# IGNOU MMT-003 Solved Assignment 2024 | M.Sc. MACS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

365.00

Access via our Android App Only

Details For MMT-003 Solved Assignment

## IGNOU MMT-003 Assignment Question Paper 2024

mmt-003-assignment-question-paper-1f02de37-022a-430d-b961-44d586763fd8

# mmt-003-assignment-question-paper-1f02de37-022a-430d-b961-44d586763fd8

1. Which of the following statements are true and which are false? Give reasons for your answer.
(a) If a finite group $G$$G$GG$G$ acts on a finite set $S$$S$SS$S$, then ${G}_{{s}_{1}}={G}_{{s}_{2}}$${G}_{{s}_{1}}={G}_{{s}_{2}}$G_(s_(1))=G_(s_(2))G_{s_1}=G_{s_2}${G}_{{s}_{1}}={G}_{{s}_{2}}$ for all ${s}_{1},{s}_{2}\in X$${s}_{1},{s}_{2}\in X$s_(1),s_(2)in Xs_1, s_2 \in X${s}_{1},{s}_{2}\in X$.
(b) There are exactly 8 elements of order 3 in ${S}_{4}$${S}_{4}$S_(4)S_4${S}_{4}$.
(c) If $F=\mathbb{Q}\left(\sqrt[5]{2},\sqrt[3]{5}\right)$$F=\mathbb{Q}\left(\sqrt[5]{2},\sqrt[3]{5}\right)$F=Q(root(5)(2),root(3)(5))F=\mathbb{Q}(\sqrt[5]{2}, \sqrt[3]{5})$F=\mathbb{Q}\left(\sqrt[5]{2},\sqrt[3]{5}\right)$, then $\left[F:\mathbb{Q}\right]=8$$\left[F:\mathbb{Q}\right]=8$[F:Q]=8[F: \mathbb{Q}]=8$\left[F:\mathbb{Q}\right]=8$.
(d) ${\mathbf{F}}_{7}\left(\sqrt{3}\right)={\mathbf{F}}_{7}\left(\sqrt{5}\right)$${\mathbf{F}}_{7}\left(\sqrt{3}\right)={\mathbf{F}}_{7}\left(\sqrt{5}\right)$F_(7)(sqrt3)=F_(7)(sqrt5)\mathbf{F}_7(\sqrt{3})=\mathbf{F}_7(\sqrt{5})${\mathbf{F}}_{7}\left(\sqrt{3}\right)={\mathbf{F}}_{7}\left(\sqrt{5}\right)$.
(e) For any $\alpha \in {\mathbb{F}}_{{2}^{5}}^{\ast },\alpha \ne 1,{\mathbb{F}}_{{2}^{5}}={\mathbb{F}}_{2}\left[\alpha \right]$$\alpha \in {\mathbb{F}}_{{2}^{5}}^{\ast },\alpha \ne 1,{\mathbb{F}}_{{2}^{5}}={\mathbb{F}}_{2}\left[\alpha \right]$alpha inF_(2^(5))^(**),alpha!=1,F_(2^(5))=F_(2)[alpha]\alpha \in \mathbb{F}_{2^5}^*, \alpha \neq 1, \mathbb{F}_{2^5}=\mathbb{F}_2[\alpha]$\alpha \in {\mathbb{F}}_{{2}^{5}}^{\ast },\alpha \ne 1,{\mathbb{F}}_{{2}^{5}}={\mathbb{F}}_{2}\left[\alpha \right]$.
2. (a) Consider the natural action of $G{L}_{2}\left(\mathbb{R}\right)$$G{L}_{2}\left(\mathbb{R}\right)$GL_(2)(R)G L_2(\mathbb{R})$G{L}_{2}\left(\mathbb{R}\right)$ on ${\mathbf{M}}_{2}\left(\mathbb{R}\right)$${\mathbf{M}}_{2}\left(\mathbb{R}\right)$M_(2)(R)\mathbf{M}_2(\mathbb{R})${\mathbf{M}}_{2}\left(\mathbb{R}\right)$, the set of $2×2$$2×2$2xx22 \times 2$2×2$ real matrices, by left multiplication.
(i) Under this action, if $\mathrm{det}\left(\mathbf{x}\right)\ne \mathbf{0}$$\mathrm{det}\left(\mathbf{x}\right)\ne \mathbf{0}$det(x)!=0\operatorname{det}(\mathbf{x}) \neq \mathbf{0}$\mathrm{det}\left(\mathbf{x}\right)\ne \mathbf{0}$, show that the stabiliser of $\mathbf{x}\in {M}_{2}\left(\mathbb{R}\right)$$\mathbf{x}\in {M}_{2}\left(\mathbb{R}\right)$xinM_(2)(R)\mathbf{x} \in M_2(\mathbb{R})$\mathbf{x}\in {M}_{2}\left(\mathbb{R}\right)$ is $\left\{\mathbf{I}\right\}$$\left\{\mathbf{I}\right\}${I}\{\mathbf{I}\}$\left\{\mathbf{I}\right\}$, where $\mathbf{I}$$\mathbf{I}$I\mathbf{I}$\mathbf{I}$ is the $2×2$$2×2$2xx22 \times 2$2×2$ identity matrix.
