\documentclass{article} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{enumerate} \usepackage{ifthen} \usepackage{graphicx} \usepackage{wasysym} \usepackage{hyperref} \usepackage{ulem} % for sout to strikethrough \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{\varspace}[2]{\mbox{\parbox[c][#1]{#2}{\hfill}}} \newcommand{\diagquotient}[2]{\displaystyle{{#1 \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop #2}}} \newcommand{\force}{{$\mbox{\;}$}} \begin{document} %\pagestyle{empty} \noindent \parbox{2in}{\bf Math 3110} \hfill {\Large \bf Test \#3} \hfill \parbox{2in}{\bf \hfill November $2^{\mathrm{nd}}$, 2020} \vspace{0.1in} \noindent {\large\bf Name:} \underline{\Large\color{red}\sc\quad Answer Key \quad} \hfill {\bf Be sure to show your work!} \vspace{0.1in} \noindent {\large 1. (15 points)} Getting things in order\dots \begin{enumerate}[(a)] \item Let $G = D_5 \times S_4$ where $D_5 = \langle x,y \;|\; x^5=1,y^2=1,xyxy=1 \rangle$ and $S_4$ is permutations on 4 things. \vspace{0.1in} The order of $G$ is $|G|=$ \underline{\quad $|D_5| \cdot |S_4| = 5(2) \cdot 4! = 10 \cdot 24 =$ \fbox{$240$} \quad} \vspace{0.1in} What is the largest element order in $D_5 \times S_4$? Give an example of such an element and explain why it has the largest possible order. \vspace{0.05in} The elements in $D_5$ have orders $1,5$ (identity and rotations like $x$) and $2$ (reflections). The elements in $S_4$ have orders $1,2,3,$ and $4$ (e.g., the identity, transpositions or stuff like $(12)(34)$, 3-cycles, and 4-cycles like $(1234)$). Recall that $|(a,b)| = \mathrm{lcm}(|a|,|b|)$ so thinking of various lcm's, the largest we can get is $\mathrm{lcm}(5,4)=20$. The largest element order in $D_5 \times S_4$ is \fbox{$|(x,(1234))| = \mathrm{lcm}(|x|,|(1234)|) = \mathrm{lcm}(5,4)=20$}. \vspace{0.05in} \item Let $G$ be a group of order 36 with subgroup $H$ of order 12. Let $K$ be some other subgroup of $G$. \vspace{0.05in} \noindent Can $H \cap K$ have order 9? Why or why not? \vspace{0.05in} \fbox{No.} The order of $H \cap K$ must divide both the order of $H$ and the order of $K$. But $9$ does not divide $12=|H|$. \vspace{0.05in} \noindent If $H \cap K$ has order 6, what are the possible orders of $K$? \vspace{0.05in} From Lagrange's theorem, we have that the order of $K$ must divide the order of $G$ (i.e., $|K|$ divides 36), but also since $H \cap K$ is a subgroup of $K$, we must have that $|H\cap K|=6$ divides the order of $K$. Therefore, $|K|$ is a divisor of 36 and a multiple of 6: \fbox{$|K|=6,12,18,$ or \sout{$36$}}. {\it (Mostly) Ridiculous Notes:} We can rule out $36$ since that would require $K=G$ and then $H\cap K=H \cap G=H$ would be of order 12 (not 6). In fact, much more can be said. It turns out that the only orders that actually occur in this situation are $|K|=6,12,$ and $18$. But this is way beyond what I was looking for here. The following \href{https://www.gap-system.org/}{\color{blue}\underline{GAP}} code cycles through all groups of order 36 (up to isomorphism) and looks at all subgroups of order 12 intersected with subgroups $K$ such that the intersection has order 6. It then prints out orders of such $K$'s: \verb|for n in [1..14] do | \verb| G := SmallGroup(36,n);| \verb| L := AllSubgroups(G); | \verb| | \verb| L12:=[]; | \verb| for H in L do | \verb| if Size(H)=12 then Append(L12,[H]); fi; | \verb| od;| \verb| | \verb| for H in L12 do | \verb| for K in L do | \verb| if Size(Intersection(H,K))=6 then | \verb| Print(Size(K)); Print(", "); | \verb| fi;| \verb| od;| \verb| od;| \verb|od;| \vspace{0.05in} \end{enumerate} \newpage \noindent {\large 2. (10 points)} Let $H$ and $K$ be normal subgroups of $G$. Prove that $H \cap K$ is a normal subgroup of $G$. \vspace{0.05in} First, notice that $H \cap K$ is a subset of $G$ since both $H$ and $K$ are subsets of $G$. Both $H$ and $K$ contain the identity $1$ of $G$ since they are subgroups of $G$. Therefore, $1 \in H \cap K$. Thus $H \cap K$ is a non-empty subset of $G$. Next, suppose $a,b \in H \cap K$. Then $a,b \in H$ and $a,b \in K$. Therefore, $ab \in H$ (since $H$ is a subgroup, it is closed under the operation). Likewise, $ab \in K$. Thus since $ab \in H$ and $ab \in K$, we have $ab \in H \cap K$. Also, given $a \in H \cap K$ so that $a \in H$ and $a \in K$, we have $a^{-1} \in H$ and $a^{-1} \in K$ ($H$ and $K$ are subgroups and thus closed under inverses). Therefore, $a^{-1} \in H \cap K$. We have now established that $H \cap K$ is a subgroup of $G$. Finally, let $g \in G$ and $a \in H \cap K$. Then $a \in H$ and since $H$ is normal, we have $gag^{-1} \in H$ (closed under conjugation). Likewise, $gag^{-1} \in K$. Therefore, $gag^{-1} \in H \cap K$. Thus $H \cap K \triangleleft G$. \vspace{0.05in} \noindent {\large 3. (15 points)} Consider $H = \{ 1,x^2 \}$ in $D_4 = \langle x,y \;|\; x^4=1,y^2=1,xyxy=1 \rangle = \{ 1,x,x^2,x^3, y,xy,x^2y, x^3y\}$. Note that $H$ is a normal subgroup of $D_4$. \begin{enumerate}[(a)] \item Quick questions about $\diagquotient{D_4}{H}$. \vspace{0.05in} The order of $\diagquotient{D_4}{H}$ is \underline{\quad $|D_4|/|H| = 8/2=$ \fbox{4} \quad}. \vspace{0.05in} The identity of $\diagquotient{D_4}{H}$ is \underline{\quad $1H=$ \fbox{$H$}\quad}. \hfill $(xH)^{-1}=$ \underline{\quad $x^{-1}H =x^3H=$ \fbox{$xH$} \quad}. \vspace{0.05in} The order of $xH$ in $\diagquotient{D_4}{H}$ is \underline{\quad \fbox{2} since $xH$ is its own inverse \quad}. \hfill The size of the set $xH$ is \underline{\quad the same as $H$: \fbox{2} \quad}. \vspace{0.05in} A few notes: The order of a quotient is just the index of the subgroup. By Lagrange, this is the order of the group divided by the order of the subgroup. Next, the identity of any quotient group is just the subgroup you are quotienting by. To find an inverse, you take the inverse of the representative. Notice that $xH =\{x\cdot 1, x\cdot x^2\}=\{x,x^3\}=x^3H$. Also, since $(xH)^{-1}=xH$ (and $xH$ isn't the identity) it must have order 2. Alternatively, we compute: $xH \not= H$ and $(xH)^2=x^2H=H$. Thus order 2. Finally, all of the cosets have the same size. This is the same as the size of the subgroup itself. In our case, that's 2. \vspace{0.05in} \item (Still referring to part (a).) $H=Z(D_4)$ (our $H$ above is actually the center of $D_4$). Is $D_4/H$ cyclic? Abelian? Identify this quotient group (it has a special name). \vspace{0.05in} By the G/Z-Theorem: $G/Z(G)$ is cyclic if and only if $G$ is Abelian. Since $D_4$ is not Abelian, $D_4/H = D_4/Z(D_4)$ must not be cyclic. However, notice that $\left|\diagquotient{D_4}{H}\right|=4$. We know that there are only two groups of order 4 (up to isomorphism): the cyclic group of order 4 and the Klein 4-group. Therefore, \fbox{$D_4/H$ is not cyclic, but it is Abelian. It is the Klein 4-group.