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\begin{document}

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\noindent
\parbox{2.25in}{\bf Math 4010/5530-101}
\hfill {\Large \bf Homework \#3} \hfill
\parbox{2.25in}{\bf \hfill Due: Tue., Feb. $4^\mathrm{th}$, 2014}

\vspace{0.2in}

\begin{enumerate}
\item {}[1.7 \#16] Let $G$ be a group. It is often useful to consider a group acting on itself. One important action of $G$ on itself is {\it conjugation}. Define our action as follows (for all $g \in G$ and $x \in X=G$) $g \cdot x = gxg^{-1}$. Prove that this is in fact a group action. [{\it Note:} One can also define $x \cdot g = g^{-1}xg$. This yields a {\it right} group action. Many group theorists prefer to act on the right and thus define conjugation this way.]

\vspace{0.1in}

\item {}[2.2 \#5] Show that for the specified group $G$ and subgroup $H$ that $C_G(H)=H$ and $N_G(H)=G$.

\begin{enumerate}[(a)]
\item $G=S_3$ and $H=\langle (123) \rangle = \{ (1), (123), (132) \}$
\item $G=D_{2 \cdot 4} = \langle r,s \;|\; r^4=s^2=rsrs=1 \rangle$ and $H = \{1,s,r^2,r^2s \}$
\item $G=D_{2 \cdot 5} = \langle r,s \;|\; r^5=s^2=rsrs=1 \rangle$ and $H = \langle r \rangle = \{1,r,r^2,r^3,r^4 \}$
\end{enumerate}

Recall that $C_G(A) = \{ g \in G \;|\; ga=ag \mbox{ for all } a \in A \}$ is the {\it centralizer} of $A$ in $G$\\ and
$N_G(A) = \{ g \in G \;|\; gAg^{-1}=A \}$ is the {\it normalizer} of $A$ in $G$.

\vspace{0.1in}

\item {}[2.3 \#16] Let $x,y \in G$ and assume $|x|=n$ and $|y|=m$ ($x$ and $y$ have finite orders $n$ and $m$ respectively). Suppose that $xy=yx$ (they commute). 

\begin{enumerate}[(a)]
\item Prove that $|xy|$ divides $\mathrm{lcm}(n,m) = \ell$. 
\item Need this be true if $x$ and $y$ fail to commute?
\item Give an example (where $x$ and $y$ commute) such that $|xy|$ is not  equal to the $\mathrm{lcm}(n,m)$.
\end{enumerate}

\vspace{0.1in}

\item {}[{\bf Grad Problem} 2.5 \#11] Consider the group $QD_{16} = QD_{2 \cdot 8} = \langle \sigma, \tau \;|\; \sigma^8=\tau^2=1, \sigma\tau=\tau\sigma^3 \rangle$. This group is called the {\it quasidihedral} or {\it semidihedral} group of order 16. This group has three subgroups of order 8: $\langle \tau,\sigma^2 \rangle \cong D_{2 \cdot 4}$, $\langle \sigma \rangle \cong \mathbb{Z}_8$, and $\langle \sigma^2,\sigma\tau \rangle \cong Q_8$ (the quaternions) and every proper subgroup is contained in one of these three subgroups.

\vspace{0.05in}

Fill in the missing subgroups in the lattice of all subgroups of the quasidihedral group (shown below). Exhibit each subgroup with no more than 2 generators. [This is an example of a nonplanar lattice.]

\vspace{0.1in}

\end{enumerate}

\begin{tikzpicture}[node distance=2cm]
 % Quasidihedral group of order 16
 \node(grp){$QD_{16}$};
 % First row of proper subgroups:  sg11 sg12 sg13
 \node(sg12)[below of = grp]{$\langle \sigma \rangle$};
 \node(sg11)[left of = sg12]{$\langle \sigma^2,\tau \rangle$};
 \node(sg13)[right of = sg12]{$\langle \sigma^2,\tau\sigma \rangle$};
 % Second row of proper subgroups:  sg21 sg22 ... sg25
 \node(sg23)[below of = sg12]{$\langle ??? \rangle$};
 \node(sg22)[left of = sg23]{$\langle \sigma^4,\tau \rangle$};
 \node(sg21)[left of = sg22]{$\langle ??? \rangle$};
 \node(sg24)[right of = sg23]{$\langle ??? \rangle$};
 \node(sg25)[right of = sg24]{$\langle ??? \rangle$};
 % Third row of proper subgroups:  sg31 sg32 ... sg35
 \node(sg35)[below of = sg23]{$\langle \sigma^4 \rangle$};
 \node(sg34)[left of = sg35]{$\langle \tau \rangle$};
 \node(sg33)[left of = sg34]{$\langle ??? \rangle$};
 \node(sg32)[left of = sg33]{$\langle ??? \rangle$};
 \node(sg31)[left of = sg32]{$\langle \tau\sigma^2 \rangle$};
 % Trivial subgroup 
 \node(1)[below of = sg35]{$\{1\}$};

 % Lines between QD_16 and the first row 
 \draw(grp)--(sg11);
 \draw(grp)--(sg12);
 \draw(grp)--(sg13);
 % Lines between 1st and 2nd rows
 \draw(sg11)--(sg21);
 \draw(sg11)--(sg22);
 \draw(sg11)--(sg23);
 \draw(sg12)--(sg23);
 \draw(sg13)--(sg23);
 \draw(sg13)--(sg24);
 \draw(sg13)--(sg25);
 % Lines between 2nd and 3rd rows
 \draw(sg21)--(sg31);
 \draw(sg21)--(sg32);
 \draw(sg21)--(sg35);
 \draw(sg22)--(sg33);
 \draw(sg22)--(sg34);
 \draw(sg22)--(sg35);
 \draw(sg23)--(sg35);
 \draw(sg24)--(sg35);
 \draw(sg25)--(sg35);
 % Lines between 3rd row and the trivial subgroup
 \draw(sg31)--(1);
 \draw(sg32)--(1);
 \draw(sg33)--(1);
 \draw(sg34)--(1);
 \draw(sg35)--(1);
\end{tikzpicture}
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