\documentclass{article} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{enumerate} \usepackage{ifthen} \usepackage{graphicx} \usepackage{color} \setlength{\unitlength}{0.1in} \setlength{\oddsidemargin}{-0.25in} \setlength{\evensidemargin}{-0.25in} \setlength{\topmargin}{-0.25in} \setlength{\headheight}{0.0in} \setlength{\headsep}{0.0in} \setlength{\topskip}{0.0in} \setlength{\textheight}{9.4in} \setlength{\textwidth}{7in} \newcommand{\comp}{ \,{\scriptstyle \stackrel{\circ}{}}\, } \newcommand{\nullset}{\mathrm{O}\!\!\!\!\big/\,} \newcommand{\divides}{\,\Big|\,} \newcommand{\myspace}{\mbox{\parbox[c][0.35in]{0.35in}{\hfill}}} \newcommand{\varspace}[2]{\mbox{\parbox[c][#1]{#2}{\hfill}}} \begin{document} %\pagestyle{empty} \noindent \parbox{2in}{\bf Math 3110} \hfill {\Large \bf Test \#2} \hfill \parbox{2in}{\bf \hfill October $26^{\mathrm{th}}$, 2011} \vspace{0.3in} \noindent {\large\bf Name:} \underline{\Large\sc\color{red} Answer Key} \hfill {\bf Be sure to show your work!} \vspace{0.2in} \noindent {\large 1. (20 points)} Cyclic \begin{enumerate}[(a)] \item Let $G = \langle g \rangle$ where $g$ has order 20. \vspace{0.1in} Don't forger $g$ has order 20, so its exponents can be reduced modulo 20. \vspace{0.05in} $\langle g^8 \rangle = \{ g^8, g^{16}, g^{24}, \dots \} = \{ g^8, g^{16}, g^{4}, g^{12}, e \} = \langle g^{4} \rangle$ \vspace{0.1in} Is $g^{102} \in \langle g^8 \rangle$? Why or why not? \vspace{0.1in} No. 102 mod 20 is 2. But $g^{102}=g^2 \not\in \langle g^8 \rangle$. OR $g^{102} \not\in \langle g^8 \rangle$ because $\mathrm{gcd}(8,20)=4$ does not divide 102. \vspace{0.1in} \item Suppose $G$ is a cyclic group with at least one element of order $6$. \vspace{0.1in} What can you say about the order of $G$? \vspace{0.1in} Since $G$ has an element of order 6, it's order must be a multiple of 6. $G$ could be a group of order 6, 12, 18, etc. \vspace{0.1in} How many elements of order 6 can $G$ have? Is there more than one possibility? \vspace{0.1in} If $g \in G$ is an element of order 6, then $\langle g \rangle$ is a subgroup of order 6. However, $G$ is cyclic. Therefore, it has a {\bf unique} subgroup of order 6. This subgroup in turn has unique subgroups of orders 1, 2, 3, and 6. Its subgroup of order 1 just contains the identity. Its subgroup of order 2 has elements of order 1 and 2, but there is only 1 element of order 1. Therefore, there are exactly $2-1=1$ element of order 2. Next, the subgroup of order 3 has elements of order 1 and 3 (the divisors of 3). There is only 1 element of order 1, thus there are $3-1=2$ elements of order 3. Finally, the subgroup of order 6 has elements of order 1, 2, 3, and 6. There is 1 element of order 1, 1 element of order 2, and 2 elements of order 3. Thus there are $6-2-1-1=2$ elements of order 6. Any {\bf cyclic} group whose order is divisible by 6 will have {\bf exactly} 2 elements of order 6 (and 2 of order 3, 1 of order 2, and 1 of order 1). This essentially follows from the fact that subgroups of cyclic groups are cyclic and there is a unique subgroup corresponding to each divisor of the order of that group. \vspace{0.1in} {\it Note:} This is not true of non-cyclic groups. For example. In $U(8)=\{1,3,5,7\}$, all