Solved assignment for IGNOU MMT-003 for the 2024 session.

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

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

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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 GGG acts on a finite set S S SSS, 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}Gs1=Gs2 for all s 1 , s 2 X s 1 , s 2 X s_(1),s_(2)in Xs_1, s_2 \in Xs1,s2X.
    (b) There are exactly 8 elements of order 3 in S 4 S 4 S_(4)S_4S4.
    (c) If F = Q ( 2 5 , 5 3 ) F = Q ( 2 5 , 5 3 ) F=Q(root(5)(2),root(3)(5))F=\mathbb{Q}(\sqrt[5]{2}, \sqrt[3]{5})F=Q(25,53), then [ F : Q ] = 8 [ F : Q ] = 8 [F:Q]=8[F: \mathbb{Q}]=8[F:Q]=8.
    (d) F 7 ( 3 ) = F 7 ( 5 ) F 7 ( 3 ) = F 7 ( 5 ) F_(7)(sqrt3)=F_(7)(sqrt5)\mathbf{F}_7(\sqrt{3})=\mathbf{F}_7(\sqrt{5})F7(3)=F7(5).
    (e) For any α F 2 5 , α 1 , F 2 5 = F 2 [ α ] α F 2 5 , α 1 , F 2 5 = F 2 [ α ] 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]αF25,α1,F25=F2[α].
  2. (a) Consider the natural action of G L 2 ( R ) G L 2 ( R ) GL_(2)(R)G L_2(\mathbb{R})GL2(R) on M 2 ( R ) M 2 ( R ) M_(2)(R)\mathbf{M}_2(\mathbb{R})M2(R), the set of 2 × 2 2 × 2 2xx22 \times 22×2 real matrices, by left multiplication.
    (i) Under this action, if det ( x ) 0 det ( x ) 0 det(x)!=0\operatorname{det}(\mathbf{x}) \neq \mathbf{0}det(x)0, show that the stabiliser of x M 2 ( R ) x M 2 ( R ) xinM_(2)(R)\mathbf{x} \in M_2(\mathbb{R})xM2(R) is { I } { I } {I}\{\mathbf{I}\}{I}, where I I I\mathbf{I}I is the 2 × 2 2 × 2 2xx22 \times 22×2 identity matrix.
    (ii) Suppose that det ( x ) = 0 det ( x ) = 0 det(x)=0\operatorname{det}(\mathbf{x})=\mathbf{0}det(x)=0 in the remaining parts of this exercise. We will show that the stabiliser of x x x\mathbf{x}x is infinite. If x = 0 x = 0 x=0\mathbf{x}=\mathbf{0}x=0, the stabiliser of x x x\mathbf{x}x is G L 2 ( R ) G L 2 ( R ) GL_(2)(R)G L_2(\mathbb{R})GL2(R). So suppose x 0 x 0 x!=0\mathbf{x} \neq \mathbf{0}x0. Let us write x = [ a c b d ] x = a      c b      d x=[[a,c],[b,d]]\mathbf{x}=\left[\begin{array}{ll}a & c \\ b & d\end{array}\right]x=[acbd]. Then, [ a b ] = λ [ c d ] a b = λ c d [[a],[b]]=lambda[[c],[d]]\left[\begin{array}{l}a \\ b\end{array}\right]=\lambda\left[\begin{array}{l}c \\ d\end{array}\right][ab]=λ[cd] for non-zero λ R λ R lambda inR\lambda \in \mathbb{R}λR. Why ?
    (iii) Let [ a b ] a b [[a^(‘)],[b^(‘)]]\left[\begin{array}{l}a^{\prime} \\ b^{\prime}\end{array}\right][ab] be a vector that is not a scalar multiple of [ a b ] a b [[a],[b]]\left[\begin{array}{l}a \\ b\end{array}\right][ab]. Show that there is a matrix b = b = b=\mathbf{b}=b= [ u v w z ] u      v w      z [[u,v],[w,z]]\left[\begin{array}{ll}u & v \\ w & z\end{array}\right][uvwz] such that b [ a b ] = 0 b a b = 0 b[[a],[b]]=0\mathbf{b}\left[\begin{array}{l}a \\ b\end{array}\right]=\mathbf{0}b[ab]=0 and b [ a b ] = α [ a b ] b a b = α a b 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]b[ab]=α[ab].(Hint: Set up two sets of simultaneous equations in two unknowns and argue why they have a solution.)
