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\begin{document}
{\bf Math 177 HW 2 Due Wednesday, January 31 }
\begin{enumerate}
\setcounter{enumi}{16}

\item \label{symmetries of square}The eight symmetries of the square are shown below.  The rotation is denoted $r$ and the ``flips'' are called $v, h, d1$ and $d2$ for ``vertical'', ``horizontal'', ``diagonal (one)'' and ``diagonal (two)'', respectively.

    \begin{center}
 \scalebox{.7}{\includegraphics{symmetriesSquare}}
    \end{center}

\begin{enumerate}
\item 
Complete the following multiplication table of the symmetry group of the square. This groups is called the {\it dihedral group of order 8} and is denoted $D_4$. More generally, the symmetry group of a regular $n$-gon is called the dihedral group $D_n$, and has $2n$ elements.



\begin{center}
$
\begin{array}{l||llllllll}
  & I & r & r^2 & r^2 & v & h & \mbox{d1} & \mbox{d2} \\
   \hline
   \hline
 I  \\

 r  \\

r^2 \\

r^3 \\

 v \\

 h  \\

 \mbox{d1}  \\

 \mbox{d2} 
\end{array}
$
\end{center}


\item Is the symmetry group of the square commutative? 
\end{enumerate}

\item Continuing with the symmetry group of the square, show that $r$ and $v$ will generate the group. That is, show that every element of the group can be  expressed in terms of $r$ and $v$ alone. In fact, show that every element of the group if equal to one of the following: $I, r, r^2, r^3, v, vr, vr^2$ and $vr^3$. (In general, for any $n$, the dihedral group $D_n$ can be generated by the rotation and a single flip across one axis of bilateral symmetry.)

\item Using $r$ and $v$ as generators, draw the Cayley graph of $D_4$. Use two different colors for the edges, one color for $r$ and one color for $v$. HINT: It will be helpful to use the multiplication table that you made for $D_4$, and even more helpful if you replace $h, d1$ and $d2$ in that table with their representations as $vr$, or $vr^2$, or $vr^3$.
\end{enumerate}

{\bf Permutations}

A {\it permutation} of $n$ things is a bijection $\sigma:\{1, 2, \dots, n\} \to \{1, 2, \dots, n\}$. There are $n!$ possible permutations. There are various ways to record a permutation. Since a permutation is a function, we can just list its values on each of the inputs. So, for example with $n=3$, we can list $\{\sigma(1), \sigma(2), \sigma(3)\}$. The six possible permutations are now $\{1,2,3\}$, $\{1,3,2\}$, $\{2,1,3\}$, $\{2,3,1\}$, $\{3,1,2\}$, and $\{3,2,1\}$. The first of these is the identity function since $\sigma(i)=i$ for $i=1,2,3$. Since permutations are functions, we can compose them, and it is easy to check that doing so turns the set of all permutations into a group, called the {\it symmetric} group of $n$ things and denoted by $S_n$.

Permutations can also be depicted graphically by drawing {\it braids}. The following braid diagram depicts the permutation $\sigma=\{4,2,1,3\}$. The numbers 1, 2, 3, 4 are written across the top and the bottom of the braid and the function is thought of as taking the 4 numbers on top to the 4 numbers on the bottom by following the braid ``strings'' from top to bottom. The string that starts at 1 on the top ends at 4 on the bottom, so $\sigma(1)=4$. To multiply two permutations, $\sigma \tau$ we simply stack the braid for $\sigma$ on top of the braid for $\tau$.
 
    \begin{center}
 \scalebox{.2}{\includegraphics{braid.pdf}}
    \end{center}

A better way to record a permutation is to write it as a product of {\it cycles}. In the example in the braid figure we see that 1 is taken to 4, 4 is taken to 3, and 3 is taken back to 1. Separately, 2 is taken to 2, which means that 2 is held fixed. We can depict this as the {\it 3-cycle} shown below:

    \begin{center}
 \scalebox{.4}{\includegraphics{cycle.pdf}}
    \end{center}

We don't use 2 in the drawing because 2 is not moved. We denote the cycle as $(1\,4\,3)$ and read it as ``1 goes to 4 which goes to 3 which goes back to 1.''  We could also denote it as $(4\,3\,1)$ (which is ``4 goes to 3 goes to 1 goes to 4'') or $(3\,1\,4)$ (which is ``3 goes to 1 goes to 4 goes to 1''). Note that if we reverse the arrows, then we would write $(1\, 3\,4)$ or $(3\, 4\,1)$ or $(4\, 1\, 3)$, all of which denote the same permutation. Moreover, the permutation $(1\, 3\,4)$ is the inverse to $(1\, 4\,3)$ since every number is sent back to where it came from. The permutation $(1\, 3\,4)$ is called a {\it 3-cycle} because the cycle has three numbers in it.