non-identity elements: 3, 5, and 7 have order 2. In $S_5$ there are many elements of order 6: $(12)(345)$, $(12)(354)$, $(13)(245)$, etc. On the other hand, $A_4$ has order 12 (which is divisible by 6), but it has ZERO elements of order 6. In fact, it doesn't even have a subgroup of order 6. \vspace{0.1in} \item List the possible orders of elements in $\mathbb{Z}_{33}$. Then determine the number of elements of each order. \vspace{0.1in} $\mathbb{Z}_{33}$ is cyclic, the divisors of 33 are 1, 3, 11, and 33, so there are elements of orders 1, 3, 11, and 33. There is 1 element of order 1 (the identity is the only element of order 1 in any group). There are $3-1=2$ elements of order 3. Next, 11 is divisible by 1 and 11 so the (unique) cyclic subgroup of order 11 (which contains all of the elements of order 11) there are $11-1=10$ elements of order 11. Finally after knocking out the elements of orders 1, 3, and 11 we are left with $33-10-2-1=20$ elements of order 33. We also need to count the number of elements of various orders for $D_n$. $D_n$'s subgroup of rotations is a cyclic subgroup of order $n$, thus isomorphic to $\mathbb{Z}_n$. So the orders of various rotations can be counted just like we count orders for $\mathbb{Z}_n$. In addition $D_n$ has $n$ reflections. A reflection is its own inverse, so its order is 2. Thus the only difference in the tables for $\mathbb{Z}_n$ and $D_n$ is that $D_n$ has an additional $n$ elements of order 2. \vspace{0.1in} \begin{tabular}{ll} For $\mathbb{Z}_{33}$: & For $D_{33}$: \\ \hspace*{3.4in} & \hspace*{3.4in} \\ \begin{tabular}{c||c|c|c|c|c|c} Order = & 1 & 3 & 11 & 33 \\ \hline Number of elements = & 1 & 2 & 10 & 20 \end{tabular} & \begin{tabular}{c||c|c|c|c|c|c} Order = & 1 & 2 & 3 & 11 & 33 \\ \hline Number of elements = & 1 & 33 & 2 & 10 & 20 \end{tabular} \end{tabular} \end{enumerate} \noindent {\large 2. (20 points)} The following pairs of groups are {\bf not} isomorphic. Prove this is the case. \begin{enumerate}[(a)] \item $\mathrm{GL}_3(\mathbb{R}) \not\cong A_{500}$ \vspace{0.1in} $\mathrm{GL}_3(\mathbb{R})$ ($2\times 2$ invertible matrices over the reals) is an infinite group. $A_{500}$ is finite (although, it's order is quite large: $500!/2$). Isormophic groups must have the same size. Thus these groups are not isomorphic. \vspace{0.1in} \item $U(8)=\{1,3,5,7\} \not\cong \mathbb{Z}_4$ \vspace{0.1in} Notice that $3^2=9=1$, $5^2=25=1$, and $7^2=49=1$ (mod $8$). Thus $U(8)$ has 3 elements of order 2 while $\mathbb{Z}_4$ only has 1 element of order 2. Therefore, they cannot be isormophic. Alternatively, $U(8)$ has no elements of order 4. Thus $U(8)$ is not cyclic. However, $\mathbb{Z}_4$ is cyclic. Therefore, they cannot be isomorphic. \vspace{0.1in} \item $\mathrm{GL}_2(\mathbb{Z}) \not\cong \mathbb{Q}$ \vspace{0.1in} $\mathrm{GL}_2(\mathbb{Z})$ ($2 \times 2$ invertible integer matrices) is a non-abelian group (in general, matrix multiplication is not commutative). However, $\mathbb{Q}$ (rational numbers under addition) is an abelian group. Therefore, they cannot be isomorphic. \vspace{0.1in} \item $S_4 \not\cong D_{12}$ \vspace{0.1in} Both groups are non-abelian (and thus also not cyclic) and both are groups of