    (iv) Check that I b I b I-b\mathbf{I}-\mathbf{b}Ib is in the stabiliser of x x x\mathbf{x}x. Also, show that there are infinitely many choices of α α alpha\alphaα for which I b I b I-b\mathbf{I}-\mathbf{b}Ib is invertible.
    (b) Let H H HHH be a finite group and, for some prime p p ppp, let P P PPP be a p-Sylow subgroup of H H HHH which is normal in H H HHH. Suppose H H HHH is normal in K K KKK, where K K KKK is a finite group. Then, show that P P PPP is normal in K K KKK.
    (c) Find the elementary divisors and invariant factors of Z 8 × Z 12 × Z 15 Z 8 × Z 12 × Z 15 Z_(8)xxZ_(12)xxZ_(15)\mathbb{Z}_8 \times \mathbb{Z}_{12} \times \mathbb{Z}_{15}Z8×Z12×Z15.
  3. Describe the set of primes p p ppp for which x 2 11 x 2 11 x^(2)-11x^2-11x211 splits into linear factors over Z p Z p Z_(p)\mathbb{Z}_pZp.
  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+21+1+2+2+2+2+2+2 ?
    (c) Compute the following:
    a) ( 173 211 ) 173 211 ((173)/(211))\left(\frac{173}{211}\right)(173211)
    b) ( 167 239 ) 167 239 ((167)/(239))\left(\frac{167}{239}\right)(167239).
  5. (a) Let F ( α ) F ( α ) F(alpha)F(\alpha)F(α) be a finite extension F F FFF of odd degree(greater than 1). Show that F ( α 2 ) = F ( α ) F α 2 = F ( α ) F(alpha^(2))=F(alpha)F\left(\alpha^2\right)=F(\alpha)F(α2)=F(α).
    (b) Let F K F K F sub KF \subset KFK and let α , β K α , β K alpha,beta in K\alpha, \beta \in Kα,βK be algebraic over F F F\mathrm{F}F of degree m m m\mathrm{m}m and n n n\mathrm{n}n, respectively. Show that [ F ( α , β ) : F ] m n [ F ( α , β ) : F ] m n [F(alpha,beta):F] <= mn[F(\alpha, \beta): F] \leq m n[F(α,β):F]mn. What can you say about [ F ( α , β ) : F ] [ F ( α , β ) : F ] [F(alpha,beta):F][F(\alpha, \beta): F][F(α,β):F] if m m m\mathrm{m}m and n n n\mathrm{n}n are coprime?
    (c) Find [ Q ( 2 3 , ω ) : Q ] [ Q ( 2 3 , ω ) : Q ] [Q(root(3)(2),omega):Q][\mathbb{Q}(\sqrt[3]{2}, \omega): \mathbb{Q}][Q(23,ω):Q] where ω 3 = 1 , ω 1 ω 3 = 1 , ω 1 omega^(3)=1,omega!=1\omega^3=1, \omega \neq 1ω3=1,ω1.
  6. (a) If char ( F ) 2 char ( F ) 2 char(F)!=2\operatorname{char}(F) \neq 2char(F)2, show that a polynomial a x 2 + b x + c a x 2 + b x + c ax^(2)+bx+ca x^2+b x+cax2+bx+c is irreducible iff b 2 4 a c F 2 b 2 4 a c F 2 b^(2)-4ac!inF^(**2)b^2-4 a c \notin \mathbb{F}^{* 2}b24acF2 where F 2 F 2 F^(**2)\mathbb{F}^{* 2}F2 is the group of squares in F F F^(**)\mathbb{F}^*F.
    (b) By looking at the factorisation of x 9 x F 3 [ x ] x 9 x F 3 [ x ] x^(9)-x inF_(3)[x]x^9-x \in \mathbb{F}_3[x]x9xF3[x] guess the number of irreducible polynomials of degree 2 over F 3 F 3 F_(3)\mathbb{F}_3F3. Find all the irreducible polynomials of degree 2 over F 3 F 3 F_(3)\mathbb{F}_3F3.