} \vspace{0.05in} \end{enumerate} \noindent {\large 4. (10 points)} Consider $\diagquotient{\mathbb{Z}_{20}}{H}$ where $H=\langle 4 \rangle = \{0,4,8,12,16\}$. List all of the cosets (and their contents) of $H$ in $\mathbb{Z}_{20}$. Then make a Cayley table for this quotient group. \vspace{0.05in} Cosets: $H=\{0,4,8,12,16\}$, $1+H=\{1,5,9,13,17\}$, $2+H=\{2,6,10,14,18\}$, and $3+H=\{3,7,11,15,19\}$. \begin{center} \begin{tabular}{c||c|c|c|c|} $\mathbb{Z}_{20}/H$ & $H$ & $1+H$ & $2+H$ & $3+H$ \\ \hline\hline $H$ & $H$ & $1+H$ & $2+H$ & $3+H$ \\ \hline $1+H$ & $1+H$ & $2+H$ & $3+H$ & $H$ \\ \hline $2+H$ & $2+H$ & $3+H$ & $H$ & $1+H$ \\ \hline $3+H$ & $3+H$ & $H$ & $1+H$ & $2+H$ \\ \hline \end{tabular} \end{center} \vspace{0.05in} What is the order of $2+H$ in $\mathbb{Z}_{20}/H$? \vspace{0.05in} $2+H \not= H$, $(2+H)+(2+H) = (2+2)+H=4+H=H$. Thus $2+H$ has \fbox{order 2}. Alternatively, we could just note that $2+H$ is not the identity but it is its own inverse (as evidenced in the above Cayley table). Thus its order is 2. \vspace{0.05in} \newpage \noindent {\large 5. (15 points)} Something is terribly, horribly wrong! \begin{enumerate}[(a)] \item Let $H \triangleleft S_4$. Why is $\diagquotient{S_4}{H} \cong \mathbb{Z}_{18}$ impossible? \vspace{0.05in} If there were such a quotient group, its order would be $18$. However, the order of $S_4/H$ must divide the order of the group (since the order of the quotient times the order of $H$ must be the order of $S_4$) and $|S_4|=4!=24$. But \fbox{18 does not divide 24}. Therefore, there can be no such quotient of $S_4$. \vspace{0.05in} \item Let $\varphi:S_4 \to G$ be a homomorpihsm into some group $G$. Let $H=\langle (12) \rangle = \{(1),(12)\}$. Why is $\mathrm{ker}(\varphi)=H$ impossible? \vspace{0.05in} Notice that $(13)H = \{ (13)(1),(13)(12) \} =\{(13),(123) \}$ whereas $H(13) = \{(1)(13),(12)(13) \} = \{(13),(132)\}$. Therefore, since $(13)H \not= H(13)$, \fbox{$H$ is not a normal subgroup} of $S_4$. Therefore, $H$ cannot be a kernel (kernels are always normal subgroups). \vspace{0.05in} \item Why can't I have an epimorphism (i.e., onto homomorphism) $\psi:D_4 \times Q \to \mathbb{Z}_{15}$ (where $Q$ is the quaternion group)? \vspace{0.05in} As a consequence of the first isomorphism theorem, we had that the order of the domain of a homomorphism is equal to the order of its kernel times the order of its range (=image). Notice that the order of the domain of $\psi$ is $|D_4 \times Q|=|D_4|\cdot |Q| =8\cdot 8 =64$ whereas the order of the image (since $\psi$ is onto) is $|\mathbb{Z}_{15}|=15$. But \fbox{15 does not divide 64} (the order of the kernel would have to be $64/15$ -- which is absurd). {\it Note:} Actually, much more can be said. Suppose $\psi:D_4 \times Q \to \mathbb{Z}_{15}$ is a homomorphism. Then the order of the range must divide both $|D_4\times Q|=64$ and $|\mathbb{Z}_{15}|=15$ (since the