order 24. Let's try looking at the orders of various elements. $D_{12}$ has a cyclic subgroup of order 12 (i.e. it's subgroup of rotations). Thus $D_{12}$ has at least one element of order 12. On the other hand $S_4$ does not have an element of order 12. It would take either a 12 cycle or a pair of disjoint 3 and 4 cycles to get an element of order 12 in a permutation group. So while $S_7$ has elements of order 12 (i.e. $(123)(4567)$), $S_4$ does not. Therefore, they cannot be isomorphic. Alternatively, the elements of order 2 in $S_4$ are $(12),(13), (14),(23), (24),(34), (12)(34),(13)(24), (14)(23)$ (there are 9). On the other hand, $D_{12}$ has 12 reflections and the $180^\circ$ rotation and no other elements of order 2. So $D_{12}$ has 13 (not 9) elements of order 2. Thus these groups cannot be isomorphic. Or we could look at elements of order 3. The only elements of order 3 in $D_6$ are rotations (reflections have order 2). Since the rotations form a cyclic group, there are only $3-1=2$ elements of order 3. On the other hand, $S_4$ has many elements of order 3 (there are a lot of 3-cycles: $(123),(132),(124),\dots$). Thus these groups cannot be isomorphic. One last reason they are not isomorphic. The center of $S_n$ is trivial for all $n \geq 3$, so $Z(S_4)=\{(1)\}$. On the other hand, the center of $D_n$ is trivial for odd $n$, but contains the $180^\circ$ rotation for even $n$, so $Z(D_{12})=\{R_0, R_{180} \}$. Since the centers are not isomorphic (they have different orders), the groups cannot be isomorphic. \vspace{0.1in} \end{enumerate} \noindent {\large 3. (20 points)} Isomorphisms \begin{enumerate}[(a)] \item Prove that $U(5) \cong \mathbb{Z}_4$. \vspace{0.1in} Consider $2 \in U(5)$. $\langle 2 \rangle = \{ 1, 2, 4, 8, \dots \} = \{ 1,2,4,3 \} = U(5)$. Thus $U(5)$ is a cyclic group of order 4. However, $\mathbb{Z}_4$ is a cyclic group of order 4. We proved in class that any two cyclic groups of the same order are isomorphic. Therefore, $U(5) \cong \mathbb{Z}_4$. \vspace{0.1in} \item Let $G$ be an {\bf Abelian} group. Define the map $\varphi:G \rightarrow G$ by $\varphi(g)=g^{-1}$. Prove that $\varphi$ is an isomorphism (actually $\varphi$ is an automorphism since its domain and codomain are equal). \vspace{0.1in} First, note that $\varphi \circ \varphi (g) = \varphi(\varphi(g)) = \varphi(g^{-1}) = (g^{-1})^{-1} = g$. Thus $\varphi^{-1}=\varphi$ ($\varphi$ is its own inverse). Therefore, $\varphi$ is a bijection (i.e. one-to-one and onto). Next, $\varphi(xy)=(xy)^{-1}=y^{-1}x^{-1} =x^{-1}y^{-1}=\varphi(x)\varphi(y)$ where we know $y^{-1}x^{-1} =x^{-1}y^{-1}$ because $G$ is abelian. Thus $\varphi$ is a homomorphism. Therefore, $\varphi$ is an isomorphism (also, an automorphism since its domain and codomain are both $G$). \vspace{0.1in} {\it Note:} Instead of showing $\varphi$ is its own inverse. We could prove it is one-to-one and onto directly. This would go something like: Suppose $\varphi(x)=\varphi(y)$ so that $x^{-1}=y^{-1}$ and thus $x=y$. Therefore, $\varphi$ is one-to-one. Next, $\varphi(x^{-1})=(x^{-1})^{-1}=x$. Therefore, $\varphi$ is onto. \vspace{0.1in} Is $\varphi$ an automorphism if $G$ is not Abelian? Why or why not? \vspace{0.1in} No. Since $G$ is not abelian, there are element $a,b \in G$ such that $ab \not= ba$ and so $a^{-1}b^{-1} \not= b^{-1}a^{-1}$. Thus we have $\varphi(ab)=\varphi(b)\varphi(a) \not= \varphi(a)\varphi(b)$ (the homomorphism property fails). \vspace{0.1in} {\it Note:} A bijective mapping for which $\varphi(ab)=\varphi(b)\varphi(a)$ (the order is reversed) is often called an {\bf anti-isomorphism}. \vspace{0.1in} \end{enumerate} \noindent {\large 4. (20 points)} Zombie Apocalypse! Does anyone actually read the directions to these problems?\\ {}\mbox{\;} \hfill [{\color{red}Answer:} Apparently, ``Yes.''] \begin{enumerate}[(a)] \item Let $\sigma = (2453)(1346)(126) \in S_6$ Write $\sigma$ as a product of disjoint cycles \vspace{0.1in} Think of each cycle as a map. For example:\\ \mbox{\;}\hfill $((2453) \circ (1346) \circ (126))[1] = (2453)((1346)((126)[1])) = (2453)((1346)[2])=(2453)[2]=4$ So $\sigma = (1462)(35)$ \vspace{0.1in} Find $\sigma^{-1}$ \hfill Simply write the permutation backwards. Then rewrite it with ``good manners''. $\sigma^{-1} = (53)(2641) = (35)(1264) = (1264)(35)$. Note that $(53)=(35)$ (bumping 3 to the front) and $(2641)=(1264)$ (bumping 1 to the front). We can switch the order of the 4-cycle $(1264)$ and the transposition $(35)$ because disjoint cycles commute. \vspace{0.1in} What is the order of $\sigma$? \underline{$\mathrm{lcm}(4,2)=4$}\\ {}[If a permutation is written as a product of disjoint cycles, its order is the least common multiple of the lengths of the cycles.] \vspace{0.1in} Is $\sigma$ even or odd? \underline{$\sigma$ is even.} \vspace{0.1in} To see that $\sigma$ is even we can note that any even length cycle is odd and any odd length cycle is even. $\sigma$ is made up of two even length cycles. So $\sigma$ is odd with odd = even. Alternatively, $\sigma = (1462)(35)=(12)(16)(14)(35)$ (four transpositions = even). \vspace{0.1in} \item For convenience: $A_4 = \{ (1), (123), (132), (124), (142), (134), (143), (234), (243), (12)(34), (13)(24), (14)(23) \}$ Consider the subgroup $H= \{ (1), (12)(34), (13)(24), (14)(23) \}$. Find the left cosets of $H$ in $A_4$. \vspace{0.1in} Since $|A_4|=12$ and $|H|=4$ we must have $[A_4:H] = |A_4|/|H|= 12/4=3$ cosets. \begin{itemize} \item $H= \{ (1), (12)(34), (13)(24), (14)(23) \}$ \item $(123)H = \{ (123)(1), (123)(12)(34), (123)(13)(24), (123)(14)(23) \} = \{ (123), (134), (243), (142) \}$ \item $(132)H = \{ \mathrm{left-overs} \} = \{ (132), (143), (234), (124) \}$ \end{itemize} Notice that a careful student only needed to perform {\large\bf 3} (non-trivial) permutation multiplications to get the correct answer! \item Let $\sigma=(14)(23) \in S_4$. What is the order of $\sigma$? \underline{$\mathrm{lcm}(2,2)=2$} \vspace{0.1in} Compute $\sigma^{999}=\sigma=(14)(23)$ (since $999\equiv 1 \;(\mathrm{mod} 2)$ and the fact that exponents operate ``mod'' the order of the element). \vspace{0.1in} \item Find an element of order 15 in $S_8$. \vspace{0.1in} We need lcm(???) = 15 and ??? adds up to 8 or less. 3 and 5 do the trick. So $(123)(45678) \in S_8$ has order $\mathrm{lcm}(3,5)=15$. \vspace{0.1in} \end{enumerate} \newpage \noindent {\large 5. (20 points)} \begin{enumerate}[(a)] \item Let $G$ be a group, $H$ be a subgroup of $G$, and $a,b \in G$. Prove $aH=bH$ implies that $a^{-1}b \in H$.