(ii) Suppose that $\mathrm{det}\left(\mathbf{x}\right)=\mathbf{0}$$\mathrm{det}\left(\mathbf{x}\right)=\mathbf{0}$det(x)=0\operatorname{det}(\mathbf{x})=\mathbf{0}$\mathrm{det}\left(\mathbf{x}\right)=\mathbf{0}$ in the remaining parts of this exercise. We will show that the stabiliser of $\mathbf{x}$$\mathbf{x}$x\mathbf{x}$\mathbf{x}$ is infinite. If $\mathbf{x}=\mathbf{0}$$\mathbf{x}=\mathbf{0}$x=0\mathbf{x}=\mathbf{0}$\mathbf{x}=\mathbf{0}$, the stabiliser of $\mathbf{x}$$\mathbf{x}$x\mathbf{x}$\mathbf{x}$ is $G{L}_{2}\left(\mathbb{R}\right)$$G{L}_{2}\left(\mathbb{R}\right)$GL_(2)(R)G L_2(\mathbb{R})$G{L}_{2}\left(\mathbb{R}\right)$. So suppose $\mathbf{x}\ne \mathbf{0}$$\mathbf{x}\ne \mathbf{0}$x!=0\mathbf{x} \neq \mathbf{0}$\mathbf{x}\ne \mathbf{0}$. Let us write $\mathbf{x}=\left[\begin{array}{ll}a& c\\ b& d\end{array}\right]$$\mathbf{x}=\left[\begin{array}{l}a c\\ b d\end{array}\right]$x=[[a,c],[b,d]]\mathbf{x}=\left[\begin{array}{ll}a & c \\ b & d\end{array}\right]$\mathbf{x}=\left[\begin{array}{ll}a& c\\ b& d\end{array}\right]$. Then, $\left[\begin{array}{l}a\\ b\end{array}\right]=\lambda \left[\begin{array}{l}c\\ d\end{array}\right]$$\left[\begin{array}{l}a\\ b\end{array}\right]=\lambda \left[\begin{array}{l}c\\ d\end{array}\right]$[[a],[b]]=lambda[[c],[d]]\left[\begin{array}{l}a \\ b\end{array}\right]=\lambda\left[\begin{array}{l}c \\ d\end{array}\right]$\left[\begin{array}{l}a\\ b\end{array}\right]=\lambda \left[\begin{array}{l}c\\ d\end{array}\right]$ for non-zero $\lambda \in \mathbb{R}$$\lambda \in \mathbb{R}$lambda inR\lambda \in \mathbb{R}$\lambda \in \mathbb{R}$. Why ?
(iii) Let $\left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ {b}^{\mathrm{\prime }}\end{array}\right]$$\left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ {b}^{\mathrm{\prime }}\end{array}\right]$[[a^(‘)],[b^(‘)]]\left[\begin{array}{l}a^{\prime} \\ b^{\prime}\end{array}\right]$\left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ {b}^{\mathrm{\prime }}\end{array}\right]$ be a vector that is not a scalar multiple of $\left[\begin{array}{l}a\\ b\end{array}\right]$$\left[\begin{array}{l}a\\ b\end{array}\right]$[[a],[b]]\left[\begin{array}{l}a \\ b\end{array}\right]$\left[\begin{array}{l}a\\ b\end{array}\right]$. Show that there is a matrix $\mathbf{b}=$$\mathbf{b}=$b=\mathbf{b}=$\mathbf{b}=$ $\left[\begin{array}{ll}u& v\\ w& z\end{array}\right]$$\left[\begin{array}{l}u v\\ w z\end{array}\right]$[[u,v],[w,z]]\left[\begin{array}{ll}u & v \\ w & z\end{array}\right]$\left[\begin{array}{ll}u& v\\ w& z\end{array}\right]$ such that $\mathbf{b}\left[\begin{array}{l}a\\ b\end{array}\right]=\mathbf{0}$$\mathbf{b}\left[\begin{array}{l}a\\ b\end{array}\right]=\mathbf{0}$b[[a],[b]]=0\mathbf{b}\left[\begin{array}{l}a \\ b\end{array}\right]=\mathbf{0}$\mathbf{b}\left[\begin{array}{l}a\\ b\end{array}\right]=\mathbf{0}$ and $\mathbf{b}\left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ {b}^{\mathrm{\prime }}\end{array}\right]=\alpha \left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ {b}^{\mathrm{\prime }}\end{array}\right]$$\mathbf{b}\left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ {b}^{\mathrm{\prime }}\end{array}\right]=\alpha \left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ {b}^{\mathrm{\prime }}\end{array}\right]$b[[a^(‘)],[b^(‘)]]=alpha[[a^(‘)],[b^(‘)]]\mathbf{b}\left[\begin{array}{l}a^{\prime} \\ b^{\prime}\end{array}\right]=\alpha\left[\begin{array}{l}a^{\prime} \\ b^{\prime}\end{array}\right]$\mathbf{b}\left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ {b}^{\mathrm{\prime }}\end{array}\right]=\alpha \left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ {b}^{\mathrm{\prime }}\end{array}\right]$.(Hint: Set up two sets of simultaneous equations in two unknowns and argue why they have a solution.)
