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\begin{document}

\title{Mathematical Scratchpad }

\section{\protect\vspace{1pt}Mathematics 433-A\ \hfill Fall\ 2006}

\subsection{\hfill Exam 1}

\subsubsection*{September 26, 2006}

$\hfill \overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Name\ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ }$

\paragraph{\protect\vspace{1pt}Directions:\ \ }

\qquad Be sure to include in-line citations, including page numbers if
appropriate, every time you use a text or notes or technology. Include a
careful sketch of any graph obtained by technology in solving a problem. \ 
\textbf{Only write on one side of each page.}

\subsection*{\protect\vspace{1pt}The Problems}

\begin{enumerate}
\item ($10,10$ points)\ 

\begin{enumerate}
\item Use one of the principles of mathematical induction to prove if $a,b,c$
are elements in a group $G$ for which $b=cac^{-1},$ then $b^{n}=ca^{n}c^{-1}$
is true for all positive integers $n.$

\begin{enumerate}
\item If $n=1$ then $b^{1}=ca^{1}c^{-1}$ is given.

\item Assume the formula holds for $n=k.$

\item Then $b^{k+1}=b^{k}b=\left( ca^{k}c^{-1}\right) \left( cac^{-1}\right)
=ca^{k}ac^{-1}=ca^{k+1}c^{-1}$ 

So the formula holds for every positive integer by weak induction.
\end{enumerate}

\item Prove that $b^{n}=ca^{n}c^{-1}$ is also true for all \textbf{negative}
integers $n$. 

\begin{enumerate}
\item Note that $b^{-n}=\left( b^{n}\right) ^{-1}=\left( ca^{n}c^{-1}\right)
^{-1}=ca^{-n}c^{-1}$ so the formula holds for negative integers. (It holds
for $n=0$ trivially.)
\end{enumerate}

\item Prove that if $\phi :G\rightarrow H$ is a group homomorphism then $%
\phi \left( a^{-1}\right) =\left( \phi \left( a\right) \right) ^{-1}$ for
every $a\in G$

\begin{enumerate}
\item $e^{\prime }=\phi \left( e\right) =\phi \left( aa^{-1}\right) =\phi
\left( a\right) \phi \left( a^{-1}\right) $ and $e^{\prime }=\phi \left(
e\right) =\phi \left( a^{-1}a\right) =\phi \left( a^{-1}\right) \phi \left(
a\right) $ so $\phi \left( a^{-1}\right) $ acts like the inverse of $\phi
\left( a\right) .$ Since inverses are unique $\phi \left( a^{-1}\right) =%
\left[ \phi \left( a\right) \right] ^{-1}$
\end{enumerate}
\end{enumerate}

\item ($10,10$ points) Let $G$ be a group and $\phi :G\rightarrow G$ be the
map $\phi \left( x\right) =x^{-1}.$

\begin{enumerate}
\item Prove that $\phi $ is a bijection 

\begin{enumerate}
\item If $\phi \left( x\right) =\phi \left( y\right) $ then $x^{-1}=y^{-1}$
so $xx^{-1}y=xy^{-1}y$ which tells us $y=x$ and $\phi $ is one-to-one.

\item Let $y$ be an element in the codomain, $G$. Then $y^{-1}$ is in $G$
(the domain) and $\phi \left( y^{-1}\right) =\left( y^{-1}\right) ^{-1}=y$
so is onto. 
\end{enumerate}

\item Prove that $\phi $ is an automorphism if and only if $G$ is abelian.

\begin{enumerate}
\item If $G$ is abelian then $\phi \left( ab\right) =\left( ab\right)
^{-1}=b^{-1}a^{-1}=a^{-1}b^{-1}=\phi \left( a\right) \phi \left( b\right) $
so $\phi $ respects the group operation.

\item If $\phi $ is an automorphism, then $a^{-1}b^{-1}=\phi \left( a\right)
\phi \left( b\right) =\phi \left( ab\right) =\left( ab\right)
^{-1}=b^{-1}a^{-1}$. Taking inverses of both ends of this equation yields $%
ba=ab$ so every pair of elements in $G$ commute and $G$ is abelian. 
\end{enumerate}
\end{enumerate}

\item ($15$ points each) Do four $\left( 4\right) $ of the following
problems.

\begin{enumerate}
\item Prove that every subgroup of a cyclic group is cyclic.

\begin{enumerate}
\item In the text and in class notes.
\end{enumerate}

\item Prove that a group in which every element except the identity has
order $2$ is abelian.

\begin{enumerate}
\item For any elements $a,b$ in $G$,  $ab=ab\left( ba\right)
^{2}=abbaba=ab^{2}aba=a^{2}ba=ba.$
\end{enumerate}

\item Find all automorphisms of the group $\left( Z,+\right) $ of integers
under the operation of addition. [Recall that every subgroup of $\left(
Z,+\right) $ has the form $bZ$.]

\begin{enumerate}
\item Let $\phi $ be an automorphism of the group $G=\left( Z,+\right) $.
Then $1$ maps to an integer, say $m$. And since $\phi $ is onto, there is an
integer $m$ that maps to $1$. 

