0] node[above] {$\pi$} (4.5,\Hf);
    \end{scope}
    %% Q=V_4
    \begin{scope}[shift={(14.5,0)}]
      \node[Blue] (G) at (0,\Hf) {$C_3$};
      \node[Blue] (1) at (0,\Hd) {$C_1$};
      \draw[N] (G)--(1); 
      \draw [->,thick] (.75,\Hf) to [bend left=0] node[above] {} (2.75,\Hf);
    \end{scope}
    %% 1
    \begin{scope}[shift={(18,0)}]
      \node[Blue] (Q8) at (0,\Hf) {$C_1$};
    \end{scope}
  \end{tikzpicture}
 \]

 \begin{enumerate}
 \item Show that every split exact sequence of abelian groups is left split. 
 \item Show that if a short exact sequence is left split, then it is (right) split. Must $G$ (the middle term) be abelian in this case? Justify your answer. 
 \item Give an example of a right split exact sequence that is not left split. 
 \end{enumerate}
 

%\newpage



\item Let $C:=\{aba^{-1}b^{-1}\mid a,b\in G\}$ be the set of commutators of a group $G$. Prove the following facts about the \emph{commutator subgroup}, $G'=\<C\>$.  \vspace{-2mm}
  \begin{enumerate}
  \item $\displaystyle G'=\bigcap_{C\subseteq N\normaleq\,G} N$.
  \item $G'\normaleq G$ and $G/G'$ is abelian.
  \item If $G'\leq H\leq G$, then $H\normaleq G$.
  \item If $N\normaleq G$, then $N'\normaleq G$. 
  \item If $\phi\colon G\onto H$, then $H$ is abelian if and only if $G'\leq\Ker(\phi)$. \smallskip
  \end{enumerate}


  
  
\item Prove the following, where $G$ is a finite group, $p$ a prime divisor of $|G|$, and 
\[
G'(p):=\bigcap\big\{N\normaleq G\mid G/N \mbox{ is an abelian $p$-group}\big\}.
\]
\begin{enumerate}
\item $G/G'(p)$ is an abelian $p$-group.
\item $G/G'(p)$ is isomorphic to the unique Sylow $p$-subgroup of $G/G'$.
\item $p\nmid [G'(p):G']$.
\item If $P$ is any Sylow $p$-subgroup of $G$, then $PG'(p)=G$. \smallskip
\end{enumerate}