(iv) Check that $\mathbf{I}-\mathbf{b}$$\mathbf{I}-\mathbf{b}$I-b\mathbf{I}-\mathbf{b}$\mathbf{I}-\mathbf{b}$ is in the stabiliser of $\mathbf{x}$$\mathbf{x}$x\mathbf{x}$\mathbf{x}$. Also, show that there are infinitely many choices of $\alpha$$\alpha$alpha\alpha$\alpha$ for which $\mathbf{I}-\mathbf{b}$$\mathbf{I}-\mathbf{b}$I-b\mathbf{I}-\mathbf{b}$\mathbf{I}-\mathbf{b}$ is invertible.
(b) Let $H$$H$HH$H$ be a finite group and, for some prime $p$$p$pp$p$, let $P$$P$PP$P$ be a p-Sylow subgroup of $H$$H$HH$H$ which is normal in $H$$H$HH$H$. Suppose $H$$H$HH$H$ is normal in $K$$K$KK$K$, where $K$$K$KK$K$ is a finite group. Then, show that $P$$P$PP$P$ is normal in $K$$K$KK$K$.
(c) Find the elementary divisors and invariant factors of ${\mathbb{Z}}_{8}×{\mathbb{Z}}_{12}×{\mathbb{Z}}_{15}$${\mathbb{Z}}_{8}×{\mathbb{Z}}_{12}×{\mathbb{Z}}_{15}$Z_(8)xxZ_(12)xxZ_(15)\mathbb{Z}_8 \times \mathbb{Z}_{12} \times \mathbb{Z}_{15}${\mathbb{Z}}_{8}×{\mathbb{Z}}_{12}×{\mathbb{Z}}_{15}$.
3. Describe the set of primes $p$$p$pp$p$ for which ${x}^{2}-11$${x}^{2}-11$x^(2)-11x^2-11${x}^{2}-11$ splits into linear factors over ${\mathbb{Z}}_{p}$${\mathbb{Z}}_{p}$Z_(p)\mathbb{Z}_p${\mathbb{Z}}_{p}$.
4. (a) Determine, up to isomorphism, all the finite groups with exactly 2 conjugacy classes.
(b) Is there a finite group with class equation $1+1+2+2+2+2+2+2$$1+1+2+2+2+2+2+2$1+1+2+2+2+2+2+21+1+2+2+2+2+2+2$1+1+2+2+2+2+2+2$ ?
(c) Compute the following:
a) $\left(\frac{173}{211}\right)$$\left(\frac{173}{211}\right)$((173)/(211))\left(\frac{173}{211}\right)$\left(\frac{173}{211}\right)$
b) $\left(\frac{167}{239}\right)$$\left(\frac{167}{239}\right)$((167)/(239))\left(\frac{167}{239}\right)$\left(\frac{167}{239}\right)$.
5. (a) Let $F\left(\alpha \right)$$F\left(\alpha \right)$F(alpha)F(\alpha)$F\left(\alpha \right)$ be a finite extension $F$$F$FF$F$ of odd degree(greater than 1). Show that $F\left({\alpha }^{2}\right)=F\left(\alpha \right)$$F\left({\alpha }^{2}\right)=F\left(\alpha \right)$F(alpha^(2))=F(alpha)F\left(\alpha^2\right)=F(\alpha)$F\left({\alpha }^{2}\right)=F\left(\alpha \right)$.
(b) Let $F\subset K$$F\subset K$F sub KF \subset K$F\subset K$ and let $\alpha ,\beta \in K$$\alpha ,\beta \in K$alpha,beta in K\alpha, \beta \in K$\alpha ,\beta \in K$ be algebraic over $\mathrm{F}$$\mathrm{F}$F\mathrm{F}$\mathrm{F}$ of degree $\mathrm{m}$$\mathrm{m}$m\mathrm{m}$\mathrm{m}$ and $\mathrm{n}$$\mathrm{n}$n\mathrm{n}$\mathrm{n}$, respectively. Show that $\left[F\left(\alpha ,\beta \right):F\right]\le mn$$\left[F\left(\alpha ,\beta \right):F\right]\le mn$[F(alpha,beta):F] <= mn[F(\alpha, \beta): F] \leq m n$\left[F\left(\alpha ,\beta \right):F\right]\le mn$. What can you say about $\left[F\left(\alpha ,\beta \right):F\right]$$\left[F\left(\alpha ,\beta \right):F\right]$[F(alpha,beta):F][F(\alpha, \beta): F]$\left[F\left(\alpha ,\beta \right):F\right]$ if $\mathrm{m}$$\mathrm{m}$m\mathrm{m}$\mathrm{m}$ and $\mathrm{n}$$\mathrm{n}$n\mathrm{n}$\mathrm{n}$ are coprime?
