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\pagestyle{fancy} \lhead{{\sf Homework 5 $|$ Due February 24 (Friday)
}} \rhead{\thepage} \cfoot{{\sf Math 4130 $|$ Visual Algebra II $|$
    Spring 2023 $|$ M.~Macauley}}

\begin{document}

$\;$ \vspace{-6mm}

\begin{enumerate}

  
\item Consider the commutative diagram of exact sequences shown at left.
  \[
  \begin{tikzpicture}[scale=1.75]
    \begin{scope}[shift={(0,0)}]
      \node (1NW) at (0,1) {$1$};
      \node (N1) at (1,1) {$N_1$};
      \node (G1) at (2,1) {$G_1$};
      \node (Q1) at (3,1) {$Q_1$};
      \node (1NE) at (4,1) {$1$};
      \node (1SW) at (0,0) {$1$};
      \node (N2) at (1,0) {$N_2$};
      \node (G2) at (2,0) {$G_2$};
\node (Q2) at (3,0) {$Q_2$};
\node (1SE) at (4,0) {$1$};
\draw[->] (1NW) -- (N1) node[midway,above]{};
\draw[->] (N1) -- (G1) node[midway,above]{\footnotesize $\iota_1$};
\draw[->] (G1) -- (Q1) node[midway,above]{\footnotesize $\pi_1$};
\draw[->] (Q1) -- (1NE) node[midway,above]{};
%%
\draw[->] (1SW) -- (N2) node[midway,above]{};
\draw[->] (N2) -- (G2) node[midway,above]{\footnotesize $\iota_2$};
\draw[->] (G2) -- (Q2) node[midway,above]{\footnotesize $\pi_2$};
\draw[->] (Q2) -- (1SE) node[midway,above]{};
%%
\draw[->] (N1) -- (N2) node[midway,right]{\footnotesize $\nu$};
\draw[->] (G1) -- (G2) node[midway,right]{\footnotesize $\gamma$};
\draw[->] (Q1) -- (Q2) node[midway,right]{\footnotesize $\kappa$};
\end{scope}
    \begin{scope}[shift={(4.85,.5)}]
      \node (1L) at (0,0) {$1$};
      \node (N) at (1,0) {$N$};
      \node (G1) at (2,.707) {$G$};
      \node (G2) at (2,-.707) {$G$};
      \node (Q) at (3,0) {$Q$};
      \node (1R) at (4,0) {$1$};
      \draw[->] (1L) -- (N) node[midway,above]{};
      \draw[right hook->] (N) -- (G1) node[midway,above left]{\footnotesize $\iota$};
      \draw[right hook->] (N) -- (G2) node[midway,below left]{\footnotesize $\iota'$};
      \draw[->>] (G1) -- (Q) node[midway,above right]{\footnotesize $\pi$};
      \draw[->>] (G2) -- (Q) node[midway,below right]{\footnotesize $\pi'$};
      \draw[->] (Q) -- (1R);
      \draw[->,dashed] (G1) -- (G2) node[midway, right]{\footnotesize $\gamma$};
    \end{scope}
\end{tikzpicture}
\]
\begin{enumerate}
\item Prove that if $\nu$ and $\kappa$ are isomorphisms, then $\gamma$
  is as well.
\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
  above (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 Consider a sequence of homomorphisms between groups:
  \[
  \begin{tikzpicture}[scale=1.75]
    \begin{scope}[shift={(0,0)}]
      \node (1-L) at (0,0) {$1$};
      \node (A) at (1,0) {$A$};
      \node (B) at (2,0) {$B$};
      \node (C) at (3,0) {$C$};
      \node (1-R) at (4,0) {$1$};
      \draw[->] (1-L) -- (A) node[midway,above]{};
      \draw[->] (A) -- (B) node[midway,above]{\footnotesize $f$};
      \draw[->] (B) -- (C) node[midway,above]{\footnotesize $g$};
      \draw[->] (C) -- (1-R) node[midway,above]{};
    \end{scope}
  \end{tikzpicture}
  \]
  This sequence is said to be \emph{exact at $B$} if
  $\Image(f)=\Ker(g)$. We can define exactness at $A$ and $C$
  analogously. The entire sequence is exact if it is exact at $A$,
  $B$, and $C$. Show that this implies that $f$ is
  injective and $g$ is surjective. \\