\item For both of the following groups shown below, find all
  compositions series (up to isomorphism), the composition factors,
  the derived series, and abelianization.
  \[
  \hspace*{-1mm}
  \begin{tikzpicture}[shorten >= -3pt, shorten <= -3pt, scale=1.6]
    %%
         \tikzstyle{every node}=[font=\small]
    \begin{scope}[shift={(0,0)}]
    \node (G) at (0,3) {$\DQ_8$};
    %%
    \node (C4xC2-1) at (-3,2) {$C_4\!\times\!C_2$}; % 010,001,011
    \node (D4-1) at (-2,2) {$D_4$}; % 100,001,101    
    \node (C4xC2-2) at (-1,2) {$C_4\!\times\!C_2$}; % 100,010,110
    \node (D4-2) at (0,2) {$D_4$};  % 100,011,111
    \node (C4xC2-3) at (1,2) {$C_4\!\times\!C_2$};  % 010,101,111
    \node (D4-3) at (2,2) {$D_4$};  % 110,001,111
    \node (Q8) at (3,2) {$Q_8$};  % 110,011,101
    %%
    \node (V4-12) at (-3,1) {$V_4$};
    \node (V4-13) at (-2,1) {$V_4$};
    \node (V4-23) at (-1,1) {$V_4$};
    \node (Z) at (0,1) {$C_4$};
    \node (C4-1) at (1,1) {$C_4$};
    \node (C4-2) at (2,1) {$C_4$};
    \node (C4-3) at (3,1) {$C_4$};
    %%
    \node (C2-11) at (-3.25,0) {$C_2$};
    \node (C2-12) at (-2.75,0) {$C_2$};
    \node (C2-21) at (-2.25,0) {$C_2$};
    \node (C2-22) at (-1.75,0) {$C_2$};
    \node (C2-31) at (-1.25,0) {$C_2$};
    \node (C2-32) at (-0.75,0) {$C_2$};    
    \node (-I) at (1,0) {$C_2$};
    %%
    \node (I) at (0,-1) {$C_1$};
    %%
    %%
    \draw[faded] (C4xC2-1) to (G); \draw[faded] (D4-1) to (G); \draw[faded] (C4xC2-2) to (G);
    \draw[faded] (D4-2) to (G); \draw[faded] (C4xC2-3) to (G); \draw[faded] (D4-3) to (G);
    \draw[faded] (Q8) to (G);
    %%
    \draw[faded] (C4xC2-1) to (Z); \draw[faded] (C4xC2-2) to (Z); \draw[faded] (C4xC2-3) to (Z);
    \draw[faded] (V4-12) to (D4-1); \draw[faded] (V4-12) to (D4-2); \draw[faded] (V4-12) to (C4xC2-3);
    \draw[faded] (V4-13) to (D4-1); \draw[faded] (V4-13) to (D4-3); \draw[faded] (V4-13) to (C4xC2-2);
    \draw[faded] (V4-23) to (D4-2); \draw[faded] (V4-23) to (D4-3); \draw[faded] (V4-23) to (C4xC2-1);
    %%
    \draw[faded] (C4-1) to (C4xC2-1); \draw[faded] (C4-1) to (D4-1);
    \draw[faded] (C4-2) to (C4xC2-2); \draw[faded] (C4-2) to (D4-2);
    \draw[faded] (C4-3) to (C4xC2-3); \draw[faded] (C4-3) to (D4-3);
    %%
    \draw[faded] (Q8) to (C4-1); \draw[faded] (Q8) to (C4-2); \draw[faded] (Q8) to (C4-3);
    %%
    \draw[faded] (V4-12) to (C2-11); \draw[faded] (V4-12) to (C2-12); 
    \draw[faded] (V4-13) to (C2-21); \draw[faded] (V4-13) to (C2-22); 
    \draw[faded] (V4-23) to (C2-31); \draw[faded] (V4-23) to (C2-32); 
    %%
    \draw[faded] (-I) to (V4-12); \draw[faded] (-I) to (V4-13); \draw[faded] (-I) to (V4-23);
    \draw[faded] (-I) to (C4-1); \draw[faded] (-I) to (C4-2); \draw[faded] (-I) to (C4-3);
    \draw[faded] (-I) to (Z);
    %%
    \draw[faded] (C2-11) to (I); \draw[faded] (C2-12) to (I); 
    \draw[faded] (C2-21) to (I); \draw[faded] (C2-22) to (I); 
    \draw[faded] (C2-31) to (I); \draw[faded] (C2-32) to (I);
    \draw[faded] (-I) to (I);
    %%
    \node[anchor=east] at (-4.05,3) {\normalsize \textbf{Order} $=16$};
    \node[anchor=east]  at (-4.05,2) {\normalsize $8$};
    \node[anchor=east]  at (-4.05,1) {\normalsize $4$};
    \node[anchor=east]  at (-4.05,0) {\normalsize $2$};
    \node[anchor=east]  at (-4.05,-1) {\normalsize $1$};
    %%
      \node[anchor=east] at (4.5,3) {\normalsize \textbf{Index} $=1$};
      \node[anchor=east]  at (4.5,2) {\normalsize $2$};
      \node[anchor=east]  at (4.5,1) {\normalsize $4$};
      \node[anchor=east]  at (4.5,0) {\normalsize $8$};
      \node[anchor=east]  at (4.5,-1) {\normalsize $16$};
  \end{scope}
  \end{tikzpicture}
  \]
  
