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\begin{document}

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\noindent
\parbox{2.25in}{\bf Math 4720 \& 5210}
\hfill {\Large \bf Homework \#4} \hfill
\parbox{2.25in}{\bf \hfill Due: Mon., Feb. $18^\mathrm{th}$, 2019}

\vspace{0.05in}

%\noindent {\bf\large Name:} \underline{\qquad\sc\large This is my fabulous name! \qquad}
%
%\vspace{0.2in}

\begin{description}
\item[\#1 A Classy Problem] Write down the class equation for $S_4$. {\bf [Grad. Problem]} Write $A_4$'s class equation.

\item[\#2 Normalizing Group Theory] We know that $G$ acts on itself via conjugation: $g \bdot x = gxg^{-1}$. This induces an action of $G$ on its powerset $\mathcal{P}(G) = \{ A \;|\; A \subseteq G \}$ where $g \bdot A = gAg^{-1} = \{ gag^{-1} \;|\; a \in A \}$.

\begin{enumerate}[(a)]

\item Show that the action of $G$ on its powerset is indeed an action (check our two axioms).  

{\it Note:} Here the stabilizers have a special name: $N_G(A) = \{ g \in G \;|\; gAg^{-1}=A\}$ is the {\it normalizer} of $A \subseteq G$ in $G$.

\item Let $\mathcal{S} = \{ H \;|\; H \mbox{ is a subgroup of } G\}$. Show that $G$ acts on $\mathcal{S}$ (via conjugation). 

{\it Hint:} You already know that the group action axioms hold (on all subsets). You essentially just need to verify a kind of closure: show $H$ a subgroup implies $gHg^{-1}$ is a subgroup.

{\it Note:} If two subgroup lie in the same orbit, we call them {\it conjugate subgroups}. That is $H$ and $K$ are conjugate subgroups iff there is some $g \in G$ such that $K = gHg^{-1}$.

\item In the previous part you showed that $G$ acts on its subgroups via conjugation. What does it mean if the orbit of a subgroup $H$ is $\{ H \}$ (a singleton orbit)?

\item Let $H$ be a subgroup of $G$. Suppose that $K$ is also a subgroup and that $H \triangleleft K$ (i.e., $H$ is normal in $K$). Show that $K \subseteq N_G(H)$.

{\it Note:} This says that the largest subgroup in which $H$ remains a normal subgroup is the normalizer of $H$.

\item Find the normalizers of all of the subgroups of $S_3$.  \qquad {\it Hint:} We always have $H \subseteq N_G(H)$.
\item Find the normalizers of all of the subgroups of $D_4$.
\item  {\bf [Grad. Problem]} Find $N_{D_8}(\langle x^4,y \rangle)$.
\end{enumerate}

\item[\#3 Permutatin' Sets] Let $X = \{ 1,2,\dots, n\}$ and $1 \leq k \leq n$. Let $A \subseteq X$ and $\sigma \in S_n$. Define $\sigma \bdot A = \{ \sigma(a) \;|\; a \in A\}$.
For example, $\sigma \bdot \{x,y,z\} = \{\sigma(x),\sigma(y),\sigma(z) \}$. 

\begin{enumerate}[(a)]
\item  {\bf [Grad. Problem]} Let $\mathcal{K} = \{ A \subseteq X \;|\; A \mbox{ has cardinality } k \}$. Show that $S_n$ acts on $\mathcal{K}$ via the operation defined above. 

{\it Note:} Don't forget to check ``closure'' as well as the other axioms.

\item Consider $X = \{1,2,3,4\}$. Describe the action of $\sigma=(12)$ and $\tau=(123)$ on the 2-element subsets of $X$ (there are 6 such subset). 

\item Find the orbit of $A = \{1,3\}$ under the action of $S_4$ on the 2-element subsets of $X$. Also, find the stabilizer of $A$.

{\it Note:} The orbit will be a {\it set of sets}.
\end{enumerate}

\end{description}

To help determine normalizers and for future reference, I have included several subgroup lattices. First, we have the subgroup lattices for $D_4 = \langle x,y \;|\; x^4=1,y^2=1,(xy)^2=1 \rangle = \{1,x,x^2,x^3,y,xy,x^2y,x^3y\}$ and $S_3$:

%\vspace{-0.2in}

\begin{center}
\begin{tikzpicture}[node distance=1.25cm]
\node(grp){$D_4$};

\node(rot)[below of = grp]{$\langle x \rangle$};
\node(sg1)[left of = rot]{$\langle x^2,y \rangle$};
\node(sg2)[right of = rot]{$\langle x^2,xy \rangle$};

