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\pagestyle{fancy} \lhead{{\sf Homework 12 $|$ Due April 18 (Friday)
}} \rhead{\thepage} \cfoot{{\sf Math 4120/6120 $|$ Visual Algebra $|$
    Spring 2025 $|$ M.~Macauley}}

\begin{document}

%$\;$

\begin{enumerate}

  %%-------------------------------------------------------------------

\item The subgroup lattice of the symmetric group $S_4$ is shown
  below. \vspace{-3mm}
  
  \[
  \hspace*{-2mm}
  \newcommand\Hg{6.5} % height of A_4 
  \newcommand\Hf{5.25} % height of A_4 
  \newcommand\He{4.5} % height of D_4
  \newcommand\Hd{3.75} % height of S_3 
  \newcommand\Hc{3} % height of C_4, V_4 
  \newcommand\Hb{2} % height of C_3
  \newcommand\Ha{1.5} % height of C_2 
  \begin{tikzpicture}[scale=.95, shorten >= -2pt, shorten <= -2pt]
    \begin{scope}[scale=1.45,shift={(0,0)},shorten >= -2pt, shorten <= -2pt]
      \node (S4) at (0,\Hg) {$S_4$};
      \node (A4) at (0,\Hf) {$A_4$};
      \node (D4-1) at (-3,\He) {$D_4$};
      \node (D4-2) at (-2.5,\He) {$D_4$};
      \node (D4-3) at (-2,\He) {$D_4$};
      %%
      \node (S3-1) at (2.25,\Hd) {$S_3$};
      \node (S3-2) at (2.75,\Hd) {$S_3$};
      \node (S3-3) at (3.25,\Hd) {$S_3$};
      \node (S3-4) at (3.75,\Hd) {$S_3$};
      %%
      \node (C4-1) at (-4.25,\Hc) {$C_4$};
      \node (C4-2) at (-3.75,\Hc) {$C_4$};
      \node (C4-3) at (-3.25,\Hc) {$C_4$};
      %%
      \node (V4-0) at (-2.5,\Hc) {$V_4$};
      \node (V4-1) at (-1.75,\Hc) {$V_4$};
      \node (V4-2) at (-1.25,\Hc) {$V_4$};
      \node (V4-3) at (-.75,\Hc) {$V_4$};
      %%
      \node (C3-1) at (2.25,\Hb) {$C_3$};
      \node (C3-2) at (2.75,\Hb) {$C_3$};
      \node (C3-3) at (3.25,\Hb) {$C_3$};
      \node (C3-4) at (3.75,\Hb) {$C_3$};
      %%
      \node (C2-1) at (-3,\Ha) {$C_2$};
      \node (C2-2) at (-2.5,\Ha) {$C_2$};
      \node (C2-3) at (-2,\Ha) {$C_2$};
      %%
      \node (C2-4) at (-1.25,\Ha) {$C_2$};
      \node (C2-5) at (-.75,\Ha) {$C_2$};
      \node (C2-6) at (-.25,\Ha) {$C_2$};
      \node (C2-7) at (.25,\Ha) {$C_2$};
      \node (C2-8) at (.75,\Ha) {$C_2$};
      \node (C2-9) at (1.25,\Ha) {$C_2$};
      \node (1) at (0,0) {$C_1$};
      %%
      \draw[f] (S4) to (A4);
      \draw[f] (S4) to (D4-1); \draw[f] (S4) to (D4-2);
      \draw[f] (S4) to (D4-3);
      \draw[f] (S4) to (S3-1); \draw[f] (S4) to (S3-2);
      \draw[f] (S4) to (S3-3); \draw[f] (S4) to (S3-4);
      %%
      \draw[f] (D4-1) to (V4-0); \draw[f] (D4-2) to (V4-0);
      \draw[f] (D4-3) to (V4-0);
      \draw[f] (D4-1) to (C4-1); \draw[f] (D4-2) to (C4-2);
      \draw[f] (D4-3) to (C4-3);
      \draw[f] (D4-1) to (V4-1); \draw[f] (D4-2) to (V4-2);
      \draw[f] (D4-3) to (V4-3);
      %%
      \draw[f] (C4-1) to (C2-1); \draw[f] (C4-2) to (C2-2);
      \draw[f] (C4-3) to (C2-3);
      %%
      \draw[f] (V4-0) to (C2-1); \draw[f] (V4-0) to (C2-2);
      \draw[f] (V4-0) to (C2-3);
      \draw[f] (V4-1) to (C2-1); \draw[f] (V4-2) to (C2-2);
      \draw[f] (V4-3) to (C2-3);
      \draw[f] (V4-1) to (C2-4); \draw[f] (V4-2) to (C2-6);