\item The exact sequence below demonstrates the construction of a group $G$ of order $16$, as a central (and hence abelian), nonsplit extension of $Q=Q_8$ by $N=C_2$.
 \[
  %\hspace*{-3em}
  \begin{tikzpicture}[shorten >= -2pt, shorten <= -2pt,scale=1.4]
    \tikzstyle{E} = [draw, gray];
    \tikzstyle{V} = [gray];
    \tikzstyle{N} = [draw, very thick, Blue];
    \tikzstyle{H} = [draw, very thick, Red];
    \tikzstyle{P} = [draw, ultra thick, amethyst];
    \tikzstyle{every node}=[font=\small]
    \begin{scope}[shift={(0,0)}]
      \node[Red] (r4) at (0,1) {$C_2$};
      \node[Blue] (1) at (0,0) {$C_1$};
      \draw[H] (1) to (r4);
      \draw [right hook->, thick] (.75,1) to [bend left=0] node[above] {$\iota$} (3,1);
    \end{scope}
    %%
    \begin{scope}[shift={(5,0)},xscale=.75]
      \node[Red] (G) at (0,4) {$G$};
      \node[Red] (C4xC2-1) at (-1.5,3) {$C_4\!\times\!C_2$};
      \node[Red] (C4xC2-2) at (0,3) {$C_4\!\times\!C_2$};
      \node[Red] (C4xC2-3) at (1.5,3) {$C_4\!\times\!C_2$};
      \node[V] (C4-1) at (-3,2) {$C_4$};
      \node[V] (C4-2) at (-2,2) {$C_4$};
      \node[V] (C4-3) at (-1.33,2) {$C_4$};
      \node[V] (C4-4) at (-.67,2) {$C_4$};
      \node[V] (C4-5) at (1.5,2) {$C_4$};
      \node[V] (C4-6) at (3,2) {$C_4$};
      \node[Red] (V4) at (0,2) {$\bm{V_4}$};
      \node[V] (C2-1) at (-1.5,1) {$C_2$};
      \node[Blue] (C2-2) at (0,1) {$C_2$};
      \node[V] (C2-3) at (1.5,1) {$C_2$};
      \node[Blue] (1) at (0,0) {$C_1$};
      \draw[H] (G) to (C4xC2-1); \draw[H] (G) to (C4xC2-2); \draw[H] (G) to (C4xC2-3);
      \draw[E] (C4xC2-1) to (C4-1);
      \draw[E] (C4xC2-1) to (C4-2);
      \draw[H] (C4xC2-1) to (V4);
      \draw[E] (C4xC2-2) to (C4-3);
      \draw[E] (C4xC2-2) to (C4-4);
      \draw[H] (C4xC2-2) to (V4); 
      \draw[E] (C4xC2-3) to (C4-5);
      \draw[E] (C4xC2-3) to (C4-6);
      \draw[H] (C4xC2-3) to (V4);
      \draw[E] (C4-1) to (C2-1); \draw[E] (C4-2) to (C2-1);
      \draw[E] (C4-3) to (C2-1); \draw[E] (C4-4) to (C2-1);
      \draw[E] (V4) to (C2-1); \draw[H] (V4) to (C2-2); 
      \draw[E] (V4) to (C2-3);
      \draw[E] (C4-5) to (C2-3); \draw[E] (C4-6) to (C2-3);
      \draw[E] (C2-1) to (1); \draw[N] (C2-2) to (1); 
      \draw[E] (C2-3) to (1);
      \draw [->>, thick] (2.5,1) to [bend left=0] node[above] {$\pi$} (5.5,1);
    \end{scope}
    %%
    \begin{scope}[shift={(10,0)},xscale=.75]
      \node[Blue] (G) at (0,4) {$Q_8$};
      \node[Blue] (C4xC2-1) at (-1.5,3) {$C_4$};
      \node[Blue] (C4xC2-2) at (0,3) {$C_4$};
      \node[Blue] (C4xC2-3) at (1.5,3) {$C_4$};
      \node[Blue] (V4) at (0,2) {$\bm{C_2}$};
      \node[Blue] (C2-2) at (0,1) {$C_1$};
      \draw[N] (G) to (C4xC2-1); \draw[N] (G) to (C4xC2-2); \draw[N] (G) to (C4xC2-3);
      \draw[N] (C4xC2-1) to (V4);
      \draw[N] (C4xC2-2) to (V4); 
      \draw[N] (C4xC2-3) to (V4);
      \draw[N] (V4) to (C2-2); 
    \end{scope}
    %%
  \end{tikzpicture}
  \]
  Though there are more up to equivalence, up to isomorphism of the
  factors, the group $G$ can be constructed as an extension of $Q$
  by $N$ in six distinct ways, where neither factor is trivial. Create
  an exact sequence with subgroup lattices, like the one above, for
  the other five. For each, decide whether it is split, abelian, and
  central. Note that it can be more than one of these, or none at all.
  
\end{enumerate}

\end{document}