\node(cent)[below of = rot]{$\langle x^2 \rangle$};
\node(sg3)[left of = cent]{$\langle x^2y \rangle$};
\node(sg4)[left of = sg3]{$\langle y \rangle$};
\node(sg5)[right of = cent]{$\langle xy \rangle$};
\node(sg6)[right of = sg5]{$\langle x^3y \rangle$};

\node(triv)[below of = cent]{$\{ 1 \}$};

\draw(grp)--(sg1);
\draw(grp)--(rot);
\draw(grp)--(sg2);

\draw(rot)--(cent);

\draw(sg1)--(sg3);
\draw(sg1)--(sg4);
\draw(sg1)--(cent);

\draw(sg2)--(sg5);
\draw(sg2)--(sg6);
\draw(sg2)--(cent);

\draw(cent)--(triv);
\draw(sg3)--(triv);
\draw(sg4)--(triv);
\draw(sg5)--(triv);
\draw(sg6)--(triv);
\end{tikzpicture}
\hspace*{1in}
\begin{tikzpicture}[node distance=1.5cm]
\node(grp){$S_3$};

\node(a3)[below right of = grp, xshift=1.5cm]{$A_3 = \langle (123) \rangle$};
\node(23)[below left of = a3, xshift=-1cm]{$\langle (23) \rangle$};
\node(13)[left of = 23]{$\langle (13) \rangle$};
\node(12)[left of = 13]{$\langle (12) \rangle$};

\node(1)[below right of = 23, yshift=-0.5cm]{$\{ (1) \}$};

\draw(grp)--(a3);
\draw(grp)--(12);
\draw(grp)--(13);
\draw(grp)--(23);

\draw(a3)--(1);
\draw(12)--(1);
\draw(13)--(1);
\draw(23)--(1);
\end{tikzpicture}
\end{center}

\newpage


Next, we have the subgroup lattice for $D_8 = \langle x,y \;|\; x^8=1,y^2=1,(xy)^2=1 \rangle = \{1,x,\dots,x^7,y,xy,\dots,x^7y\}$:

\begin{center}
\begin{tikzpicture}[node distance=1.75cm]
\node(grp){$D_8$};

\node(rot)[below of = grp]{$\langle x \rangle$};
\node(sg1)[left of = rot]{$\langle x^2,y \rangle$};
\node(sg2)[right of = rot]{$\langle x^2,xy \rangle$};

\node(rot2)[below of = rot]{$\langle x^2 \rangle$};
\node(sg3)[left of = rot2]{$\langle x^4,y \rangle$};
\node(sg4)[left of = sg3]{$\langle x^4,x^2y \rangle$};
\node(sg5)[right of = rot2]{$\langle x^4,x^3y \rangle$};
\node(sg6)[right of = sg5]{$\langle x^4,xy \rangle$};

\node(cent)[below of = rot2]{$\langle x^4 \rangle$};
\node(sg7)[left of = cent]{$\langle y \rangle$};
\node(sg8)[left of = sg7]{$\langle x^4y \rangle$};
\node(sg9)[left of = sg8]{$\langle x^2y \rangle$};
\node(sg10)[left of = sg9]{$\langle x^6y \rangle$};
\node(sg11)[right of = cent]{$\langle x^3y \rangle$};
\node(sg12)[right of = sg11]{$\langle x^7y \rangle$};
\node(sg13)[right of = sg12]{$\langle xy \rangle$};
\node(sg14)[right of = sg13]{$\langle x^5y \rangle$};

\node(triv)[below of = cent]{$\{ 1 \}$};

\draw(grp)--(sg1);
\draw(grp)--(rot);
\draw(grp)--(sg2);

\draw(rot)--(rot2);

\draw(sg1)--(sg3);
\draw(sg1)--(sg4);
\draw(sg1)--(rot2);

\draw(sg2)--(sg5);
\draw(sg2)--(sg6);
\draw(sg2)--(rot2);

\draw(rot2)--(cent);
\draw(sg3)--(cent);
\draw(sg4)--(cent);
\draw(sg5)--(cent);
\draw(sg6)--(cent);

\draw(sg3)--(sg7);
\draw(sg3)--(sg8);
\draw(sg4)--(sg9);
\draw(sg4)--(sg10);

\draw(sg5)--(sg11);
\draw(sg5)--(sg12);
\draw(sg6)--(sg13);
\draw(sg6)--(sg14);


\draw(cent)--(triv);
\draw(sg7)--(triv);
\draw(sg8)--(triv);
\draw(sg9)--(triv);
\draw(sg10)--(triv);
\draw(sg11)--(triv);
\draw(sg12)--(triv);
\draw(sg13)--(triv);
\draw(sg14)--(triv);
\end{tikzpicture}
\end{center}

\end{document}