(c) Find $\left[\mathbb{Q}\left(\sqrt[3]{2},\omega \right):\mathbb{Q}\right]$$\left[\mathbb{Q}\left(\sqrt[3]{2},\omega \right):\mathbb{Q}\right]$[Q(root(3)(2),omega):Q][\mathbb{Q}(\sqrt[3]{2}, \omega): \mathbb{Q}]$\left[\mathbb{Q}\left(\sqrt[3]{2},\omega \right):\mathbb{Q}\right]$ where ${\omega }^{3}=1,\omega \ne 1$${\omega }^{3}=1,\omega \ne 1$omega^(3)=1,omega!=1\omega^3=1, \omega \neq 1${\omega }^{3}=1,\omega \ne 1$.
6. (a) If $\mathrm{char}\left(F\right)\ne 2$$\mathrm{char}\left(F\right)\ne 2$char(F)!=2\operatorname{char}(F) \neq 2$\mathrm{char}\left(F\right)\ne 2$, show that a polynomial $a{x}^{2}+bx+c$$a{x}^{2}+bx+c$ax^(2)+bx+ca x^2+b x+c$a{x}^{2}+bx+c$ is irreducible iff ${b}^{2}-4ac\notin {\mathbb{F}}^{\ast 2}$${b}^{2}-4ac\notin {\mathbb{F}}^{\ast 2}$b^(2)-4ac!inF^(**2)b^2-4 a c \notin \mathbb{F}^{* 2}${b}^{2}-4ac\notin {\mathbb{F}}^{\ast 2}$ where ${\mathbb{F}}^{\ast 2}$${\mathbb{F}}^{\ast 2}$F^(**2)\mathbb{F}^{* 2}${\mathbb{F}}^{\ast 2}$ is the group of squares in ${\mathbb{F}}^{\ast }$${\mathbb{F}}^{\ast }$F^(**)\mathbb{F}^*${\mathbb{F}}^{\ast }$.
(b) By looking at the factorisation of ${x}^{9}-x\in {\mathbb{F}}_{3}\left[x\right]$${x}^{9}-x\in {\mathbb{F}}_{3}\left[x\right]$x^(9)-x inF_(3)[x]x^9-x \in \mathbb{F}_3[x]${x}^{9}-x\in {\mathbb{F}}_{3}\left[x\right]$ guess the number of irreducible polynomials of degree 2 over ${\mathbb{F}}_{3}$${\mathbb{F}}_{3}$F_(3)\mathbb{F}_3${\mathbb{F}}_{3}$. Find all the irreducible polynomials of degree 2 over ${\mathbb{F}}_{3}$${\mathbb{F}}_{3}$F_(3)\mathbb{F}_3${\mathbb{F}}_{3}$.
(c) If $\mathbb{F}$$\mathbb{F}$F\mathbb{F}$\mathbb{F}$ is a finite field show that there is always an irreducible polynomial of the form ${x}^{3}-x+a$${x}^{3}-x+a$x^(3)-x+ax^3-x+a${x}^{3}-x+a$ where $a\in F$$a\in F$a in Fa \in F$a\in F$.(Hint: Show that $x↦{x}^{3}-x$$x↦{x}^{3}-x$x|->x^(3)-xx \mapsto x^3-x$x↦{x}^{3}-x$ is not a surjective map.)
7. (a) Suppose that $M=\left[\begin{array}{ll}A& B\\ C& D\end{array}\right]$$M=\left[\begin{array}{l}A B\\ C D\end{array}\right]$M=[[A,B],[C,D]]M=\left[\begin{array}{ll}A & B \\ C & D\end{array}\right]$M=\left[\begin{array}{ll}A& B\\ C& D\end{array}\right]$ is $2n×2n$$2n×2n$2n xx2n2 n \times 2 n$2n×2n$ matrix where $A,B,C$$A,B,C$A,B,CA, B, C$A,B,C$ and $D$$D$DD$D$ are $n×n$$n×n$n xx nn \times n$n×n$ matrices. Show that $M$$M$MM$M$ is symplectic if and only if the following conditions are satisfied:
$\begin{array}{rl}& {A}^{t}D-{C}^{t}B=\mathbf{I}\\ & {A}^{t}C-{C}^{t}A=\mathbf{0}\\ & {B}^{t}D-{D}^{t}B=\mathbf{0}\end{array}$$\begin{array}{r}{A}^{t}D-{C}^{t}B=\mathbf{I}\\ {A}^{t}C-{C}^{t}A=\mathbf{0}\\ {B}^{t}D-{D}^{t}B=\mathbf{0}\end{array}${:[A^(t)D-C^(t)B=I],[A^(t)C-C^(t)A=0],[B^(t)D-D^(t)B=0]:}\begin{aligned} & A^t D-C^t B=\mathbf{I} \\ & A^t C-C^t A=\mathbf{0} \\ & B^t D-D^t B=\mathbf{0} \end{aligned}$\begin{array}{rl}& {A}^{t}D-{C}^{t}B=\mathbf{I}\\ & {A}^{t}C-{C}^{t}A=\mathbf{0}\\ & {B}^{t}D-{D}^{t}B=\mathbf{0}\end{array}$
(Hint: Use block matrix multiplication.)