      \draw[f] (V4-3) to (C2-8);
      \draw[f] (V4-1) to (C2-5); \draw[f] (V4-2) to (C2-7);
      \draw[f] (V4-3) to (C2-9);
      %%
      \draw[f] (C2-1) to (1); \draw[f] (C2-2) to (1); \draw[f] (C2-3) to (1);
      \draw[f] (C2-4) to (1); \draw[f] (C2-5) to (1); \draw[f] (C2-6) to (1);
      \draw[f] (C2-7) to (1); \draw[f] (C2-8) to (1); \draw[f] (C2-9) to (1);
      %%
      \draw[f] (A4) to (V4-0); \draw[f] (A4) to (C3-1); \draw[f] (A4) to (C3-2);
      \draw[f] (A4) to (C3-3); \draw[f] (A4) to (C3-4);
      %%
      \draw[f] (S3-1) to (C3-1); \draw[f] (S3-2) to (C3-2);
      \draw[f] (S3-3) to (C3-3); \draw[f] (S3-4) to (C3-4);
      %%
      \draw[f] (C3-1) to (1); \draw[f] (C3-2) to (1);
      \draw[f] (C3-3) to (1); \draw[f] (C3-4) to (1);
      %%
      \draw[f] (S3-1) to (C2-4); \draw[f] (S3-1) to (C2-6);
      \draw[f] (S3-1) to (C2-8);
      \draw[f] (S3-2) to (C2-5); \draw[f] (S3-2) to (C2-6);
      \draw[f] (S3-2) to (C2-9);
      \draw[f] (S3-3) to (C2-4); \draw[f] (S3-3) to (C2-7);
      \draw[f] (S3-3) to (C2-9);
      \draw[f] (S3-4) to (C2-5); \draw[f] (S3-4) to (C2-7);
      \draw[f] (S3-4) to (C2-8);
      %%
      \node[anchor=east] at (5,\Hg) {\textbf{Order} $=24$};
      \node[anchor=east] at (5,\Hf) {$12$};
      \node[anchor=east]  at (5,\He) {$8$};
      \node[anchor=east]  at (5,\Hd) {$6$};
      \node[anchor=east]  at (5,\Hc) {$4$};
      \node[anchor=east]  at (5,\Hb) {$3$};
      \node[anchor=east]  at (5,\Ha) {$2$};
      \node[anchor=east]  at (5,0) {$1$};
      %%
      \node[anchor=east] at (-5,\Hg) {\textbf{Index} $=1$};
      \node[anchor=east]  at (-5,\Hf) {$2$};
      \node[anchor=east]  at (-5,\He) {$3$};
      \node[anchor=east]  at (-5,\Hd) {$4$};
      \node[anchor=east]  at (-5,\Hc) {$6$};
      \node[anchor=east]  at (-5,\Hb) {$8$};
      \node[anchor=east]  at (-5,\Ha) {$12$};
      \node[anchor=east]  at (-5,0) {$24$};
    \end{scope}
  \end{tikzpicture}
  \]
  \begin{enumerate}
  \item Partition the subgroups into conjugacy classes. Carefully and
    completely justify your answers using the Sylow theorems, without
    making reference to cycle type.
  \item For each conjugacy class $\cl_G(H)$, find the isomorphism type of the
    normalizer $N_G(H)$.
  \item Using the GroupNames website, make a table of all 15 groups of
    order $24$, the number of subgroups, and basic information
    about their Sylow $p$-subgroups (number and isomorphism
    type). Write down at least one observation that you find
    interesting.
  \item Which groups are \emph{not} an internal
    direct or semidirect product of Sylow subgroups?
  \item None of the following groups are among the 15 listed on
    GroupNames: $D_6\times C_2$, $C_6\rtimes C_4$, $C_6\rtimes C_2^2$,
    $C_4\rtimes C_6$, $C_3\rtimes C_2^3$, $C_2^3\rtimes C_3$,
    $C_2^2\rtimes C_6$, $C_3\rtimes Q_8$, $Q_8\rtimes C_3$,
    $C_4\rtimes S_3$. Find which of the 15 each is isomorphic to, and add
    this this to your table under the ``alias(es)'' column. \bigskip
  \end{enumerate}