    (c) If F F F\mathbb{F}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+ax3x+a where a F a F a in Fa \in FaF.(Hint: Show that x x 3 x x x 3 x x|->x^(3)-xx \mapsto x^3-xxx3x is not a surjective map.)
  7. (a) Suppose that M = [ A B C D ] M = A      B C      D M=[[A,B],[C,D]]M=\left[\begin{array}{ll}A & B \\ C & D\end{array}\right]M=[ABCD] is 2 n × 2 n 2 n × 2 n 2n xx2n2 n \times 2 n2n×2n matrix where A , B , C A , B , C A,B,CA, B, CA,B,C and D D DDD are n × n n × n n xx nn \times nn×n matrices. Show that M M MMM is symplectic if and only if the following conditions are satisfied:
A t D C t B = I A t C C t A = 0 B t D D t B = 0 A t D C t B = I A t C C t A = 0 B t D D t B = 0 {:[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}AtDCtB=IAtCCtA=0BtDDtB=0
(Hint: Use block matrix multiplication.)
Also, check that the matrix [ 0 A A 0 ] 0 A A 0 [[0,-A],[A,0]]\left[\begin{array}{cc}\mathbf{0} & -A \\ A & \mathbf{0}\end{array}\right][0AA0], where A A AAA is a n × n n × n n xx nn \times nn×n orthogonal matrix, is a symplectic matrix.
(b) The aim of this exercise is to show that S P 2 ( R ) S P 2 ( R ) SP_(2)(R)S P_2(\mathbb{R})SP2(R) acts transitively on R 2 { 0 } R 2 { 0 } R^(2)\\{0}\mathbb{R}^2 \backslash\{0\}R2{0}.
(c) Show that
(i) Show that a matrix [ a b c d ] G L 2 ( R ) a      b c      d G L 2 ( R ) [[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})[abcd]GL2(R) is symplectic if and only if a d b c = 1 a d b c = 1 ad-bc=1a d-b c=1adbc=1.
(ii) Show that, to prove that S P 2 ( R ) S P 2 ( R ) SP_(2)(R)\mathrm{SP}_2(\mathbb{R})SP2(R) acts transitively on G L 2 ( R ) G L 2 ( R ) GL_(2)(R)G L_2(\mathbb{R})GL2(R), it is enough to show that, for any vector [ a b ] 0 R 2 a b 0 R 2 [[a],[b]]!=0inR^(2)\left[\begin{array}{l}a \\ b\end{array}\right] \neq \mathbf{0} \in \mathbb{R}^2[ab]0R2, there is a 2 × 2 2 × 2 2xx22 \times 22×2 symplectic matrix with [ a b ] a b [[a],[b]]\left[\begin{array}{l}a \\ b\end{array}\right][ab] as the first column. (Hint: For any matrix A A AAA, what is A [ 1 0 ] A 1 0 A[[1],[0]]A\left[\begin{array}{l}1 \\ 0\end{array}\right]A[10] ?)
(iii) Complete the proof by showing that, given any non-zero vector [ a b ] a b [[a],[b]]\left[\begin{array}{l}a \\ b\end{array}\right][ab], there is always a non-zero vector [ a b ] a b [[a^(‘)],[b^(‘)]]\left[\begin{array}{l}a^{\prime} \\ b^{\prime}\end{array}\right][ab] such that [ a a b b ] a      a b      b [[a,a^(‘)],[b,b^(‘)]]\left[\begin{array}{ll}a & a^{\prime} \\ b & b^{\prime}\end{array}\right][aabb] is symplectic.
8. In this exercise, we ask you to find the Sylow p p ppp-subgroups of the dihedral group
D n = x , y : x n , y 2 , y x y x , n N , n 2 . D n = x , y : x n , y 2 , y x y x , n N , n 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 .Dn=x,y:xn,y2,yxyx,nN,n2.
(a) Let p p ppp be an odd prime that divides n , n = p r l , p l n , n = p r l , p l n,n=p^(r)l,p∤ln, n=p^r l, p \nmid ln,n=prl,pl. Suppose C = x l C = x l C=(:x^(l):)C=\left\langle x^l\right\rangleC=xl. Show that C C CCC is the unique Sylow p p ppp-subgroup of D n D n D_(n)D_nDn.