  \[
  \hspace*{-1mm}
  \begin{tikzpicture}[shorten >= -3pt, shorten <= -3pt, scale=1.4,yscale=1.15]
    \begin{scope}[shift={(0,0)}]
    \tikzstyle{every node}=[font=\small]
      \node[anchor=east] at (-4.75,4) {\normalsize \textbf{Order} $=40$};
      \node[anchor=east]  at (-4.75,3.3) {\normalsize $20$};
      \node[anchor=east]  at (-4.75,2.5) {\normalsize $10$};
      \node[anchor=east]  at (-4.75,2.25) {\normalsize $8$};
      \node[anchor=east]  at (-4.75,1.75) {\normalsize $5$};
      \node[anchor=east]  at (-4.75,1.5) {\normalsize $4$};
      \node[anchor=east]  at (-4.75,.7) {\normalsize $2$};
      \node[anchor=east]  at (-4.75,.0) {\normalsize $1$};
      %%
      \node[anchor=east] at (5,4) {\normalsize \textbf{Index} $=1$};
      \node[anchor=east]  at (5,3.3) {\normalsize $2$};
      \node[anchor=east]  at (5,2.5) {\normalsize $4$};
      \node[anchor=east]  at (5,2.25) {\normalsize $5$};
      \node[anchor=east]  at (5,1.75) {\normalsize $8$};
      \node[anchor=east]  at (5,1.5) {\normalsize $10$};
      \node[anchor=east]  at (5,.7) {\normalsize $20$};
      \node[anchor=east]  at (5,.0) {\normalsize $40$};
      %%
      \node (G) at (0,4) {$C_4\times D_5$};
      \node (D6) at (-2.2,3.3) {$D_{10}$};
      \node (C12) at (0,3.3) {$C_{20}$};
      \node (Dic6) at (2.5,3.3) {$\Dic_{10}$};
      \node (C4xC2) at (2.2,2.25) {$\!\!\!{}_{\color{midgrey} 5}C_4\!\times\!C_2$};
      %%
      \node (D3-1) at (-3.0,2.5) {$D_5$};
      \node (D3-2) at (-2.2,2.5) {$D_5$};
      \node (C6) at (-1.4,2.5) {$C_{10}$};
      %%
      %%
      \node (V4) at (1.4,1.5) {${}_{\color{midgrey} 5}V_4$};
      \node (C4-1) at (2.2,1.5) {$C_4$};
      \node (C4-2) at (3.0,1.5) {${}_{\color{midgrey} 5}C_4$};
      %%
      \node (C3) at (-2.2,1.75) {$C_5$};
      \node (C2-1) at (-2.5,.7) {${}_{\color{midgrey} 5}C_2$};
      \node (C2-3) at (0,.7) {${}_{\color{midgrey} 5}C_2$};
      \node (C2-2) at (2.2,.7) {$C_2$};
      \node (1) at (0,0) {$C_1$};
      %%
      %%
      \draw[faded] (G) to (D6); \draw[faded] (G) to (C12); 
      \draw[faded] (G) to (Dic6); \draw[faded] (G) to (C4xC2); 
      %%
      \draw[faded] (D6) to (D3-1); \draw[faded] (D6) to (D3-2); 
      \draw[faded] (D6) to (C6); \draw[faded] (D6) to (V4);
      %%
      \draw[faded] (C12) to (C6); \draw[faded] (C12) to (C4-1);
      %%
      \draw[faded] (Dic6) to (C6); \draw[faded] (Dic6) to (C4-2);
      %%
      \draw[faded] (C4xC2) to (V4); \draw[faded] (C4xC2) to (C4-1);
      \draw[faded] (C4xC2) to (C4-2);
      %%
      \draw[faded] (D3-1) to (C2-1); \draw[faded] (D3-1) to (C3);
      \draw[faded] (D3-2) to (C2-3); \draw[faded] (D3-2) to (C3);
      %%
      \draw[faded] (C6) to (C3); \draw[faded] (C6) to (C2-2); 
      %%
      \draw[faded] (C4-1) to (C2-2); \draw[faded] (C4-2) to (C2-2);
      %%
      \draw[faded] (V4) to (C2-1); \draw[faded] (V4) to (C2-2); \draw[faded] (V4) to (C2-3);
      %%
      \draw[faded] (C2-1) to (1); \draw[faded] (C2-2) to (1); 
      \draw[faded] (C2-3) to (1); \draw[faded] (C3) to (1);
      %%
    \end{scope}
    \end{tikzpicture}
  \]

\end{enumerate}

\end{document}
