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IGNOU MMT-003 Solved Assignment 2024 | M.Sc. MACS
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IGNOU MMT-003 Assignment Question Paper 2024
mmt-003-assignment-question-paper-1f02de37-022a-430d-b961-44d586763fd8
- Which of the following statements are true and which are false? Give reasons for your answer.
(a) If a finite groupG G acts on a finite setS S , thenG_(s_(1))=G_(s_(2)) G_{s_1}=G_{s_2} for alls_(1),s_(2)in X s_1, s_2 \in X .
(b) There are exactly 8 elements of order 3 inS_(4) S_4 .
(c) IfF=Q(root(5)(2),root(3)(5)) F=\mathbb{Q}(\sqrt[5]{2}, \sqrt[3]{5}) , then[F:Q]=8 [F: \mathbb{Q}]=8 .
(d)F_(7)(sqrt3)=F_(7)(sqrt5) \mathbf{F}_7(\sqrt{3})=\mathbf{F}_7(\sqrt{5}) .
(e) For anyalpha 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] . - (a) Consider the natural action of
GL_(2)(R) G L_2(\mathbb{R}) onM_(2)(R) \mathbf{M}_2(\mathbb{R}) , the set of2xx2 2 \times 2 real matrices, by left multiplication.
(i) Under this action, ifdet(x)!=0 \operatorname{det}(\mathbf{x}) \neq \mathbf{0} , show that the stabiliser ofxinM_(2)(R) \mathbf{x} \in M_2(\mathbb{R}) is{I} \{\mathbf{I}\} , whereI \mathbf{I} is the2xx2 2 \times 2 identity matrix.
(ii) Suppose thatdet(x)=0 \operatorname{det}(\mathbf{x})=\mathbf{0} in the remaining parts of this exercise. We will show that the stabiliser ofx \mathbf{x} is infinite. Ifx=0 \mathbf{x}=\mathbf{0} , the stabiliser ofx \mathbf{x} isGL_(2)(R) G L_2(\mathbb{R}) . So supposex!=0 \mathbf{x} \neq \mathbf{0} . Let us writex=[[a,c],[b,d]] \mathbf{x}=\left[\begin{array}{ll}a & c \\ b & d\end{array}\right] . Then,[[a],[b]]=lambda[[c],[d]] \left[\begin{array}{l}a \\ b\end{array}\right]=\lambda\left[\begin{array}{l}c \\ d\end{array}\right] for non-zerolambda inR \lambda \in \mathbb{R} . Why ?
(iii) Let[[a^(‘)],[b^(‘)]] \left[\begin{array}{l}a^{\prime} \\ b^{\prime}\end{array}\right] be a vector that is not a scalar multiple of[[a],[b]] \left[\begin{array}{l}a \\ b\end{array}\right] . Show that there is a matrixb= \mathbf{b}= [[u,v],[w,z]] \left[\begin{array}{ll}u & v \\ w & z\end{array}\right] such thatb[[a],[b]]=0 \mathbf{b}\left[\begin{array}{l}a \\ b\end{array}\right]=\mathbf{0} andb[[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] .(Hint: Set up two sets of simultaneous equations in two unknowns and argue why they have a solution.)
(iv) Check thatI-b \mathbf{I}-\mathbf{b} is in the stabiliser ofx \mathbf{x} . Also, show that there are infinitely many choices ofalpha \alpha for whichI-b \mathbf{I}-\mathbf{b} is invertible.
(b) LetH H be a finite group and, for some primep p , letP P be a p-Sylow subgroup ofH H which is normal inH H . SupposeH H is normal inK K , whereK K is a finite group. Then, show thatP P is normal inK K .
(c) Find the elementary divisors and invariant factors ofZ_(8)xxZ_(12)xxZ_(15) \mathbb{Z}_8 \times \mathbb{Z}_{12} \times \mathbb{Z}_{15} . - Describe the set of primes
p p for whichx^(2)-11 x^2-11 splits into linear factors overZ_(p) \mathbb{Z}_p . - (a) Determine, up to isomorphism, all the finite groups with exactly 2 conjugacy classes.
(b) Is there a finite group with class equation1+1+2+2+2+2+2+2 1+1+2+2+2+2+2+2 ?
(c) Compute the following:
a)((173)/(211)) \left(\frac{173}{211}\right)
b)((167)/(239)) \left(\frac{167}{239}\right) . - (a) Let
F(alpha) F(\alpha) be a finite extensionF F of odd degree(greater than 1). Show thatF(alpha^(2))=F(alpha) F\left(\alpha^2\right)=F(\alpha) .