\item Thus, $1=\phi \left( m\right) =m\phi \left( 1\right) =nm.$ 

\item Since both $n,m$ are integers they are either both $1$ or both $-1.$ 

\item So $\phi \left( n\right) =n$ and $\psi \left( n\right) =-n$ are the
only possible automorphisms.

\item It is now easy to check they are both automorphisms.
\end{enumerate}

\item ($15$ points) Let $\phi ,\psi $ be two homomorphisms from a group $G$
to another group $G^{\prime }$ and let $H\subset G$ be the subset of $G$
given by $H=\left\{ x\in G:\phi \left( x\right) =\psi \left( x\right)
\right\} .$Prove or disprove, $H$ is a subgroup of $G$.

\begin{enumerate}
\item Let $a,b$ be elements in $H.$ Then $\phi \left( ab\right) =\phi \left(
a\right) \phi \left( b\right) =\psi \left( a\right) \psi \left( b\right)
=\psi \left( ab\right) $ so $H$ is closed.

\item Let $a$ be an element of $H$. Then $\phi \left( a\right) =\psi \left(
a\right) $ tells us $\phi \left( a^{-1}\right) =\left[ \phi \left( a\right) %
\right] ^{-1}=\left[ \psi \left( a\right) \right] ^{-1}=\psi \left(
a^{-1}\right) $ so $H$ contains the inverse of each of its elements.

\item Thus $H$ is a subgroup of $G$.
\end{enumerate}

\item Let $H$ be a subgroup of a group $G.$ Prove that the relation defined
by the rule $a\symbol{126}b$ if and only if $b^{-1}a\in H$ is an equivalence
relation on $G$.

\begin{enumerate}
\item Since $a^{-1}a=e\in H,$ then $a\symbol{126}a$

\item If $a\symbol{126}b$ then $b^{-1}a\in H$ and taking inverses, $\left(
b^{-1}a\right) ^{-1}=a^{-1}b\in H$ so $b\symbol{126}a$

\item If $a\symbol{126}b$ and $b\symbol{126}c$ then $b^{-1}a,$ $c^{-1}b\in H$
so $\left( c^{-1}b\right) \left( b^{-1}a\right) =c^{-1}a\in H$ and $a\symbol{%
126}c$.
\end{enumerate}

\item The orders of the elements in $U\left( 20\right) $ and $U\left(
24\right) $ are given in the tables below. Prove that these two groups are
not isomorphic by proving that if $\phi :G\rightarrow H$ is an isomorphism,
then the order of $a$ must equal the order of $\phi \left( a\right) $, $%
\left\vert a\right\vert =\left\vert \phi \left( a\right) \right\vert $.%
\[
\begin{tabular}{lllllllll}
$U\left( 20\right) $ & 1 & 3 & 7 & 9 & 11 & 13 & 17 & 19 \\ 
Order & 1 & 4 & 4 & 2 & 2 & 4 & 4 & 2 \\ 
&  &  &  &  &  &  &  &  \\ 
$U\left( 24\right) $ & 1 & 5 & 7 & 11 & 13 & 17 & 19 & 23 \\ 
Order & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2%
\end{tabular}%
\]

\begin{enumerate}
\item Let $a$ be an element of $G$ of order $n$ and denote the order of $%
\phi \left( a\right) $ by $m$.  Then $e=\phi \left( e\right) =\phi \left(
a^{n}\right) =\left[ \phi \left( a\right) \right] ^{n}$ tells us that $m$
must divide $n.$ 

\item We know that $\phi ^{-1}$ is also an isomorphism and the order of $%
\phi \left( a\right) $ is some integer, say $m.$Then $e=\phi ^{-1}\left(
e\right) =\phi ^{-1}\left( \left[ \phi \left( a\right) \right] ^{m}\right)
=\phi ^{-1}\left( \phi \left( a\right) \right) ^{m}=a^{m}$ which tells us 
that $n$ must divide $m$. The only way two positive integers $m$ and $n$ can
divide each other is if they are equal. 

\item Note that this proof tells us that either $a$ and $\phi \left(
a\right) $ both have infinite order or both have finite order. 
\end{enumerate}
\end{enumerate}
\end{enumerate}

\subsection{Definitions you should know.}

\begin{enumerate}
\item The \textbf{general linear} group of order $n$ over the real numbers $%
GL\left( n,\mathbf{R}\right) .$

\item The \textbf{center, }$Z\left( G\right) ,$ of a group $G.$

\item The \textbf{centralizer, }$C\left( a\right) ,$\textbf{\ }of an element 
$a$ in a group $G.$

\item A\ \textbf{normal} subgroup $N$ of a group $G.$

\item A\ \textbf{homomorphism} from the group $G\;$to the group $G\,^{\prime
}.$
\end{enumerate}

\end{document}
