\documentclass{article} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{enumerate} \usepackage{ifthen} \usepackage{graphicx} \usepackage{color} \setlength{\unitlength}{0.1in} \setlength{\oddsidemargin}{-0.50in} \setlength{\evensidemargin}{-0.50in} \setlength{\topmargin}{-0.50in} \setlength{\headheight}{0.0in} \setlength{\headsep}{0.0in} \setlength{\topskip}{0.0in} \setlength{\textheight}{9.4in} \setlength{\textwidth}{7.5in} \newcommand{\comp}{ \,{\scriptstyle \stackrel{\circ}{}}\, } \newcommand{\nullset}{\mathrm{O}\!\!\!\!\big/\,} \newcommand{\divides}{\,\Big|\,} \newcommand{\myspace}{\mbox{\parbox[c][0.35in]{0.35in}{\hfill}}} \newcommand{\myspacea}{\mbox{\parbox[c][0.35in]{0.65in}{\hfill}}} \newcommand{\varspace}[2]{\mbox{\parbox[c][#1]{#2}{\hfill}}} \newcommand{\force}{{$\mbox{\;}$}} \begin{document} %\pagestyle{empty} \noindent \parbox{1.5in}{\bf Math 3110} \hfill {\Large \bf Test \#2} \hfill \parbox{1.5in}{\bf \hfill October $9^{\mathrm{th}}$, 2020} \vspace{0.05in} \noindent {\large\bf Name:} \underline{\Large\color{red}\sc\qquad Answer Key \qquad} \hfill {\bf Be sure to show your work!} \vspace{0.1in} \noindent{\bf\large 1. (20 points)} Random Group Stuff --- Fill out the following table: \vspace{0.1in} \mbox{} \hspace{-0.4in} \begin{tabular}{|c||c|c|c|c|c|c|} \hline $G=$ & \parbox[c][0.5in]{0.85in}{What is the\\ identity of $G$?} & \parbox[c][0.5in]{1in}{What is the\\ order of ...?} & \parbox[c][0.5in]{1.25in}{Does $G$ have an\\ element of order 5?} & \parbox[c][0.5in]{0.85in}{Is $G$ abelian?} & \parbox[c][0.5in]{0.85in}{Is $G$ cyclic?} \\ \hline \hline \hline $\mathbb{Z}_{80}$ & $0$ & \parbox[c][0.6in]{1in}{$|25|=$\\ $80/\mathrm{gcd}(80,25)=$\\ $80/5=16$} & Yes, $|80/5|=|16|=5$. & Yes. & \parbox[c][0.4in]{1in}{Yes, it can be generated by $1$.} \Large\strut \\ \hline $U(9)$ & $1$ & $|2|=6$ & No, 5 doesn't divide 6. & Yes. & Yes, $2$ is a generator. \Large\strut \\ \hline $D_{12}$ & $1=R_{0^\circ}$ & $|x^8|=3$ & No, 5 doesn't divide 24. & No, $xy\not=yx$. & No, it's not even Abelian. \Large\strut \\ \hline $S_9$ & $(1)$ & \parbox[c][0.6in]{1.2in}{$|(1234)(56)(789)|=$ \vspace{0.05in} $\mathrm{lcm}(4,2,3)$ $=12$} & Yes, $|(12345)|=5$. & \parbox[c][0.5in]{1in}{No, $(12)(13)$\\ \force\quad $\not=(13)(12).$} & No, it's not even Abelian. \\ \hline \end{tabular} \vspace{0.05in} {\bf Recall:} $D_{12} = \{ 1, x, \dots, x^{11}, y, xy, \dots, x^{11}y \}$ where $x^{12}=1$, $y^2=1$, and $xyxy=1$. %\vspace{0.1in} \begin{itemize} \item $\mathbb{Z}_{80}$ is a group under addition modulo $80$ thus its identity is $0$. We know $\langle 25 \rangle = \langle \mathrm{gcd}(25,80) \rangle = \langle 5 \rangle$, so $|25|=|5|=80/5=16$. Notice that for any divisor $k$, $80/k$ is an element of order $k$. So, $80/5=16$ has order $5$. $\mathbb{Z}_{80}$ is generated by every element in $U(80)$ -- in particular, it's generated by $1$. Therefore, this is a cyclic group (and thus Abelian). \item $U(9)$ is a group under multiplication modulo $9$ thus its identity is $1$. $U(9)=\{ k \;|\; \mathrm{gcd}(k,9)=1 \} = \{ 1,2,4,5,7,8\}$, so this is a group of order $6$. Notice that $2^1=2$, $2^2=4$, $2^3=8$, $2^4=16=7$, $2^5=32=5$, and $2^6=64=1$. Thus $|2|=6$. This also shows that $2$ generates all of $U(6)$. Thus $U(6)$ is cyclic (and thus Abelian). Finally, since $|U(9)|=6$ and $5$ doesn't divide $6$, there are no elements of order $5$. Alternatively, we could just verify the order of each element (we would see they have orders 1, 2, 3, and 6 but not 5). \item $D_{12}$ is a group under function composition if thought of as a symmetry group. Otherwise, its operation just an abstract generator/relation group multiplication. Thus the identity of $D_{12}$ is denoted by either $R_{0^\circ}$ (the zero degree rotation) or just plain old $1$. Note that $|x|=12$, so $\langle x^8 \rangle = \langle x^{\mathrm{gcd}(8,12)} \rangle = \langle x^4 \rangle$. Thus $|x^8|=|x^4|=12/4=3$. Alternatively, $(x^8)^1=x^8$, $(x^8)^2=x^{16}=x^4$, $(x^8)^3=x^{24}=1$, so $|x^8|=3$. We know that $D_{12}$ has rotations of orders 1, 2, 3, 4, 6, and 12 (i.e., divisors of $12$) and its reflections have order $2$. Thus it has no elements of order $5$. This is not an Abelian group since $xy\not=yx=x^{-1}y=x^{11}y$. Finally, it isn't cyclic since cyclic groups must be Abelian. \item $S_9$ is a group under function composition with the identity transformation, $(1)$, as its identity element. Orders of permutations are just least common multiples of the length of the permutation's cycles if it is written in terms of disjoint cycles. Thus $|(1234)(56)(789)|=\mathrm{lcm}(4,2,3)=12$. Any 5-cycle gives us an element of order $5$, so $|(12345)|=5$. We know that $S_n$ is not Abelian for $n\geq 3$. In particular, we have $(12)(13)=(132) \not= (123)=(13)(12)$. Since $S_9$ isn't Abelian, it cannot be cyclic. \end{itemize} \noindent {\bf\large 2. (20 points)} Cyclic Stuff \begin{enumerate}[(a)] \item Let $G$ be a finite group and $g \in G$. Suppose that $|g|=49$. \begin{enumerate}[i.] \item What is the order of $g^{28}$? List the distinct elements in $\langle g^{28} \rangle$. $|g^{28}| = \dfrac{49}{\mathrm{gcd}(28,49)} = \dfrac{49}{7}=$ \fbox{$7$}. We have $\langle g^{28} \rangle = \langle g^{\mathrm{gcd}(28,49)} \rangle = \langle g^{7} \rangle =$ \fbox{$\{ 1,g^7,g^{14},g^{21},g^{28},g^{35},g^{42} \}$} (7 elements). \item Is $g^{6} \in \langle g^{14} \rangle$? \quad {\large\bf Yes \ \ / \ \ \fbox{No}} \hfill $\langle g^{14} \rangle = \langle g^{\mathrm{gcd}(14,49)} \rangle = \langle g^{7} \rangle$ (as above), so $g^6 \not\in \langle g^{14} \rangle$. \end{enumerate} \item How many elements of order 8 does $\mathbb{Z}_{40}$ have? What are they? Notice that $8$ divides $40$, so $\mathbb{Z}_{40}$ does in fact have elements of order $8$. In any cyclic group whose order is divisible by $8$, we have $1$ element of order $1$, $2-1=1$ element of order $2$, $4-1-1=2$ elements of order $4$, and $8-2-1-1=4$ elements of order $8$. To get a single element of order $8$, we divide: $40/8=5$. To get the rest of the elements of order $8$, we multiply this element, $5$, by elements of $U(8)=\{1,3,5,7\}$. Thus, there are \fbox{4} elements of order 8. In particular, they are \fbox{5, 15, 25, and 35}. \item List the orders of elements in $\mathbb{Z}_{77}$. Then determine the number of elements of each order. There are elements of order $k$ for each divisor $k$ of 77. Also, we use the fact that