range is a subgroup of the codomain). Thus the order of the range must divide $\mathrm{gcd}(64,15)=1$. This means that the {\it only} homomorphism between $D_4 \times Q$ and $\mathbb{Z}_{15}$ is the trivial morphism: $\psi(x)=0$ for all $x\in D_4 \times Q$. In other words, $\psi$ can't come anywhere close to being onto! \vspace{0.05in} \end{enumerate} \noindent {\large 6 (10 points)} Let $H$ and $K$ be groups. Define $\pi:H \times K \to K$ by $\pi((h,k))=k$. Show $\pi$ is a homomorphism. What does the first isomorphism theorem tell us in this particular situation? \vspace{0.05in} Suppose $(a,b),(x,y) \in H \times K$. Then $\pi((a,b)(x,y))=\pi((ax,by))=by=\pi((a,b))\pi((x,y))$. Therefore, $\pi$ is a homomorphism. Notice that $\pi$ is onto since given any $k \in K$ we have $\pi((1_H,k))=k$ (where $1_H$ is the identity of $H$). Next, \vspace{-0.1in} $$\mathrm{ker}(\pi) = \{ (h,k) \in H \times K \;|\; \pi((h,k))=1_K \}=\{ (h,k) \in H \times K \;|\; k=1_K \} = \{ (h,1_K) \;|\; h \in H\} = H \times \{1_K\}$$ \vspace{-0.05in} (where $1_K$ is the identity of $K$). Therefore, by the first isomorphism theorem: $\diagquotient{H \times K}{\mathrm{ker}(\pi)} \cong \mathrm{im}(\pi)$. In particular, \fbox{$\diagquotient{H \times K}{H \times \{1_K\}} \cong K$}. \vspace{0.05in} \noindent {\large 7. (25 points)} Finite Abelian Groups \begin{enumerate}[(a)] \item List all of the non-isomorphic abelian groups of order $72 =2^3 3^2$. Circle any that are cyclic. \vspace{0.05in} $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{3}$, \ $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{9}$, \ $\mathbb{Z}_{2} \times \mathbb{Z}_{4} \times \mathbb{Z}_{3} \times \mathbb{Z}_{3}$, \ $\mathbb{Z}_{2} \times \mathbb{Z}_{4} \times \mathbb{Z}_{9}$, \ $\mathbb{Z}_{8} \times \mathbb{Z}_{3} \times \mathbb{Z}_{3}$, and \fbox{$\mathbb{Z}_{8} \times \mathbb{Z}_{9}$} \vspace{0.05in} \item How many non-isomorphic abelian groups of order 55,635,800 are there? {\it Note:} 55,635,800 $=2^3 \cdot 5^2 \cdot 11^4 \cdot 19$ and there are 5 non-isomorphic abelian groups of order 14,641 $=11^4$. \smiley \vspace{0.05in} There are 3 ways to split up $2^3$, 2 ways to split of $5^2$, 5 ways to split of $11^4$ and nothing to bust up $19^1$. Therefore, there are $3 \cdot 2 \cdot 5 =$ \fbox{30} Abelian groups of order 55,635,800 (up to isomorphism). \vspace{0.05in} \item Are the groups $\mathbb{Z}_{10} \times \mathbb{Z}_{25} \times \mathbb{Z}_{36}$ and $\mathbb{Z}_{100} \times \mathbb{Z}_{90}$ isomorphic? Explain your answer. \vspace{0.05in} \fbox{Yes.} \quad We can split/combine if moduli are relatively prime: $\mathbb{Z}_{10} \times \mathbb{Z}_{25} \times \mathbb{Z}_{36} \cong \mathbb{Z}_{10} \times \mathbb{Z}_{25} \times \mathbb{Z}_{4} \times \mathbb{Z}_9 \cong \mathbb{Z}_{100} \times \mathbb{Z}_{90}$ \vspace{0.05in} \item What is the largest order among elements of $\mathbb{Z}_{35} \times \mathbb{Z}_{10} \times \mathbb{Z}_{14}$? Explain you answer. \vspace{0.05in} The biggest order we can get is the least common multiple of the largest orders in our three component groups: $\mathrm{lcm}(35,10,14) = \mathrm{lcm}(5\cdot 7, 2 \cdot 5, 2 \cdot 7) = 2 \cdot 5 \cdot 7=$ \fbox{70}. \vspace{0.05in} \end{enumerate} \end{document}