   %%--------------------------------------------------------------------
  
\item Show that there are no simple groups of the following order. 
  \begin{multicols}{4}
    \begin{enumerate}[(i)]
    \item $45=3^2\cdot 5$
    \item $56=2^3\cdot 7$
    \item $108=2^2\cdot 3^3$
    \item $p^n\quad (n>1)$.
    \end{enumerate}
  \end{multicols}
      [\emph{Hint}: For Part~(d), first let $G$ act on iteslf by
    conjugation, and deduce that $|Z(G)|>1$.] \bigskip
  
  %%--------------------------------------------------------------------

\item After $A_5$, the next smallest nonabelian simple group is
  $G=\GL_3(\Z_2)$, the invertible $3\times 3$ binary matrices. It has
  order $168=2^3\cdot 3\cdot 7$, and its subgroup diagram is shown below. 
  \[
  \hspace*{-8mm}
  \begin{tikzpicture}
    \newcommand\Hk{6} % height of GL_3(Z_2): 168 
    \newcommand\Hj{4.7} % height of S_4: 24
    \newcommand\Hi{4.4} % height of C_7 x| C_3: 21
    \newcommand\Hh{3.7} % height of A_4: 12  
    \newcommand\Hg{2.8} % height of D_4: 8
    \newcommand\Hf{2.55} % height of C_7: 7 
    \newcommand\He{2.25} % height of S_3: 6 
    \newcommand\Hd{1.7} % height of C_4: 4
    \newcommand\Hc{1.4} % height of C_3: 3
    \newcommand\Hb{.75} % height of C_2: 2
    \newcommand\Ha{0} % height of C_1   
    \tikzstyle{gr} = [draw, f]
    %%
    \begin{scope}[shift={(8,0)},scale=1.55,yscale=1.1]
      %\tikzstyle{every node}=[font=\small]
      \node(G) at (0,\Hk) {$\GL_3(\Z_2)$};
      \node(C7xC3) at (-3,\Hi) {${}_{\color{midgray} 8}C_7\rtimes C_3$};
      \node(S4-1) at (0,\Hj) {${}_{\color{midgray} 7}S_4$};
      \node(S4-2) at (3,\Hj) {${}_{\color{midgray} 7}S_4$};
      \node(A4-1) at (0,\Hh) {${}_{\color{midgray} 7}A_4$};
      \node(A4-2) at (3.5,\Hh) {${}_{\color{midgray} 7}A_4$};
      \node(D4) at (2.5,\Hg) {${}_{\color{midgray} 21}D_4$};
      \node(C7) at (-3.5,\Hf) {${}_{\color{midgray} 8}C_7$};
      \node(S3) at (-1.4,\He) {${}_{\color{midgray} 28}S_3$};
      \node(C4) at (0,\Hd) {${}_{\color{midgray} 21}C_4$};
      \node(V4-1) at (1.5,\Hd) {${}_{\color{midgray} 7}V_4$};
      \node(V4-2) at (3,\Hd) {${}_{\color{midgray} 7}V_4$};
      \node(C3) at (-1.4,\Hc) {${}_{\color{midgray} 28}C_3$};
      \node(C2) at (0,\Hb) {${}_{\color{midgray} 21}C_2$};
      \node(1) at (0,\Ha) {$C_1$};
      \draw[gr](G) -- (C7xC3); 
      \draw[gr](G) -- (S4-1); 
      \draw[gr](G) -- (S4-2);
      \draw[gr](C7xC3) -- (C7); \draw[gr](C7xC3) -- (C3);
      \draw[gr](S4-1) -- (S3); \draw[gr](S4-1) -- (A4-1); 
      \draw[gr](S4-1) -- (D4); \draw[gr](S4-2) -- (D4);
      \draw[gr](S4-2) -- (S3); \draw[gr](S4-2) -- (A4-2); 
      \draw[gr](C7) -- (1);
      \draw[gr](S3) -- (C3); \draw[gr](S3) -- (C2); 
      \draw[gr](A4-1) -- (C3); \draw[gr](A4-1) -- (V4-1);  
      \draw[gr](A4-2) -- (C3); \draw[gr](A4-2) -- (V4-2);  
      \draw[gr](D4) -- (C4);
      \draw[gr](D4) -- (V4-1); \draw[gr](D4) -- (V4-2);
      \draw[gr](C4) -- (C2); 
      \draw[gr](V4-1) -- (C2); \draw[gr](V4-2) -- (C2);
      \draw[gr](C3) -- (1); \draw[gr](C2) -- (1);  
      \node[anchor=east] at (4.75,\Hk) {\textbf{Order} $=168$};
      \node[anchor=east] at (4.75,\Hj) {$24$};
      \node[anchor=east] at (4.75,\Hi) {$21$};
      \node[anchor=east] at (4.75,\Hh) {$12$};
      \node[anchor=east] at (4.75,\Hg) {$8$};
      \node[anchor=east] at (4.75,\Hf) {$7$};
      \node[anchor=east] at (4.75,\He) {$6$};
      \node[anchor=east] at (4.75,\Hd) {$4$};
      \node[anchor=east] at (4.75,\Hc) {$3$};
      \node[anchor=east] at (4.75,\Hb) {$2$};
      \node[anchor=east] at (4.75,\Ha) {$1$};
      %%
      \node[anchor=east] at (-4.25,\Hk) {\textbf{Index} $=1$};
      \node[anchor=east] at (-4.25,\Hj) {$7$};
      \node[anchor=east] at (-4.25,\Hi) {$8$};
      \node[anchor=east] at (-4.25,\Hh) {$14$};
      \node[anchor=east] at (-4.25,\Hg) {$21$};
      \node[anchor=east] at (-4.25,\Hf) {$24$};
      \node[anchor=east] at (-4.25,\He) {$28$};
      \node[anchor=east] at (-4.25,\Hd) {$42$};
      \node[anchor=east] at (-4.25,\Hc) {$56$};
      \node[anchor=east] at (-4.25,\Hb) {$84$};
      \node[anchor=east] at (-4.25,\Ha) {$168$};
    \end{scope}
  \end{tikzpicture}
  \]  
  \begin{enumerate}
  \item Color-code the $p$-subgroups, then draw arrows
    from each $\cl(H)$ to $\cl(N(H))$.
  \item Show that there is a non-trivial homomorphism $\phi\colon\!
    \GL_3(\Z_2)\to S_8$.
  \item Show that this homomorphism must be an embedding, and conclude
    that the order-40320 group $S_8$ has at least one subgroup
    isomorphic to $\GL_3(\Z_2)$.
  \item Show that every such subgroup of $S_8$ additionally must be
    contained in $A_8$. \medskip
  \end{enumerate}
  