\\ {}[You may {\bf not} use any theorems about cosets.] \vspace{0.1in} I'll give two similar proofs. The second is a little cleaner but involves a bit more forethought. \vspace{0.1in} \begin{description} \item[Proof 1:] Since $aH=bH$ they share all elements, pick one: $x \in aH=bH$. Now since $x \in aH$, there exists some $h \in H$ such that $x=ah$. Likewise since $x \in bH$, there exists some $h' \in H$ such that $x=bh'$. Thus we have $ah=x=bh'$. Multiply on the left by $a^{-1}$ and on the right by $(h')^{-1}$ and get: $a^{-1}ah(h')^{-1}=a^{-1}bh'(h')^{-1}$ so that $h(h')^{-1}=a^{-1}b$. Now $h,h' \in H$ thus $h(h')^{-1} \in H$. Therefore, we have that $a^{-1}b \in H$. \item[Proof 2:] Let $e \in G$ be the identity. Then $e \in H$ (the identity belongs to every subgroup) so $b=be \in bH$. Thus $b \in aH$ (since $aH=bH$). Therefore, there is some $h\in H$ such that $b=ah$. Thus $a^{-1}b=h\in H$. \end{description} \vspace{0.1in} \item My friend is computing some cosets of $K$ which is a subgroup of $S_4$. He claims has found a left coset $L = \{ (243), (142), (123), (134) \}$. \vspace{0.1in} Assuming my friend didn't make a mistake, what is the order of $K$? \underline{$|L|=4$} since all cosets have the same size and the subgroup itself is a coset. \vspace{0.1in} How many cosets will $K$ have in $S_4$? \underline{$[S_4:K]=|S_4|/|K|=24/4=6$} by Lagrange's theorem. \vspace{0.1in} My friend then starts computing right cosets and finds a coset $R = \{ (1234), (24), (1432), (13), (1324) \}$. I know he must have made a mistake. Why? \vspace{0.1in} $|R|=5$ but all cosets (left and right) must have the same size. Assuming the original coset is ok, this one has one too many elements. Note $R$ is still definitely wrong even if we don't know about the other coset's size. Why? The size of a coset must divide is order of the group (since coset size = subgroup size = divisor of the group's order). But $5=|R|$ does not divide $|S_4|=4!=24$. \vspace{0.1in} \item $H$ and $K$ are subgroups of $G$ such that $H \subsetneq K \subsetneq G$ (they are proper subsets of each other). $G$ has order 50. $K$ has order 5. What are the possible orders for $H$? \vspace{0.1in} {\it Note:} This problem as stated contains a typo. First, I will work it as it is written. We have that $H$ and $K$ are subgroups of $G$. So their orders must divide $|G|=50$. Thus $|H|,|K|$ could be $1,2,5,10,25,$ or $50$. Next, $|K|=5$. So $|H|$ must divide 5. Thus $|H|=1$ or $5$. But we are also told that $H\not= K$ and $K\not=G$. Thus $|H|\not=|K|=5$. Therefore, $|H|=1$. [So $H$ must be the trivial subgroup: $H=\{e\}$.] {\bf The Intended Problem:} Suppose $H \subsetneq K \subsetneq G$, $G$ has order 50. $H$ has order 5. What are the possible orders for $K$? (I mixed up $H$ and $K$ when typing the original test.) In this case, we must have $|K|=1,2,5,10,25,$ or $50$ since it's a subgroup of $G$. Then $|K| \not= 50$ because $K\not=G$. Next, $H$ is a subgroup of $K$ so $5=|H|$ must divide $|K|$. This rules out $1$ and $2$. Finally, $H \not= K$ so $|K| \not= |H|=5$. Therefore, $|K|$ is either 10 or 25. \end{enumerate} \end{document}