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\Large
Math 113 Homework 4. Due 9/29
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\noindent \textbf{Reading corresponding to lectures:} \\
Tuesday 9/22: DF 1.7, (and a small part of 4.2) \\
Thursday 9/23 DF 3.1 \textit{Read it very carefully!}
\\
Tuesday 9/32 DF 3.1, 3.2
\medskip
\noindent \textbf{Problems to hand in:}
\begin{enumerate}[1.]
\item Normal subgroups:
\begin{enumerate}
\item
(Refer to the definition of \emph{normal} from problem set 3, and the definition of coset from lecture) \\
Let $G$ be a group, and $H$ a subgroup of $G$. Prove that $gH = Hg$ for all $g \in G$ if and only if $H$ is normal.
\medskip
\item Let $A$ and $B$ be groups. Show that the set $\{(a, 1_B) \mid a \in A\}$ is a normal subgroup of $A \times B$. (hint for a possible approach: is it the kernel of a homomorphism?)
\end{enumerate}
\bigskip
\item Fibers of homomorphisms (to be done after Thursday's lecture). Do these problems from DF Section 3.1: 8, 9, 12
\vspace{.7cm}
\noindent \hspace{-.7cm} \parbox{6 in} {The remaining problems are meant to introduce you to an important concept: the \emph{sign} of a permutation. You will use familiar tools (from chapters 1 and 2): actions, homomorphisms, the symmetric group, matrices... }
\bigskip
\item
Define an action of $S_n$ on $\R^n$ by
$$\sigma \cdot (x_1, x_2, \ldots x_n) = (x_{\sigma^{-1}(1)}, x_{\sigma^{-1}(2)}, \ldots x_{\sigma^{-1}(n)})$$
\begin{enumerate}
\item If $\sigma = (1234)$ in $S^4$, what is $\sigma(x_1, x_2, x_3, x_4)$? In general (for any permutation $\sigma$), in what new position does the action of $\sigma$ put $x_k$?
\item Prove that this satisfies the two axioms in the definition of \emph{action}.
\end{enumerate}
\bigskip
\item
\begin{enumerate}
\item Let $(\R)_nLG$ denote the set of $n \times n$ invertible matrices with real entries, with a binary operation $\ast$ defined by
$$A \ast B = BA$$
Here the right hand side is the usual matrix multiplication. \\
Prove that $(\R)_nLG$ is a group \footnote{You may have noticed that the name for this group is $GL_n(\R)$ backwards. The group is $GL_n(\R)$ with a ``backwards" version of matrix multiplication! The reason for doing this instead of the usual $GL_n(\R)$ has to do with the fact that function composition is a ``backwards" kind of multiplication and how our action works -- analogous to how we read braids from bottom to top when we thought of them as permutations. You can read about the problem here: \url{http://en.wikipedia.org/wiki/Permutation_matrix}}
\medskip
\item
Define a function $\phi: S_3 \to (\R)_3 LG$ by:\\
$\phi(e) = \left( \begin{smallmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 &1 \end{smallmatrix} \right)$ \hspace{.5cm}
$\phi((12)) = \left( \begin{smallmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 &1 \end{smallmatrix} \right)$ \hspace{.5cm}
$\phi((23))= \left( \begin{smallmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 &0 \end{smallmatrix} \right)$ \\
$\phi((13)) = \left( \begin{smallmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 &0 \end{smallmatrix} \right)$ \hspace{.5cm}
$\phi((123)) = \left( \begin{smallmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 &0 \end{smallmatrix} \right)$ \hspace{.5cm}
$\phi((132)) = \left( \begin{smallmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 &0 \end{smallmatrix} \right)$
\medskip
Show that $\phi$ is a homomorphism. (you may either check this directly, or explain using facts about elementary matrices)
\end{enumerate}
\bigskip
\item A \emph{permutation matrix} is a $n\times n$ matrix, where each entry is either 0 or 1, and the number 1 appears exactly once in each column and once in each row. For example,
$\left( \begin{smallmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 &0 \end{smallmatrix} \right)$ is a $3 \times 3$ permutation matrix
but $\left( \begin{smallmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{smallmatrix} \right)$ is not.
\begin{enumerate} \medskip
\item How many $n \times n$ permutation matrices are there? (give a formula involving $n$ and explain)
\item Show that $\phi$ from the question above is an isomorphism onto the set of $3 \times 3$ permutation matrices inside of $(\R)_3LG$. (and hence the permutation matrices form a group). \medskip
\item Show that, for each transposition $\sigma$ in $S_3$ the action of $\sigma$ on $\R^3$ specified in question 3 agrees with the action defined by
$$\sigma \cdot (x_1, x_2, x_3) = \phi(\sigma) \left( \begin{smallmatrix} x_1\\ x_2 \\ x_3\end{smallmatrix} \right)$$
\footnotesize(on the right hand side, you are multiplying matrices and vectors in the usual way) \vspace{.2cm}\\
\normalsize
[optional bonus: use the properties of actions and homomorphisms and the fact that transpositions generate $S^3$ to prove that this holds not just for transpositions, but for all elements of $S^3$].
\end{enumerate}
\medskip
\item By a similar procedure, we can define a homomorphism $\phi: S_n \to (\R)_nLG$ by specifying $\phi(\sigma)$ to be the matrix that maps $(x_1, x_2, \ldots x_n)$ to $(x_{\sigma(1)}, x_{\sigma(2)}, \ldots x_{\sigma(n)})$. \\
\footnotesize[you do not need to prove that this is a homomorphism, but you should think about why it is] \\
\normalsize
\begin{enumerate}
\item Describe, in words, the matrices that correspond to \emph{transpositions} (the definition of transposition is on a previous homework). \medskip
\item Show that a matrix corresponding to a transposition always has determinant -1. (you may use standard facts about elementary matrices to make this very easy. Or you can prove it yourself) \medskip
\item Deduce that an element $\sigma \in S_n$ can be written as a product of an even number of transpositions if and only if $\det (\phi(\sigma)) = 1$. \medskip
\end{enumerate}
\medskip
\bigskip
\noindent \textbf{Important note:}
$\det (\phi(\sigma))$ is called the \emph{sign} of the permutation $\sigma$. \\
The map $\epsilon: S_n \to \R$, defined by $\epsilon(\sigma) = det(\phi(\sigma))$ is a homomorphism from $S_n$ to the cyclic group with two elements, here thought of as $\{ -1, +1\}$ with multiplication. The kernel of this homomorphism is called the \emph{alternating group} $A_n$. \medskip
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\noindent \textbf{Astounding fact!} A single permutation can be written as a product of transpositions in many ways. For example, $(123) = (13)(12) = (12)(13)(12)(13)$. What you just showed is that the number of transpositions is either always even or always odd!
\medskip
\noindent This material is presented in Section 3.5 of DF, although with a slightly different approach.
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\end{enumerate}
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