  %%--------------------------------------------------------------------
  
\item In this problem, you will classify all groups of order $18$. You
  are encouraged to consult the LMFDB.
  \begin{enumerate}
  \item Find all abelian groups of order $18$, up to isomorphism.
  \item Let $G$ be a nonabelian group of order $18$. Show that $G$
    must have a normal Sylow $p$-subgroup, for some $p$.
  \item Prove that $G$ must be a semidirect product of two of its
    Sylow $p$-subgroups.
  \item Find all semidirect products $G\cong P\rtimes_\theta Q$ by
    constructing all homomorphisms $\theta\colon Q\to\Aut(P)$, where
    $P$ and $Q$ are Sylow subgroups for different
    primes. [\emph{Useful fact}: According to the LMDFB, $\Aut(C_3^2)$
      is the order-48 group $\GL_2(\Z_3)$.]
  \item There are three nonabelian groups of order $18$: the dihedral
    group $D_6$, the direct product $C_3\times D_3$, and $C_3\rtimes
    D_3$. For each semidirect product in the previous part, determine
    which of these it is isomorphic to, with justification.
  \end{enumerate}

  
  \newpage
  
  %%--------------------------------------------------------------------
  
\item The alternating group $A_6$ is the third smallest nonabelian
  simple group. It has order $6!/2=360=2^3\cdot 3^2\cdot 5$, and $501$
  subgroups contained in $22$ conjugacy classes.
  