Also, check that the matrix $\left[\begin{array}{cc}\mathbf{0}& -A\\ A& \mathbf{0}\end{array}\right]$$\left[\begin{array}{cc}\mathbf{0}& -A\\ A& \mathbf{0}\end{array}\right]$[[0,-A],[A,0]]\left[\begin{array}{cc}\mathbf{0} & -A \\ A & \mathbf{0}\end{array}\right]$\left[\begin{array}{cc}\mathbf{0}& -A\\ A& \mathbf{0}\end{array}\right]$, where $A$$A$AA$A$ is a $n×n$$n×n$n xx nn \times n$n×n$ orthogonal matrix, is a symplectic matrix.
(b) The aim of this exercise is to show that $S{P}_{2}\left(\mathbb{R}\right)$$S{P}_{2}\left(\mathbb{R}\right)$SP_(2)(R)S P_2(\mathbb{R})$S{P}_{2}\left(\mathbb{R}\right)$ acts transitively on ${\mathbb{R}}^{2}\mathrm{\setminus }\left\{0\right\}$${\mathbb{R}}^{2}\mathrm{\setminus }\left\{0\right\}$R^(2)\\{0}\mathbb{R}^2 \backslash\{0\}${\mathbb{R}}^{2}\mathrm{\setminus }\left\{0\right\}$.
(c) Show that
(i) Show that a matrix $\left[\begin{array}{ll}a& b\\ c& d\end{array}\right]\in G{L}_{2}\left(\mathbb{R}\right)$$\left[\begin{array}{l}a b\\ c d\end{array}\right]\in G{L}_{2}\left(\mathbb{R}\right)$[[a,b],[c,d]]in GL_(2)(R)\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \in G L_2(\mathbb{R})$\left[\begin{array}{ll}a& b\\ c& d\end{array}\right]\in G{L}_{2}\left(\mathbb{R}\right)$ is symplectic if and only if $ad-bc=1$$ad-bc=1$ad-bc=1a d-b c=1$ad-bc=1$.
(ii) Show that, to prove that ${\mathrm{S}\mathrm{P}}_{2}\left(\mathbb{R}\right)$${\mathrm{S}\mathrm{P}}_{2}\left(\mathbb{R}\right)$SP_(2)(R)\mathrm{SP}_2(\mathbb{R})${\mathrm{S}\mathrm{P}}_{2}\left(\mathbb{R}\right)$ acts transitively on $G{L}_{2}\left(\mathbb{R}\right)$$G{L}_{2}\left(\mathbb{R}\right)$GL_(2)(R)G L_2(\mathbb{R})$G{L}_{2}\left(\mathbb{R}\right)$, it is enough to show that, for any vector $\left[\begin{array}{l}a\\ b\end{array}\right]\ne \mathbf{0}\in {\mathbb{R}}^{2}$$\left[\begin{array}{l}a\\ b\end{array}\right]\ne \mathbf{0}\in {\mathbb{R}}^{2}$[[a],[b]]!=0inR^(2)\left[\begin{array}{l}a \\ b\end{array}\right] \neq \mathbf{0} \in \mathbb{R}^2$\left[\begin{array}{l}a\\ b\end{array}\right]\ne \mathbf{0}\in {\mathbb{R}}^{2}$, there is a $2×2$$2×2$2xx22 \times 2$2×2$ symplectic matrix with $\left[\begin{array}{l}a\\ b\end{array}\right]$$\left[\begin{array}{l}a\\ b\end{array}\right]$[[a],[b]]\left[\begin{array}{l}a \\ b\end{array}\right]$\left[\begin{array}{l}a\\ b\end{array}\right]$ as the first column. (Hint: For any matrix $A$$A$AA$A$, what is $A\left[\begin{array}{l}1\\ 0\end{array}\right]$$A\left[\begin{array}{l}1\\ 0\end{array}\right]$A[[1],[0]]A\left[\begin{array}{l}1 \\ 0\end{array}\right]$A\left[\begin{array}{l}1\\ 0\end{array}\right]$ ?)
(iii) Complete the proof by showing that, given any non-zero vector $\left[\begin{array}{l}a\\ b\end{array}\right]$$\left[\begin{array}{l}a\\ b\end{array}\right]$[[a],[b]]\left[\begin{array}{l}a \\ b\end{array}\right]$\left[\begin{array}{l}a\\ b\end{array}\right]$, there is always a non-zero vector $\left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ {b}^{\mathrm{\prime }}\end{array}\right]$$\left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ {b}^{\mathrm{\prime }}\end{array}\right]$[[a^(‘)],[b^(‘)]]\left[\begin{array}{l}a^{\prime} \\ b^{\prime}\end{array}\right]$\left[\begin{array}{l}{a}^{\mathrm{\prime }}\\ {b}^{\mathrm{\prime }}\end{array}\right]$ such that $\left[\begin{array}{ll}a& {a}^{\mathrm{\prime }}\\ b& {b}^{\mathrm{\prime }}\end{array}\right]$$\left[\begin{array}{l}a {a}^{\mathrm{\prime }}\\ b {b}^{\mathrm{\prime }}\end{array}\right]$[[a,a^(‘)],[b,b^(‘)]]\left[\begin{array}{ll}a & a^{\prime} \\ b & b^{\prime}\end{array}\right]$\left[\begin{array}{ll}a& {a}^{\mathrm{\prime }}\\ b& {b}^{\mathrm{\prime }}\end{array}\right]$ is symplectic.
