\documentclass{article} \usepackage{graphicx} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{enumerate} \usepackage{ifthen} \setlength{\oddsidemargin}{0.0in} \setlength{\evensidemargin}{0.0in} \setlength{\topmargin}{0.0in} \setlength{\headheight}{0.0in} \setlength{\headsep}{0.0in} \setlength{\topskip}{0.0in} \setlength{\textheight}{8.9in} \setlength{\textwidth}{6.5in} \newcommand{\comp}{ \,{\scriptstyle \stackrel{\circ}{}}\, } \newcommand{\nullset}{\mathrm{O}\!\!\!\!\big/\,} \newcommand{\divides}{\,\Big|\,} \newcommand{\myspace}{\mbox{\parbox[c][0.3in]{0.3in}{\hfill}}} \begin{document} \pagestyle{empty} \noindent \parbox{2in}{\bf Math 3110} \hfill {\Large \bf Test \#2} \hfill \parbox{2in}{\bf \hfill October 28$^{th}$, 2009} \vspace{-0.1in} \begin{center} \Large ANSWER KEY \end{center} \vspace{0.2in} \noindent {\bf \large 1. (16 points)} Either prove $G$ is a group or explain why it is not a group. \begin{enumerate}[(a)] \item $G = \{ x \in D_5 \,|\, x \mbox{ is a reflection} \}$ (the operation is composition of symmetries). \vspace{0.1in} \noindent {\bf Answer:} No. $G$ is not a group. Two easy ways to see that $G$ is not a group are: \\ (1) $G$ lacks an identity. The identity symmetry is not a reflection (it's a rotation of zero degrees).\\ (2) The operation lacks closure. A reflection composed with a reflection is a rotation. So given any $r_1,r_2 \in G$ we have $r_1r_2 \not\in G$. \vspace{0.1in} \item $\mathbb{R}_{>0}$ (positive real numbers) where the operation is multiplication. \vspace{0.1in} \noindent {\bf Answer:} Yes. $\mathbb{R}_{>0}$ is a group under multiplication. \vspace{0.1in} Notice that $\mathbb{R}_{>0}$ is a subset of $\mathbb{R}_{\not=0}$ (which is a group under multiplication). So to show that $\mathbb{R}_{>0}$ is a group we may just show it is a subgroup of $\mathbb{R}_{\not=0}$. To do this we need to check closure under multiplication and closure under inversion. \begin{itemize} \item $x>0$ and $y>0$ $\Rightarrow$ $xy>0$ so we have closure under multiplication. \item $x >0$ $\Rightarrow$ $1/x >0$ so we have closure under inversion. \end{itemize} Therefore, $\mathbb{R}_{>0}$ is a subgroup of $\mathbb{R}_{\not=0}$ (and so it is a group). \vspace{0.1in} \end{enumerate} \noindent {\bf \large 2. (16 points)} \begin{enumerate}[(a)] \item Show $\displaystyle{ H = \left\{ \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} \,\Bigg|\, a,b \in \mathbb{R} \mbox{ and } a,b \not=0 \right\}}$ is a subgroup of $\mathrm{GL}_2(\mathbb{R})$ \vspace{0.1in} (Recall $\mathrm{GL}_2(\mathbb{R})$ --- the set of $2 \times 2$ invertible matrices --- is a group under matrix multiplication). \vspace{0.1in} First, we'll set up some notation. Let $A,B \in H$. Thus there exists some non-zero real numbers $a,b,c,d$ such that $\displaystyle{A = \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}}$ and $\displaystyle{B = \begin{bmatrix} c & 0 \\ 0 & d \end{bmatrix}}$. Notice that $\mathrm{det}(A) = ab \not=0$ because $a\not=0$ and $b\not=0$. Therefore, for all $A \in H$, we have $A \in \mathrm{GL}_2(\mathbb{R})$. Thus $H \subset \mathrm{GL}_2(\mathbb{R})$. Obviously $H$ is not empty. $\displaystyle{AB = \begin{bmatrix} ac & 0 \\ 0 & bd \end{bmatrix}}$ (which looks right -- it's diagonal). Since $a,b,c,d$ are non-zero, we have that $ac$ and $bd$ are non-zero as well. Therefore, $AB \in H$. $\displaystyle{A^{-1} = \frac{1}{ab-0^2} \begin{bmatrix} b & -0 \\ -0 & a \end{bmatrix} = \begin{bmatrix} 1/a & 0 \\ 0 & 1/b \end{bmatrix}}$ (which looks right -- it's diagonal). Since $a,b$ are non-zero, we have that $1/a$ and $1/b$ are non-zero as well. Therefore, $A^{-1} \in H$. Therefore, $H$ is a non-empty subset of $\mathrm{GL}_2(\mathbb{R})$ which is closed under matrix multiplication and inversion. Thus $H$ is a subgroup of $\mathrm{GL}_2(\mathbb{R})$. \vspace{0.1in} \item Quickly (in a few words) why are the even integers a subgroup of the integers? \vspace{0.1in} Even plus even is even and the negative of an even number is even. \vspace{0.1in} \end{enumerate} \noindent {\bf \large 3. (20 points)} Workin' mod 5. \begin{enumerate}[(a)] \item Fill out the following tables (don't worry about brackets for equivalence classes.) \begin{center} \hspace*{-0.6in} \begin{tabular}{c|c|c|c|c|c|} $(\mathbb{Z}_5,+)$ & 0 & 1 & 2 & 3 & 4 \\ \hline 0 & 0 & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 & 0 \\ \hline 2 & 2 & 3 & 4 & 0 & 1 \\ \hline 3 & 3 & 4 & 0 & 1 & 2 \\ \hline 4 & 4 & 0 & 1 & 2 & 3 \\ \hline \multicolumn{6}{c}{$\mathbb{Z}_5$ Addition Table} \end{tabular} \hspace{0.3in} \begin{tabular}{c|c|c|c|c|c|} $(\mathbb{Z}_5,\times)$ & 0 & 1 & 2 & 3 & 4 \\ \hline 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 1 & 0 & 1 & 2 & 3 & 4 \\ \hline 2 & 0 & 2 & 4 & 1 & 3 \\ \hline 3 & 0 & 3 & 1 & 4 & 2 \\ \hline 4 & 0 & 4 & 3 & 2 & 1 \\ \hline \multicolumn{6}{c}{$\mathbb{Z}_5$ Multiplication Table} \end{tabular} \end{center} \item Compute $2^{-1}(4+3)-2$ (mod 5). \vspace{0.1in} Notice that $2(3)=1$ (mod 5) so $2^{-1}=3$. Thus $2^{-1}(4+3)-2 = 3(4+3)-2 = 3(7)-2 = 3(2)-2 = 6-2 = 1-2 = -1 = 4$ (mod 5) Of course, we could wait until the very end to reduce mod 5: $2^{-1}(4+3)-2=3(4+3)-2 = 3(7)-2=21-2=19=4$ (mod 5). The important thing to remember is that addition, subtraction and multiplication of integers haven't changed. When working mod 5, we've just changed the idea of what ``='' means. The only operation that requires special care is ``division''. To compute inverses we can either guess/look up the answer on the multiplication table OR run the Euclidean algorithm. \vspace{0.1in} \item Find $\langle 3 \rangle$ (the subgroup generated by $3$) in $U(5)$ (NOT $\mathbb{Z}_5$!!!). \vspace{0.1in} Remember $U(5)$ contains (congruence classes) of integer which are relatively prime with 5. So $U(5) = \{ 1,2,3,4 \}$ (Also, remember that the operation is multiplication mod 5 NOT addition). Note: $3^1=3$, $3^2=4$, $3^3=3^2(3)=4(3)=2$, and $3^4=3^3(3)=2(3)=1$ so the order of $3$ (in $U(5)$) is 4. \vspace{-0.15in} \[ \langle 3 \rangle = \{ \dots 3^{-1}, 3^0, 3^1, 3^2, \dots \} = \{ 1, 3, 4, 2 \} = U(5) \] \item Find the orders of elements of $U(5)$. Is $U(5)$ cyclic? Why or why not? \begin{minipage}[c]{3in} \begin{itemize} \setlength{\itemsep}{1pt} \item $1^1 = 1$ \item $2^1 = 2$, $2^2 = 4$, $2^3 = 3$, $2^4 = 1$ \item $3^1 = 3$, $3^2 = 4$, $3^3 = 2$, $3^4 = 1$ \item $4^1 = 4$, $4^2 = 1$ \end{itemize} \end{minipage} \hfill \hfill \begin{minipage}[c]{3in} \begin{tabular}{r|c|c|c|c} element = & 1 & 2 & 3 & 4 \\ \hline order = & 1 & 4 & 4 & 2 \end{tabular} \end{minipage} %\vspace{0.1in} {\bf YES} $U(5)$ is cyclic because the order of $U(5)$ is 4 and we have an element (in fact 2 elements) of order 4. ...OR... $U(5)$ is cyclic because $U(5)=\langle 3\rangle$ ...OR... $U(5)$ is cyclic because $U(5) = \langle 2 \rangle$. \vspace{0.1in} \end{enumerate} \noindent {\bf \large 4. (16 points)} Quick proofs \begin{enumerate}[(a)] \item Let $G$ be a group and suppose that $x=x^{-1}$ for all $x \in G$. Prove that $G$ is Abelian.\\ Hint: $xy=(xy)^{-1}=\;\;??