  \[
  \hspace*{-10mm}
  \begin{tikzpicture}[scale=1.5,yscale=1.6]
    \newcommand\Ho{5.1} % height of A_6: 360
    \newcommand\Hn{4.4} % height of A_5: 60
    \newcommand\Hm{4.2} % height of C_3^2 x| C_4: 36
    \newcommand\Hl{4} % height of S_4: 24
    \newcommand\Hk{3.3} % height of C_3 x| S_3: 18
    \newcommand\Hj{3.1} % height of A_4: 12
    \newcommand\Hi{2.9} % height of D_5: 10
    \newcommand\Hh{2.7} % height of C_3^2: 9  
    \newcommand\Hg{2.4} % height of D_4: 8
    \newcommand\Hf{2} % height of S_3: 6 
    \newcommand\He{1.8} % height of C_5: 5 
    \newcommand\Hd{1.6} % height of C_4: 4
    \newcommand\Hc{1} % height of C_3: 3
    \newcommand\Hb{.8} % height of C_2: 2
    \newcommand\Ha{0} % height of C_1   
    \tikzstyle{gr} = [draw, faded]                 
    %%
    \begin{scope}[shift={(0,0)},scale=1]
      \node(G) at (0,\Ho) {$A_6$};
      \node(C3^2xC4) at (-3.3,\Hm) {${}_{\color{midgray} 10}C_3^2\!\rtimes\!C_4$};
      \node(A5-1) at (-1.5,\Hn) {${}_{\color{midgray} 6}A_5$};
      \node(S4-1) at (0,\Hl) {${}_{\color{midgray} 15}S_4$};
      \node(A5-2) at (1.5,\Hn) {${}_{\color{midgray} 6}A_5$};
      \node(S4-2) at (3.1,\Hl) {${}_{\color{midgray} 15}S_4$};
      \node(C3xS3) at (-2.7,\Hk) {${}_{\color{midgray} 10}C_3\!\rtimes\!S_3$};
      \node(A4-1) at (-1,\Hj) {${}_{\color{midgray} 15}A_4$};
      \node(D4) at (0,\Hg) {${}_{\color{midgray} 45}D_4$};
      \node(D5) at (1.5,\Hi) {${}_{\color{midgray} 36}D_5$};
      \node(A4-2) at (3.75,\Hj) {${}_{\color{midgray} 15}A_4$};
      \node(C4) at (-3.75,\Hd) {${}_{\color{midgray} 45}C_4$};
      \node(C3^2) at (-2.5,\Hh) {${}_{\color{midgray} 10}C_3^2$};
      \node(S3-1) at (-1.25,\Hf) {${}_{\color{midgray} 60}S_3$};
      \node(V4-1) at (0,\Hd) {${}_{\color{midgray} 15}V_4$};
      \node(C5) at (1.25,\He) {${}_{\color{midgray} 36}C_5$};
      \node(S3-2) at (2.65,\Hf) {${}_{\color{midgray} 60}S_3$};
      \node(V4-2) at (3.75,\Hd) {${}_{\color{midgray} 15}V_4$};
      \node(C3-1) at (-2.5,\Hc) {${}_{\color{midgray} 20}C_3$};
      \node(C2) at (0,\Hb) {${}_{\color{midgray} 45}C_2$};
      \node(C3-2) at (2.65,\Hc) {${}_{\color{midgray} 20}C_3$};
      \node(1) at (0,\Ha) {$C_1$};
      %%
      \draw[gr] (G) -- (C3^2xC4);
      \draw[gr] (G) -- (A5-1); \draw[gr] (G) -- (A5-2);
      \draw[gr] (G) -- (S4-1); \draw[gr] (G) -- (S4-2);
      %%
      \draw[gr] (C3^2xC4) -- (C4); \draw[gr] (C3^2xC4) -- (C3xS3);
      \draw[gr] (A5-1) -- (A4-1); \draw[gr] (A5-1) -- (S3-1);
      \draw[gr] (A5-2) -- (A4-2); \draw[gr] (A5-2) -- (S3-2);
      \draw[gr] (A5-1) -- (D5); \draw[gr] (A5-2) -- (D5);
      \draw[gr] (S4-1) -- (A4-1); \draw[gr] (S4-2) -- (A4-2);
      \draw[gr] (S4-1) -- (D4); \draw[gr] (S4-2) -- (D4);
      \draw[gr] (S4-1) -- (S3-1); \draw[gr] (S4-2) -- (S3-2);
      %%
      \draw[gr] (C3xS3) -- (C3^2); 
      \draw[gr] (C3xS3) -- (S3-1); \draw[gr] (C3xS3) -- (S3-2);
      \draw[gr] (A4-1) -- (C3-1); \draw[gr] (A4-2) -- (C3-2);
      \draw[gr] (A4-1) -- (V4-1); \draw[gr] (A4-2) -- (V4-2);   
      \draw[gr] (D4) -- (C4); \draw[gr] (D4) -- (V4-1);
      \draw[gr] (D4) -- (V4-2);
      \draw[gr] (D5) -- (C2); \draw[gr] (D5) -- (C5); 