8. In this exercise, we ask you to find the Sylow $p$$p$pp$p$-subgroups of the dihedral group
${D}_{n}=⟨x,y:{x}^{n},{y}^{2},yxyx⟩,n\in \mathbb{N},n\ge 2.$${D}_{n}=⟨x,y:{x}^{n},{y}^{2},yxyx⟩,n\in \mathbb{N},n\ge 2.$D_(n)=(:x,y:x^(n),y^(2),yxyx:),n inN,n >= 2.D_n=\left\langle x, y: x^n, y^2, y x y x\right\rangle, n \in \mathbb{N}, n \geq 2 .${D}_{n}=⟨x,y:{x}^{n},{y}^{2},yxyx⟩,n\in \mathbb{N},n\ge 2.$
(a) Let $p$$p$pp$p$ be an odd prime that divides $n,n={p}^{r}l,p\nmid l$$n,n={p}^{r}l,p\nmid l$n,n=p^(r)l,p∤ln, n=p^r l, p \nmid l$n,n={p}^{r}l,p\nmid l$. Suppose $C=⟨{x}^{l}⟩$$C=⟨{x}^{l}⟩$C=(:x^(l):)C=\left\langle x^l\right\rangle$C=⟨{x}^{l}⟩$. Show that $C$$C$CC$C$ is the unique Sylow $p$$p$pp$p$-subgroup of ${D}_{n}$${D}_{n}$D_(n)D_n${D}_{n}$.
(b) Prove the relation
${y}^{i}{x}^{j}{y}^{k}{x}^{l}=\left\{\begin{array}{ll}{y}^{i}{x}^{j+l}& \text{if}\mathrm{k}\text{is even}\\ {y}^{i+k}{x}^{l-j}& \text{if}\mathrm{k}\text{is odd.}\end{array}$${y}^{i}{x}^{j}{y}^{k}{x}^{l}=\left\{\begin{array}{l}{y}^{i}{x}^{j+l} \text{if}\mathrm{k}\text{is even}\\ {y}^{i+k}{x}^{l-j} \text{if}\mathrm{k}\text{is odd.}\end{array}\right\$y^(i)x^(j)y^(k)x^(l)={[y^(i)x^(j+l),” if “k” is even “],[y^(i+k)x^(l-j),” if “k” is odd. “]:}y^i x^j y^k x^l= \begin{cases}y^i x^{j+l} & \text { if } \mathrm{k} \text { is even } \\ y^{i+k} x^{l-j} & \text { if } \mathrm{k} \text { is odd. }\end{cases}${y}^{i}{x}^{j}{y}^{k}{x}^{l}=\left\{\begin{array}{ll}{y}^{i}{x}^{j+l}& \text{if}\mathrm{k}\text{is even}\\ {y}^{i+k}{x}^{l-j}& \text{if}\mathrm{k}\text{is odd.}\end{array}$
Further, find all the elements of order 2 in ${D}_{n}$${D}_{n}$D_(n)D_n${D}_{n}$.
(c) Find all the Sylow 2-subgroups of ${D}_{n}$${D}_{n}$D_(n)D_n${D}_{n}$ when $n$$n$nn$n$ is odd. Describe them in terms of $x$$x$xx$x$ and $y$$y$yy$y$.
(d) Suppose $n$$n$nn$n$ is even, $n={2}^{k}m$$n={2}^{k}m$n=2^(k)mn=2^k m$n={2}^{k}m$, where $2\nmid m,k\ge 2$$2\nmid m,k\ge 2$2∤m,k >= 22 \nmid m, k \geq 2$2\nmid m,k\ge 2$. Let $N=⟨{x}^{m}⟩$$N=⟨{x}^{m}⟩$N=(:x^(m):)N=\left\langle x^m\right\rangle$N=⟨{x}^{m}⟩$ and $H=⟨y⟩$$H=⟨y⟩$H=(:y:)H=\langle y\rangle$H=⟨y⟩$. Show that $HN$$HN$HNH N$HN$ is a subgroup of ${D}_{n}$${D}_{n}$D_(n)D_n${D}_{n}$. What is its order?
(e) Suppose $n$$n$nn$n$ is as in the previous part. Find all the Sylow 2-supgroups of ${D}_{n}$${D}_{n}$D_(n)D_n${D}_{n}$. Describe them in terms of $x$$x$xx$x$ and $y$$y$yy$y$.
9. (a) Let $G=⟨a,b\mid {a}^{2},{b}^{3},ab{a}^{-1}{b}^{-1}⟩$$G=⟨a,b\mid {a}^{2},{b}^{3},ab{a}^{-1}{b}^{-1}⟩$G=(:a,b∣a^(2),b^(3),aba^(-1)b^(-1):)G=\left\langle a, b \mid a^2, b^3, a b a^{-1} b^{-1}\right\rangle$G=⟨a,b\mid {a}^{2},{b}^{3},ab{a}^{-1}{b}^{-1}⟩$. Show that $G$$G$GG$G$ is the cyclic group of order six.