(b) LetF sub K F \subset K and letalpha,beta in K \alpha, \beta \in K be algebraic overF \mathrm{F} of degreem \mathrm{m} andn \mathrm{n} , respectively. Show that[F(alpha,beta):F] <= mn [F(\alpha, \beta): F] \leq m n . What can you say about[F(alpha,beta):F] [F(\alpha, \beta): F] ifm \mathrm{m} andn \mathrm{n} are coprime?
(c) Find[Q(root(3)(2),omega):Q] [\mathbb{Q}(\sqrt[3]{2}, \omega): \mathbb{Q}] whereomega^(3)=1,omega!=1 \omega^3=1, \omega \neq 1 . - (a) If
char(F)!=2 \operatorname{char}(F) \neq 2 , show that a polynomialax^(2)+bx+c a x^2+b x+c is irreducible iffb^(2)-4ac!inF^(**2) b^2-4 a c \notin \mathbb{F}^{* 2} whereF^(**2) \mathbb{F}^{* 2} is the group of squares inF^(**) \mathbb{F}^* .
(b) By looking at the factorisation ofx^(9)-x inF_(3)[x] x^9-x \in \mathbb{F}_3[x] guess the number of irreducible polynomials of degree 2 overF_(3) \mathbb{F}_3 . Find all the irreducible polynomials of degree 2 overF_(3) \mathbb{F}_3 .
(c) IfF \mathbb{F} is a finite field show that there is always an irreducible polynomial of the formx^(3)-x+a x^3-x+a wherea in F a \in F .(Hint: Show thatx|->x^(3)-x x \mapsto x^3-x is not a surjective map.) - (a) Suppose that
M=[[A,B],[C,D]] M=\left[\begin{array}{ll}A & B \\ C & D\end{array}\right] is2n xx2n 2 n \times 2 n matrix whereA,B,C A, B, C andD D aren xx n n \times n matrices. Show thatM M is symplectic if and only if the following conditions are satisfied:
Also, check that the matrix
(b) The aim of this exercise is to show that
(c) Show that
(i) Show that a matrix
(ii) Show that, to prove that
(iii) Complete the proof by showing that, given any non-zero vector
8. In this exercise, we ask you to find the Sylow
(b) Prove the relation
(c) Find all the Sylow 2-subgroups of
(d) Suppose
(e) Suppose
9. (a) Let
(b) Solve the following set of congruences:
MMT-003 Sample Solution 2024
mmt-003-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee
- Which of the following statements are true and which are false? Give reasons for your answer.
(a) If a finite groupG G acts on a finite setS S , thenG_(s_(1))=G_(s_(2)) G_{s_1}=G_{s_2} for alls_(1),s_(2)in X s_1, s_2 \in X .
e e fixes boths_(1) s_1 ands_(2) s_2 (as it must, being the identity).g g swapss_(1) s_1 ands_(2) s_2 , i.e.,g*s_(1)=s_(2) g \cdot s_1 = s_2 andg*s_(2)=s_(1) g \cdot s_2 = s_1 .
G_(s_(1))={e} G_{s_1} = \{e\} because only the identity element fixess_(1) s_1 .G_(s_(2))={e} G_{s_2} = \{e\} because only the identity element fixess_(2) s_2 .
- The symmetric group
S_(4) S_4 consists of all permutations of four elements, with a total of4!=24 4! = 24 elements. - The order of an element in a group is the smallest positive integer
n n such that the element raised to then n th power (under the group operation, which is composition in this case) equals the identity element. For permutations, this means repeating the permutationn n times returns the set to its original order.
F=Q(root(5)(2),root(3)(5)) F=\mathbb{Q}(\sqrt[5]{2}, \sqrt[3]{5}) is a field extension ofQ \mathbb{Q} that includes bothroot(5)(2) \sqrt[5]{2} androot(3)(5) \sqrt[3]{5} .- The degree of the extension
[F:Q] [F: \mathbb{Q}] is the dimension ofF F as a vector space overQ \mathbb{Q} . - The degree of a composite extension can be determined by the multiplicativity of degrees in tower extensions: if
K sube L sube M K \subseteq L \subseteq M are fields, then[M:K]=[M:L][L:K] [M:K] = [M:L][L:K] .