      %%
      \draw[gr] (C4) -- (C2); \draw[gr] (C3^2) -- (C3-1);
      \draw[gr] (C3^2) -- (C3-2);
      \draw[gr] (V4-1) -- (C2); \draw[gr] (V4-2) -- (C2);
      \draw[gr] (S3-1) -- (C3-1);  \draw[gr] (S3-2) -- (C3-2);
      \draw[gr] (S3-1) -- (C2);  \draw[gr] (S3-2) -- (C2);
      \draw[gr] (C5) -- (1);
      %%
      \draw[gr] (C3-1) -- (1); \draw[gr] (C3-2) -- (1); );
      \draw[gr] (C2) -- (1);
      %%
      \node[anchor=east] at (4.75,\Ho) {\textbf{Order} $=360$};
      \node[anchor=east] at (4.75,\Hn) {$60$};
      \node[anchor=east] at (4.75,\Hm) {$36$};
      \node[anchor=east] at (4.75,\Hl) {$24$};
      \node[anchor=east] at (4.75,\Hk) {$18$};
      \node[anchor=east] at (4.75,\Hj) {$12$};
      \node[anchor=east] at (4.75,\Hi) {$10$};
      \node[anchor=east] at (4.75,\Hh) {$9$};
      \node[anchor=east] at (4.75,\Hg) {$8$};
      \node[anchor=east] at (4.75,\Hf) {$6$};
      \node[anchor=east] at (4.75,\He) {$5$};
      \node[anchor=east] at (4.75,\Hd) {$4$};
      \node[anchor=east] at (4.75,\Hc) {$3$};
      \node[anchor=east] at (4.75,\Hb) {$2$};
      \node[anchor=east] at (4.75,\Ha) {$1$};
      %%
      \node[anchor=east] at (-4.5,\Ho) {\textbf{Index} $=1$};
      \node[anchor=east] at (-4.5,\Hn) {$6$};
      \node[anchor=east] at (-4.5,\Hm) {$10$};
      \node[anchor=east] at (-4.5,\Hl) {$15$};
      \node[anchor=east] at (-4.5,\Hk) {$20$};
      \node[anchor=east] at (-4.5,\Hj) {$30$};
      \node[anchor=east] at (-4.5,\Hi) {$36$};
      \node[anchor=east] at (-4.5,\Hh) {$40$};
      \node[anchor=east] at (-4.5,\Hg) {$45$};
      \node[anchor=east] at (-4.5,\Hf) {$60$};
      \node[anchor=east] at (-4.5,\He) {$72$};
      \node[anchor=east] at (-4.5,\Hd) {$90$};
      \node[anchor=east] at (-4.5,\Hc) {$120$};
      \node[anchor=east] at (-4.5,\Hb) {$180$};
      \node[anchor=east] at (-4.5,\Ha) {$360$};
    \end{scope}
  \end{tikzpicture}
  \]
  \begin{enumerate}
  \item Distinguish the $p$-subgroups by colors on the lattice.
  \item For each non-singleton conjugacy class $\cl(H)$, draw an arrow from it
    to $\cl(N(H))$, the conjugacy class of its normalizer. 
  \item Now, let $G$ be an unknown group of order $90=2\cdot 3^2\cdot 5$.
    \begin{enumerate}[(i)]
    \item Show that if $G$ has a non-normal Sylow $5$-subgroup, then 
      there is a non-trivial homomorphism $\phi\colon G\to S_6$.
    \item Show that if $\phi(G)$ is contained in the simple group $A_6$,
      then $\phi$ cannot be injective.
    \item Explain why this implies that $G$ cannot be simple.
    \item Find all possibilities for $n_2$, $n_3$, and $n_5$, where
      $n_p$ is the number of Sylow $p$-subgroups of $G$. Then, using
      GroupNames or LMFDB, make a list of all groups of order $90$,
      and write down the actual vaules of $n_2$, $n_3$, and $n_5$ for
      each, as well as the isomorphism type of the Sylow
      $3$-subgroup(s) -- either $C_9$ or $C_3^2$. Does anything
      surprise you about this?
    \end{enumerate}
  \end{enumerate} 
  
  %%--------------------------------------------------------------------  

\end{enumerate}

\end{document} 
    
  
 