cyclic subgroups have unique (cyclic) subgroups for each divisor order. Thus, for example, there is exactly one subgroup of order $11$. It's elements have orders $1$ and $11$ (divisors of $11$). If we toss out the (only) element of order $1$ (i.e., the identity), we are left with $11-1=10$ elements. Each of these must have order $11$. We thus recursively compute the number of elements of each order after first noting that the divisors of $77$ are 1, 7, 11, and 77. \begin{tabular}{c||c|c|c|c} Order = & 1 & 7 & 11 & 77 \Large\strut \\ \hline Number of elements = & 1 & $7-1=6$ & $11-1=10$ & $77-10-6-1=60$ \Large\strut \end{tabular} %\vspace{0.2in} \item List the orders of elements in $D_{77}$. Then determine the number of elements of each order. The rotations form a subgroup isomorphic to $\mathbb{Z}_{77}$. Thus we can use the table from part (c) to account for their orders and frequency. We also have 77 reflections. Reflections have order $2$. \vspace{-0.2in} \force \hfill \begin{tabular}{c||c|c|c|c|c} Order = & 1 & 2 & 7 & 11 & 77 \Large\strut \\ \hline Number of elements = & 1 & 77 & 6 & 10 & 60 \Large\strut \end{tabular} \end{enumerate} \vspace{-0.1in} \noindent {\bf\large 3. (22 points)} Permutations \qquad {\it Note:} Please giive simplified (``good manners'') answers. \begin{enumerate}[(a)] \item Consider $D_4 = \langle x,y \;|\; x^4=1, y^2=1, (xy)^2=1 \rangle = \{ 1,x,x^2,x^3,y,xy,x^2y,x^3y \}$. Label these elements in the order listed above: $1$ as $1$, $x$ as $2$,\dots $x^3y$ as $8$. Cayley's theorem says that $D_4$ is isomorphic to a subgroup of $S_8$. Using left multiplication maps and the labels provided, what element of $S_8$ represents $y$? Left multiplication by $y$ sends $1$ to $y$ (i.e., 1 to 5), $x$ to $yx=x^{-1}y=x^3y$ (i.e., 2 to 8), $x^2$ to $yx^2=x^{-2}y=x^2y$ (i.e., 3 to 7), and $x^3$ to $yx^3=x^{-3}y=xy$ (i.e., 4 to 6). It then sends each of these outputs back to where they came from (since $y$ is its own inverse). Thus $y$'s left multiplication is represented by \fbox{$(15)(28)(37)(46)$}. Assuming that this correspondence associates $xy$ with $(16)(25)(38)(47)$ and $x$ with $(1234)(5678)$. What permuation is associated with $x^2y$? Since Cayley's theorem gives an isomorphism between $D_4$ and some permutations in $S_8$ and since $x^2y= x \cdot xy$, we should have that $x^2y$ is represented by $x$'s permutation times $xy$'s permutation: $(1234)(5678)(16)(25)(38)(47)=$ \fbox{$(17)(26)(35)(48)$}. Notice also, $y=x^4y = x^2 \cdot x^2y$ so $y$ should be represented by $\Big((1234)(5678)\Big)^2(17)(26)(35)(48)$\\ $= (13)(24)(57)(68)(17)(26)(35)(48)=(15)(28)(37)(46)$ just like we found in part (a). \item Write $\sigma = (1235)(14)(236)$ as a product of disjoint cycles (and compute some stuff). \fbox{$\sigma = (1425)(36)$} and so $\sigma^{-1} = (63)(5241) = (5241)(63) =$ \fbox{$(1524)(36)$}. The order of $\sigma$ is $|\sigma|=\mathrm{lcm}(4,2)=$ \fbox{4}. Write $\sigma$ as a product of transpositions: $\sigma=$ \fbox{$(15)(12)(14)(36)$} or we could use the original (un-simplified non-disjoint description of $\sigma$) $\sigma=$ \fbox{$(15)(13)(12)(14)(26)(23)$}. Either way there are an \qquad $\sigma$ is \quad {\large\bf \fbox{Even} \ \ / \ \ Odd} number of transpositions used to write $\sigma$. Compute $\sigma^{102}=\sigma^{102 \text{ mod } 4} = \sigma^2 = (1425)(36)(1425)(36) =$ \fbox{$(12)(45)$} (here we use the fact that $|\sigma|=4$ to reduce the exponent). Alternatively, noting that disjoint cycles commute, we have $\sigma^{102} = \Big((1524)(36)\Big)^{102} = (1524)^{102}(36)^{102} = (1524)^{102 \text{ mod 4}}(36)^{102 \text{ mod } 2}=(1524)^2(36)^{0} = (1524)^2=(12)(45)$. \end{enumerate} \noindent {\bf\large 4. (18 points)} Explain why the following pairs of groups are not isomorphic. \begin{enumerate}[(a)] \item $\mathbb{R}^{2 \times 2} \not\cong \mathrm{GL}_2(\mathbb{R})$ \quad [$\mathbb{R}^{2 \times 2}$ is the $2 \times 2$ matrices under addition.] Both of these are infinite groups (in fact, continuum in cardinality), so size doesn't help us. However, since matrix multiplication is non-commutative, $\mathrm{GL}_2(\mathbb{R})$ is not Abelian. Therefore, these groups cannot be isomorphic given $\mathbb{R}^{2 \times 2}$ is \fbox{Abelian} (under matrix addition) whereas $\mathrm{GL}_2(\mathbb{R})$ is \fbox{not Abelian} (under matrix multiplication). Of course, there are other ways to see that these groups cannot be isomorphic. For example, if $A\not=0$, then $A+A+\cdots+A \not=0$. Thus other than the identity (i.e., the zero matrix) every element in $\mathbb{R}^{2 \times 2}$ has infinite order. On the other hand, there are elements of every possible order in $\mathrm{GL}_2(\mathbb{R})$. In particular, $B=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ has order 2 since $B^2=I_2$. \item $D_4 \not\cong Q$ \quad [$Q = \{\pm 1, \pm i, \pm j, \pm k\}$ is the group of quaternions.] Both of these are non-Abelian groups of order $8$, so no help there. However, notice that $D_4$ has $5$ elements of order $2$ (it has the $180^{\circ}$ rotation and 4 reflections). On the other hand, $Q$'s only element of order $2$ is $-1$. Thus \fbox{$D_4$ and $Q$ do not have the same number of elements of order $2$}. Thus they are not isomorphic. As always, we could see this other ways. For example, $D_4$ only has 2 elements of order 4 whereas $Q$ has 6 elements of order 4. \item $A_5 \not\cong \mathbb{Z}_{60}$ Notice that $|A_5|=5!/2=5\cdot 4\cdot 3\cdot 2\cdot 1/2 = 60$, so both are groups of order 60. However, $A_5$ is \fbox{not Abelian} whereas $\mathbb{Z}_{60}$ is \fbox{Abelian} (in fact, it's cyclic). Therefore, these groups cannot be isomorphic. Again, there are many other ways to see these groups cannot be isomorphic. We could note that $30$ is the only element of order $2$ in $\mathbb{Z}_{60}$ whereas $A_5$ has many elements of order $2$ including $(12)(34)$ and $(13)(24)$. Or notice that the largest order appearing in $S_5$ is $6$ (e.g., $(123)(45)$), but such elements are odd. So the largest order appearing in $A_5$ is $5$ (e.g., $(12345)$). On the other hand, $\mathbb{Z}_{60}$ has elements of higher orders such as 10, 12, 15, 30, and 60. \end{enumerate} \noindent {\bf\large 5. (20 points)} A few proofs \begin{enumerate}[(a)] \item Let $G = \{ 1,i,-1,-i \}$ where $i=\sqrt{-1}$. \qquad [$G$ is a subgroup of $\mathbb{C}_{\not=0}$ (nonzero complex numbers).] Prove that $G \cong U(10)$. Notice