$ \vspace{0.05in} Suppose that $x,y \in G$. Then $x^{-1}=x$, $y^{-1}=y$ and $(xy)=(xy)^{-1}$ (by assumption). Recall that in any group we have $(xy)^{-1}=y^{-1}x^{-1}$ (the ``socks shoes principle''). Therefore, $xy=(xy)^{-1} = y^{-1}x^{-1} = yx$. Thus $G$ is Abelian. \vspace{0.05in} \item Let $n$ be an integer such that $n \geq 3$. Consider $(12), (13) \in S_n$. Compute $(12)(13)$ and $(13)(12)$. Is $S_n$ cyclic? Why or why not? \vspace{0.05in} \begin{minipage}[c]{1.5in} \begin{itemize} \item $(12)(13)=(132)$ \item $(13)(12)=(123)$ \end{itemize} \end{minipage} \hfill \hfill \begin{minipage}[c]{4.5in} Since $(12)(13) \not= (13)(12)$, we can conclude that $S_n$ is {\bf not Abelian}. And since $S_n$ is not Abelian, it is {\bf not cyclic} (because cyclic implies Abelian). \end{minipage} \vspace{0.05in} \end{enumerate} \noindent {\bf \large 5. (16 points)} $G = \{1,a,a^2,b,ab,a^2b \}$ is a group. Finish filling out $G$'s Cayley table then answer some questions. \begin{center} \begin{tabular}{|c|||c|c|c|c|c|c|} \hline $G$ & $1$ & $a$ & $a^2$ & $b$ & $ab$ & $a^2b$ \\ \hline \hline \hline $1$ & $1$ & $a$ & $a^2$ & $b$ & $ab$ & $a^2b$ \\ \hline $a$ & $a$ & $a^2$ & $1$ & $a b$ & $a^2b$ & $ b$ \\ \hline $a^2$ & $a^2$ & $1 $ & $a$ & $a^2b$ & $ b$ & $a b$ \\ \hline $b$ & $b$ & $a^2b$ & $a b$ & $1 $ & $a^2 $ & $a$ \\ \hline $ab$ & $ab$ & $ b$ & $a^2b$ & $a$ & $1$ & $a^2$ \\ \hline $a^2b$ & $a^2b$ & $a b$ & $b$ & $a^2 $ & $a$ & $1 $ \\ \hline \end{tabular} \end{center} \begin{enumerate}[(a)] \item What is the order of $a^2$? Determine $\langle a^2 \rangle$ (the subgroup generated by $a^2$). \vspace{0.05in} $(a^2)^1 = a^2$, $(a^2)^2 = a$, and $(a^2)^3 = a(a^2) = 1$. Thus the order of $a^2$ is 3.\\ Also, $\langle a^2 \rangle = \{ a^2, a,1 \} = \{1,a,a^2 \}$. \vspace{0.05in} \item Is $G$ Abelian? Is $G$ cyclic? Why or why not? \vspace{0.05in} {\bf NO} $G$ is {\bf not Abelian}. Just look at the Cayley table and notice that $ab \not= ba = a^2b$. Also, since $G$ is not Abelian it is {\bf not cyclic} (since all cyclic groups are automatically Abelian). \vspace{0.05in} \end{enumerate} \noindent {\bf \large 6. (16 points)} Permutations! \begin{enumerate}[(a)] \item Write $\sigma = (125)(35)(24)(264)$ as a product of disjoint cycles. \vspace{0.05in} $\sigma = (125)(35)(24)(264) = (12653)(4) = (12563)$ \vspace{0.05in} \item What is the order of $\sigma$? \vspace{0.05in} The order of $\sigma$ is 5 (the least common multiple of the lengths of the cycles when $\sigma$ is written as a product of {\bf disjoint} cycles). \vspace{0.05in} \item Write $\sigma$ as a product of transpositions. Is $\sigma$ even or odd? \vspace{0.05in} There are infinitely many answers to this question here are two possibilities:\\ $\sigma = (15)(12)(35)(24)(24)(26) = (13)(16)(15)(12)$. Either way we have an even number of transpositions so $\sigma$ {\bf is even}. \vspace{0.05in} \item Let $\tau = (1452)(367)(89)$. What is the order of $\tau$? \vspace{0.05in} $\tau$ is written as the product of disjoint cycle so we can just take the least common multiple of the lengths of its cycles. $\mbox{lcm}(4,3,2) = 12$. So the order of $\tau$ is 12. \vspace{0.05in} \item Write $\tau^{26}$ as the product of disjoint cycles. \vspace{0.05in} Since the order of $\tau$ is 12, we have that $\tau^{12} = (1)$ so $\tau^{26} = \tau^{12}\tau^{12}\tau^2 = \tau^2 = (15)(24)(376)$. \vspace{0.05in} \item What is the order of $\tau^6$?\\ {}[Hint: You shouldn't need to compute any powers of $\tau$ to answer this question.] \vspace{0.05in} Quick way: $\tau^6 \not= (1)$ since the order of $\tau$ is greater than 6. But $(\tau^6)^2 = \tau^{12} = (1)$ since the order of $\tau$ is 12. Thus the second power of $\tau^6$ is the identity so $\tau^6$ has order 2. Calculatin' way: $\tau^6 = \left( (1452)(367)(89) \right)^6 = (1452)^6 (367)^6 (89)^6$ since disjoint cycles commute. Continuing... $= (1456)^4(1456)^2((367)^3)^2((89)^2)^3 = (1)(15)(24)(1)^2(1)^3 = (15)(24)$ so $\tau^6$ has order $\mbox{lcm}(2,2)=2$. \end{enumerate} \end{document}