(b) Solve the following set of congruences:
$\begin{array}{rlrl}x& \equiv 2& & \left(mod17\right)\\ 3x& \equiv 4& \left(mod19\right)\\ x& \equiv 7& & \left(mod23\right)\end{array}$$\begin{array}{r}x \equiv 2 \left(mod17\right)\\ 3x \equiv 4 \left(mod19\right)\\ x \equiv 7 \left(mod23\right)\end{array}${:[x,-=2,,(mod 17)],[3x,-=4,(mod 19)],[x,-=7,,(mod 23)]:}\begin{array}{rlrl} x & \equiv 2 & & (\bmod 17) \\ 3 x & \equiv 4 & (\bmod 19) \\ x & \equiv 7 & & (\bmod 23) \end{array}$\begin{array}{rlrl}x& \equiv 2& & \left(mod17\right)\\ 3x& \equiv 4& \left(mod19\right)\\ x& \equiv 7& & \left(mod23\right)\end{array}$
(c) Show that $\mathbb{Q}\left(\sqrt{-19}\right)$$\mathbb{Q}\left(\sqrt{-19}\right)$Q(sqrt(-19))\mathbb{Q}(\sqrt{-19})$\mathbb{Q}\left(\sqrt{-19}\right)$ is not a UFD by giving two different factorisations of 20 .
$$b^2=c^2+a^2-2ac\:Cos\left(B\right)$$

## MMT-003 Sample Solution 2024

mmt-003-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

# mmt-003-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

MMT-003 Solved Assignment 2024
1. Which of the following statements are true and which are false? Give reasons for your answer.
(a) If a finite group $G$$G$GG$G$ acts on a finite set $S$$S$SS$S$, then ${G}_{{s}_{1}}={G}_{{s}_{2}}$${G}_{{s}_{1}}={G}_{{s}_{2}}$G_(s_(1))=G_(s_(2))G_{s_1}=G_{s_2}${G}_{{s}_{1}}={G}_{{s}_{2}}$ for all ${s}_{1},{s}_{2}\in X$${s}_{1},{s}_{2}\in X$s_(1),s_(2)in Xs_1, s_2 \in X${s}_{1},{s}_{2}\in X$.
The statement "If a finite group $G$$G$GG$G$ acts on a finite set $S$$S$SS$S$, then ${G}_{{s}_{1}}={G}_{{s}_{2}}$${G}_{{s}_{1}}={G}_{{s}_{2}}$G_(s_(1))=G_(s_(2))G_{s_1}=G_{s_2}${G}_{{s}_{1}}={G}_{{s}_{2}}$ for all ${s}_{1},{s}_{2}\in S$${s}_{1},{s}_{2}\in S$s_(1),s_(2)in Ss_1, s_2 \in S${s}_{1},{s}_{2}\in S$" is false. Here, ${G}_{{s}_{1}}$${G}_{{s}_{1}}$G_(s_(1))G_{s_1}${G}_{{s}_{1}}$ and ${G}_{{s}_{2}}$${G}_{{s}_{2}}$G_(s_(2))G_{s_2}${G}_{{s}_{2}}$ denote the stabilizer subgroups of ${s}_{1}$${s}_{1}$s_(1)s_1${s}_{1}$ and ${s}_{2}$${s}_{2}$s_(2)s_2${s}_{2}$ in $S$$S$SS$S$, respectively. The stabilizer subgroup ${G}_{s}$${G}_{s}$G_(s)G_s${G}_{s}$ of an element $s\in S$$s\in S$s in Ss \in S$s\in S$ is defined as the set of all elements in $G$$G$GG$G$ that fix $s$$s$ss$s$, i.e., ${G}_{s}=\left\{g\in G\mid g\cdot s=s\right\}$${G}_{s}=\left\{g\in G\mid g\cdot s=s\right\}$G_(s)={g in G∣g*s=s}G_s = \{g \in G \mid g \cdot s = s\}${G}_{s}=\left\{g\in G\mid g\cdot s=s\right\}$.
Justification/Proof:
The statement claims that for any two elements ${s}_{1}$${s}_{1}$s_(1)s_1${s}_{1}$ and ${s}_{2}$${s}_{2}$s_(2)s_2${s}_{2}$ in the set $S$$S$SS$S$, their stabilizer subgroups in $G$$G$GG$G$ are equal. This would mean that every element of $G$$G$GG$G$ that fixes ${s}_{1}$${s}_{1}$s_(1)s_1${s}_{1}$ also fixes ${s}_{2}$${s}_{2}$s_(2)s_2${s}_{2}$ and vice versa. However, this is not necessarily true for all actions of $G$$G$GG$G$ on $S$$S$SS$S$. The equality of stabilizers for all elements in $S$$S$SS$S$ would imply a very rigid action, essentially saying that either the entire group fixes every element of $S$$S$SS$S$ or that the action is trivial (where $G$$G$GG$G$ acts on $S$$S$SS$S$ in such a way that every element of $G$$G$GG$G$ fixes every element of $S$$S$SS$S$).