-
Consider the extension
Q(root(5)(2)) \mathbb{Q}(\sqrt[5]{2}) overQ \mathbb{Q} . The minimal polynomial ofroot(5)(2) \sqrt[5]{2} overQ \mathbb{Q} isx^(5)-2 x^5 – 2 , which is irreducible overQ \mathbb{Q} by Eisenstein’s criterion (with primep=2 p=2 ). Thus,[Q(root(5)(2)):Q]=5 [\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q}] = 5 . -
Consider the extension
Q(root(3)(5)) \mathbb{Q}(\sqrt[3]{5}) overQ \mathbb{Q} . The minimal polynomial ofroot(3)(5) \sqrt[3]{5} overQ \mathbb{Q} isx^(3)-5 x^3 – 5 , which is also irreducible overQ \mathbb{Q} by Eisenstein’s criterion (with primep=5 p=5 ). Thus,[Q(root(3)(5)):Q]=3 [\mathbb{Q}(\sqrt[3]{5}):\mathbb{Q}] = 3 . -
To find
[F:Q] [F: \mathbb{Q}] whereF=Q(root(5)(2),root(3)(5)) F=\mathbb{Q}(\sqrt[5]{2}, \sqrt[3]{5}) , we need to consider howroot(5)(2) \sqrt[5]{2} androot(3)(5) \sqrt[3]{5} interact within the extension. If these roots were to generate independent extensions (i.e., the extensions do not intersect in a way that reduces the overall degree), then we might expect the degree ofF F overQ \mathbb{Q} to be the product of the degrees of the individual extensions. However, the actual degree of the combined extension depends on the interaction between these two extensions.
F_(7) \mathbf{F}_7 is the finite field with 7 elements, typically represented as{0,1,2,3,4,5,6} \{0, 1, 2, 3, 4, 5, 6\} with arithmetic operations performed modulo 7.F_(7)(sqrt3) \mathbf{F}_7(\sqrt{3}) denotes the smallest field extension ofF_(7) \mathbf{F}_7 containingsqrt3 \sqrt{3} , and similarly,F_(7)(sqrt5) \mathbf{F}_7(\sqrt{5}) denotes the smallest field extension ofF_(7) \mathbf{F}_7 containingsqrt5 \sqrt{5} .- In the context of finite fields, an element
a a is said to have a square root inF_(7) \mathbf{F}_7 if there exists an elementb inF_(7) b \in \mathbf{F}_7 such thatb^(2)-=amod7 b^2 \equiv a \mod 7 .
-
Square Root of 3 in
F_(7) \mathbf{F}_7 : We need to find if there exists an elementx inF_(7) x \in \mathbf{F}_7 such thatx^(2)-=3mod7 x^2 \equiv 3 \mod 7 . -
Square Root of 5 in
F_(7) \mathbf{F}_7 : Similarly, we need to find if there exists an elementy inF_(7) y \in \mathbf{F}_7 such thaty^(2)-=5mod7 y^2 \equiv 5 \mod 7 .
- For
sqrt3 \sqrt{3} : We look for an elementx x such thatx^(2)-=3mod7 x^2 \equiv 3 \mod 7 . - For
sqrt5 \sqrt{5} : We look for an elementy y such thaty^(2)-=5mod7 y^2 \equiv 5 \mod 7 .
F_(2^(5)) \mathbb{F}_{2^5} is a finite field with32 32 elements. It is the splitting field for the polynomialx^(32)-x x^{32} – x overF_(2) \mathbb{F}_2 , and it contains all roots of this polynomial.F_(2^(5))^(**) \mathbb{F}_{2^5}^* denotes the multiplicative group ofF_(2^(5)) \mathbb{F}_{2^5} , which consists of all non-zero elements ofF_(2^(5)) \mathbb{F}_{2^5} . This group has31 31 elements, and it is cyclic.F_(2)[alpha] \mathbb{F}_2[\alpha] denotes the smallest field containing bothF_(2) \mathbb{F}_2 and the elementalpha \alpha . Ifalpha \alpha is a primitive element ofF_(2^(5)) \mathbb{F}_{2^5} (i.e., a generator of the multiplicative groupF_(2^(5))^(**) \mathbb{F}_{2^5}^* ), thenF_(2)[alpha]=F_(2^(5)) \mathbb{F}_2[\alpha] = \mathbb{F}_{2^5} .- A primitive element of a finite field is an element whose powers generate all non-zero elements of the field.
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