The identity permutation does not move any numbers so  does not have any cycles. Rather than write it as nothing, or as the capital letter I, we usually write it as the 1-cycle $(1)$.
 
A permutation may have more than one cycle. For example,  $\sigma \in S_5$ might send 1 to 2, 2 to 4, 3 to 5, 4 to 1, and 5 to 3. This consists of two {\it disjoint} cycles: the 3-cycle $(1\, 2\,4)$ and the 2-cycle $(3\,5)$. The cycles are disjoint because they have no numbers in common. It is easy to see that  $\sigma$ is the product  $(1\,2\,4)(3\,5)$ or $(3\,5)(1\,2\,4)$. Thinking about braids, we see that {\sl disjoint cycles commute}.

\begin{enumerate}
\setcounter{enumi}{10}
\item Write all six elements of $S_3$, the set of permutations of 1, 2, and 3, as both braid figures, and in cycle form. Do the same for the 24 elements of $S_4$.


\item Fill in the following multiplication table of $S_3$. Here the elements are represented in cycle form. Warning: there is no particular reason to believe that multiplication of permutations is commutative, that is, independent of the order of multiplication. So don't just assume that (12) times (123) is the same as (123) times (12) and skip filling in BOTH entries in the multiplication table.

\begin{tabular}{c|c|c|c|c|c|c|}
&(1)&(12)&(13)&(23)&(123)&(132)\\
\hline
(1)&&&&&&\\
\hline
(12)&&&&&&\\
\hline
(13)&&&&&&\\
\hline
(23)&&&&&&\\
\hline
(123)&&&&&&\\
\hline
(132)&&&&&&\\
\hline
\end{tabular}


Does multiplication of permutations in $S_3$ turn out to be commutative?


\item A {\it transposition} is a permutation consisting of a single 2-cycle. In other words, a transposition is a permutations that trades two things and keeps all other things fixed. In $S_3$ there are three transpositions: $(1\, 2), (1\, 3)$, and $(2\, 3)$. What are all the transpositions in $S_4$?

\item Every permutation has an inverse because it is a bijection. If $\sigma$ is represented by a braid, then turning the braid upside down will give the braid of its inverse.

\begin{enumerate}
\item List  all 6 elements of $S_3$ together with their inverses. 
\item List  all 24 elements of $S_4$ together with their inverses.
\item What is the inverse of a transposition?
\item If a permutation is equal to its own inverse, does it have to be a transposition? (Look in $S_4$.)
\item What is the inverse of the 3-cycle $(a\, b\, c)$? What is the inverse of the 4-cycle $(a\, b\, c\, d)$? In general, what is the inverse of the $n$-cycle $(a_1\, a_2\, \dots \,a_n)$?
\end{enumerate}


\item Prove  that  every permutation is a product of transpositions.  Thus, the transpositions generate $S_n$. In fact, we only need transpositions of the form $(k\,k+1)$. So,  $S_3$ is generated by $(1\,2)$ and $(2\,3)$. We don't need $(1\,3)$. (Hint: Use the braid diagram of a permutation to see this.) Using the braid diagrams for the elements of both $S_3$ and $S_4$, write all the element of $S_3$ and $S_4$ as products of transpositions. 

\item Draw the Cayley graph of $S_3$ using the generators $(1\,2)$ and $(2\,3)$. Draw the Cayley graph of $S_4$ using the generators $(1\,2)$,  $(2\,3)$, and $(3\,4)$.

\item Show that $S_3$ is generated by $(1\,2)$ and $(1\,2\,3)$. Draw the Cayley graph using these generators.
Show that $(1\, 2\,3)$ and $(1\,2\,3\,4)$ generate $S_4$. Using these as generators, draw the Cayley graph of $S_4$.
 
\end{enumerate}
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