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\begin{center} {\Large\bf Honors Algebra \\ Assignment 3} \\ Due,
Friday, Sept 18 \end{center}
\begin{quote}
Well, you see, Haresh Chacha, its like this. First you have ten, that's
just ten, that is, ten to the first power. Then you have a hundred, which
is ten times ten, which makes it ten to the second power. Then you have a
thousand which is ten to the third power. Then you have ten thousand,
which is ten to the fourth power - but this is where the problem begins,
don't you see? We don't have a special word for that, and we really
should. \ldots But you know, said Haresh, I think there is a special
word for ten thousand. The Chinese tanners of Calcutta once told me that
they used the number ten-thousand as a standard unit of counting. What
they call it I can't remember \ldots Bhaskar was electrified. But Haresh
Chacha you must find that number for me, he said. You must find out what
they call it. I have to know, he said, his eyes burning with mystical fire
and his small frog-like features taking on an astonishing radiance.
\\ -- from A Suitable Boy by Vikran Seth
\end{quote}
\ben
\item Let $\vec{v}$ be any nonzero vector in $R^n$ and set
\[ H=\{A\in GL_n(R): A\vec{v}=\lam\vec{v} \mbox{ for some positive} \lam\}
\]
Prove that $H$ is a subgroup of $GL_n(R)$.
\item Let $\Omega\subset R^2$
be any set (e.g.: a square) and let $G$ be the group of isometries of
$A$ (by which, formally, we mean a bijective map from $\Omega$ to
itself which preserves distances) of $\Omega$. For $\vec{x}\in\Omega$
let $H_{\vec{x}}= \{A\in G: A(\vec{x})= \vec{x}\}$. Let $B\in G$ with
$B(\vec{y})=\vec{x}$. Our object (similar to Problem 4 in Assignment 2)
is to describe $BH_{\vec{x}}B^{-1}$.
\ben
\item Show that for any $A\in H_{\vec{x}}$, $B^{-1}(A(B(\vec{y}))=\vec{y}$.
\item Show that if $C(\vec{y})=\vec{y}$ then we can write $C=BAB^{-1}$ for
some $A\in H_{\vec{x}}$. (Idea: Set $A=BCB^{-1}$.)
\item Conclude that $BH_{\vec{x}}B^{-1} = H_{\vec{y}}$.
\een
Further, in the special case where $\Omega$ is a square,
labelling the corners $0,1,2,3$ in clockwise order with $0$ in the
upper right, find explicitly the four values $H_{\vec{x}}$ when $\vec{x}$
ranges over the corners.
\item Let $a,b\geq 2$ and let $H=aZ$, $K=bZ$, both subgroups of $Z$
under $+$.
\ben
\item Show that when $a=5$ and $b=7$ then $H+K=Z$
\item Argue that there exists a $c\geq 1$ so that $H+K=cZ$. (Hint:
Use the description of subgroups of $Z$ that we have already proven.)
\item Prove that $c|a$ and $c|b$. (Idea: $a=a+0\in H+K$,\ldots)
\item Prove that there exist $x,y\in Z$ with $c=ax+by$.
(Idea: $c\in cZ=H+K$.)
\een
(Note: $c$ turns out to be the greatest common divisor of $a,b$.)
\item Let $G=Z_{15}^*$ and $H=\{1,4\}$. Give a table for $G/H$.
\item By $Z^2$ we mean the set of vectors $\vec{v}=(a,b)$ with
$a,b\in Z$. (The operation is normal vector addition.) Set
\[ H = \{x(2,1)+y(-1,2): x,y\in Z\} \]
$H$ is a subgroup of $Z^2$. (A good exercise but you aren't
asked now to give the argument.)
\ben
\item Give a drawing (preferably on graph paper) of the elements
of $H$. (E.g.: $(7,1)=3(2,1)-(-1,2)$ would be marked. Mark all those
$(c,d)$ with $-10\leq c,d \leq 10$.)
\item Find elements $\vec{v_1},\vec{v_2},\vec{v_3},\vec{v_4},\vec{v_5}$
(take $\vec{v_1}=(0,0)$) so that the five cosets $H+\vec{v_i}$ give all
of $Z^2$.
\item Optional: If you have some colored pens or pencils mark the
points of $Z^2$ with five colors according to the coset.
\een
\een
\begin{quote}
He who learns but does not think is lost. He who thinks but does not
learn is in great danger. \\ -- Confucius
\end{quote}
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