that $G=\langle i \rangle$ is cyclic of order $4$. Likewise, since $3^1=3$, $3^2=9$, $3^3=27=7$, and $3^4=81=1$ (mod $10$), we have $U(10)=\{1,3,7,9\}=\langle 3 \rangle$ is also cyclic of order $4$. Therefore, since we know that cyclic groups of the same order must be isomorphic, we have that $G \cong U(10)$. \item We know $A_n$ is not abelian for $n>3$. Show this. The usual counterexample: $(12)(13)=(132)\not=(123)=(13)(12)$ that shows $S_n$ is non-Abelian for $n \geq 3$ does not work here since $(12),(13) \not\in A_n$ (they're odd). However, notice that $(123),(234) \in A_n$ as along as $n \geq 4$ (odd length cycles are even permutations). Now notice that $(123)(234)=(12)(34) \not= (13)(24)=(234)(123)$. Therefore, $A_n$ is not Abelian (as long as $n>3$). {\it Note:} $A_1$ doesn't really make sense since we need at least 2 things to form transpositions. Notice that $A_2=\{(1)\}$, so $A_2$ -- the trivial group -- is Abelian and even cyclic. Also, $A_3 = \{ (1),(123),(132) \} = \langle (123) \rangle$, so again $A_3$ is not only Abelian, it's cyclic! \item Pick {\bf ONE} of the following\dots \begin{enumerate}[I.] \item Define $\varphi:\mathbb{Z}_9 \to \mathbb{Z}_9$ by $\varphi(x)=4x$. Show $\varphi$ is an automorphism of $\mathbb{Z}_9$. What is its inverse? Why is $\psi:\mathbb{Z}_8 \to \mathbb{Z}_8$ defined by $\psi(x)=4x$ {\bf not} an automorphism of $\mathbb{Z}_8$? Let $x,y \in \mathbb{Z}_9$. Notice that $4 \cdot 7=28=1$ (mod 9). Thus $\varphi^{-1}(x)=7x$: $\varphi^{-1}(\varphi(x))=\varphi^{-1}(4x)=7(4x)=28x=x$ and $\varphi(\varphi^{-1}(x))=\varphi(7x)=4(7x)=28x=x$. This means $\varphi$ is invertible (i.e., one-to-one and onto). Next, $\varphi(x+y)=4(x+y)=4x+4y=\varphi(x)+\varphi(y)$. Thus $\varphi$ is operation preserving. Therefore, $\varphi$ is an isomorphism from $\mathbb{Z}_9$ to itself (i.e., it's an automorphism). On the other hand, $4$ has no multiplicative inverse modulo $8$ (i.e., $4 \not\in U(8)$), so $\psi$ doesn't have an inverse. In particular, notice $\psi(2)=4(2)=0=4(0)=\psi(0)$ (working mod $8$). Thus $\psi$ isn't one-to-one. Therefore, $\psi$ is not an automorphism of $\mathbb{Z}_8$. On the other hand, it is an homomorphism from $\mathbb{Z}_8$ to itself: $\psi(x+y)=4(x+y)=4x+4y=\psi(x)+\psi(y)$ (for any $x,y \in \mathbb{Z}_8$), so $\psi$ is an {\it endomorphism} of $\mathbb{Z}_8$. \item Show that if $G$ is cyclic, then it must also be abelian. Is the converse also true? Why or why not? Justify your answer(s). \end{enumerate} Suppose $G$ is cyclic. Then there exists some $g \in G$ such that $G=\langle g \rangle$. Now suppose $x,y \in G$. Then since $x,y \in \langle g \rangle$, there exists $k,\ell \in \mathbb{Z}$ such that $x=g^k$ and $y=g^\ell$. We have $xy=g^kg^\ell = g^{k+\ell}=g^{\ell+k}=g^\ell g^k=yx$. Thus $G$ is Abelian. The converse is false. Note that $U(12)=\{1,5,7,11\}$ is an Abelian group (under multiplication mod $12$). But $5^2=25=1$, $7^2=49=1$, $11^2=121=1$ (mod 12). Thus the elements in $U(12)$ have orders $1$ (i.e., the identity) and $2$ (i.e., $5,7,11$). Since $U(12)$ has no element of order $4$, it has no generator. Thus $U(12)$ is {\it not} cyclic -- even though it is Abelian. \end{enumerate} \end{document}