Counterexample:
Consider the group $G=\left\{e,g\right\}$$G=\left\{e,g\right\}$G={e,g}G = \{e, g\}$G=\left\{e,g\right\}$ where $e$$e$ee$e$ is the identity element and $g$$g$gg$g$ is an element of order 2 (meaning ${g}^{2}=e$${g}^{2}=e$g^(2)=eg^2 = e${g}^{2}=e$), acting on the set $S=\left\{{s}_{1},{s}_{2}\right\}$$S=\left\{{s}_{1},{s}_{2}\right\}$S={s_(1),s_(2)}S = \{s_1, s_2\}$S=\left\{{s}_{1},{s}_{2}\right\}$ with two elements. Define the action of $G$$G$GG$G$ on $S$$S$SS$S$ as follows:
• $e$$e$ee$e$ fixes both ${s}_{1}$${s}_{1}$s_(1)s_1${s}_{1}$ and ${s}_{2}$${s}_{2}$s_(2)s_2${s}_{2}$ (as it must, being the identity).
• $g$$g$gg$g$ swaps ${s}_{1}$${s}_{1}$s_(1)s_1${s}_{1}$ and ${s}_{2}$${s}_{2}$s_(2)s_2${s}_{2}$, i.e., $g\cdot {s}_{1}={s}_{2}$$g\cdot {s}_{1}={s}_{2}$g*s_(1)=s_(2)g \cdot s_1 = s_2$g\cdot {s}_{1}={s}_{2}$ and $g\cdot {s}_{2}={s}_{1}$$g\cdot {s}_{2}={s}_{1}$g*s_(2)=s_(1)g \cdot s_2 = s_1$g\cdot {s}_{2}={s}_{1}$.
For this action, we have:
• ${G}_{{s}_{1}}=\left\{e\right\}$${G}_{{s}_{1}}=\left\{e\right\}$G_(s_(1))={e}G_{s_1} = \{e\}${G}_{{s}_{1}}=\left\{e\right\}$ because only the identity element fixes ${s}_{1}$${s}_{1}$s_(1)s_1${s}_{1}$.
• ${G}_{{s}_{2}}=\left\{e\right\}$${G}_{{s}_{2}}=\left\{e\right\}$G_(s_(2))={e}G_{s_2} = \{e\}${G}_{{s}_{2}}=\left\{e\right\}$ because only the identity element fixes ${s}_{2}$${s}_{2}$s_(2)s_2${s}_{2}$.
In this specific example, ${G}_{{s}_{1}}$${G}_{{s}_{1}}$G_(s_(1))G_{s_1}${G}_{{s}_{1}}$ and ${G}_{{s}_{2}}$${G}_{{s}_{2}}$G_(s_(2))G_{s_2}${G}_{{s}_{2}}$ happen to be equal because the non-identity element of $G$$G$GG$G$ does not fix either of the elements of $S$$S$SS$S$ but rather swaps them. However, the action of $g$$g$gg$g$ clearly differentiates ${s}_{1}$${s}_{1}$s_(1)s_1${s}_{1}$ and ${s}_{2}$${s}_{2}$s_(2)s_2${s}_{2}$, showing that the nature of the action can lead to different behaviors for different elements of $S$$S$SS$S$, and it’s easy to construct examples where ${G}_{{s}_{1}}\ne {G}_{{s}_{2}}$${G}_{{s}_{1}}\ne {G}_{{s}_{2}}$G_(s_(1))!=G_(s_(2))G_{s_1} \neq G_{s_2}${G}_{{s}_{1}}\ne {G}_{{s}_{2}}$ by adjusting the action.
For a more direct counterexample to the statement, consider a group $G$$G$GG$G$ acting on a set $S=\left\{{s}_{1},{s}_{2},{s}_{3}\right\}$$S=\left\{{s}_{1},{s}_{2},{s}_{3}\right\}$S={s_(1),s_(2),s_(3)}S = \{s_1, s_2, s_3\}$S=\left\{{s}_{1},{s}_{2},{s}_{3}\right\}$ where $G$$G$GG$G$ fixes ${s}_{1}$${s}_{1}$s_(1)s_1${s}_{1}$ and ${s}_{2}$${s}_{2}$s_(2)s_2${s}_{2}$ but not ${s}_{3}$${s}_{3}$s_(3)s_3${s}_{3}$, or any scenario where the action is not uniform across all elements of $S$$S$SS$S$. This shows that the original statement is false.
(b) There are exactly 8 elements of order 3 in ${S}_{4}$${S}_{4}$S_(4)S_4${S}_{4}$.
To determine the truth of the statement "There are exactly 8 elements of order 3 in ${S}_{4}$${S}_{4}$S_(4)S_4${S}_{4}$," we need to understand the structure of the symmetric group ${S}_{4}$${S}_{4}$S_(4)S_4${S}_{4}$ and the concept of an element’s order.
• The symmetric group ${S}_{4}$${S}_{4}$S_(4)S_4${S}_{4}$ consists of all permutations of four elements, with a total of $4!=24$$4!=24$4!=244! = 24$4!=24$ elements.
• The order of an element in a group is the smallest positive integer $n$$n$nn