(b) Prove the relation
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 = 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)={[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}yixjykxl={yixj+l if k is even yi+kxlj if k is odd.
Further, find all the elements of order 2 in D n D n D_(n)D_nDn.
(c) Find all the Sylow 2-subgroups of D n D n D_(n)D_nDn when n n nnn is odd. Describe them in terms of x x xxx and y y yyy.
(d) Suppose n n nnn is even, n = 2 k m n = 2 k m n=2^(k)mn=2^k mn=2km, where 2 m , k 2 2 m , k 2 2∤m,k >= 22 \nmid m, k \geq 22m,k2. Let N = x m N = x m N=(:x^(m):)N=\left\langle x^m\right\rangleN=xm and H = y H = y H=(:y:)H=\langle y\rangleH=y. Show that H N H N HNH NHN is a subgroup of D n D n D_(n)D_nDn. What is its order?
(e) Suppose n n nnn is as in the previous part. Find all the Sylow 2-supgroups of D n D n D_(n)D_nDn. Describe them in terms of x x xxx and y y yyy.
9. (a) Let G = a , b a 2 , b 3 , a b a 1 b 1 G = a , b a 2 , b 3 , a b 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\rangleG=a,ba2,b3,aba1b1. Show that G G GGG is the cyclic group of order six.
(b) Solve the following set of congruences:
x 2 ( mod 17 ) 3 x 4 ( mod 19 ) x 7 ( mod 23 ) x      2           ( mod 17 ) 3 x      4      ( mod 19 ) x      7           ( mod 23 ) {:[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}x2(mod17)3x4(mod19)x7(mod23)
(c) Show that Q ( 19 ) Q ( 19 ) Q(sqrt(-19))\mathbb{Q}(\sqrt{-19})Q(19) 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 GGG acts on a finite set S S SSS, 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}Gs1=Gs2 for all s 1 , s 2 X s 1 , s 2 X s_(1),s_(2)in Xs_1, s_2 \in Xs1,s2X.
Answer:
The statement "If a finite group G G GGG acts on a finite set S S SSS, 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}Gs1=Gs2 for all s 1 , s 2 S s 1 , s 2 S s_(1),s_(2)in Ss_1, s_2 \in Ss1,s2S" is false. Here, G s 1 G s 1 G_(s_(1))G_{s_1}Gs1 and G s 2 G s 2 G_(s_(2))G_{s_2}Gs2 denote the stabilizer subgroups of s 1 s 1 s_(1)s_1s1 and s 2 s 2 s_(2)s_2s2 in S S SSS, respectively. The stabilizer subgroup G s G s G_(s)G_sGs of an element s S s S s in Ss \in SsS is defined as the set of all elements in G G GGG that fix s s sss, i.e., G s = { g G g s = s } G s = { g G g s = s } G_(s)={g in G∣g*s=s}G_s = \{g \in G \mid g \cdot s = s\}Gs={gGgs=s}.
Justification/Proof:
The statement claims that for any two elements s 1 s 1 s_(1)s_1s1 and s 2 s 2 s_(2)s_2s2 in the set S S SSS, their stabilizer subgroups in G G GGG are equal. This would mean that every element of G G GGG that fixes s 1 s 1 s_(1)s_1s1 also fixes s 2 s 2 s_(2)s_2s2 and vice versa. However, this is not necessarily true for all actions of G G GGG on S S SSS. The equality of stabilizers for all elements in S S SSS would imply a very rigid action, essentially saying that either the entire group fixes every element of S S SSS or that the action is trivial (where G G GGG acts on S S SSS in such a way that every element of G G GGG fixes every element of S S SSS).
Counterexample:
Consider the group G = { e , g } G = { e , g } G={e,g}G = \{e, g\}G={e,g} where e e eee is the identity element and g g ggg is an element of order 2 (meaning g 2 = e g 2 = e g^(2)=eg^2 = eg2=e), acting on the set S = { s 1 , s 2 } S = { s 1 , s 2 } S={s_(1),s_(2)}S = \{s_1, s_2\}S={s1,s2} with two elements. Define the action of G G GGG on S S SSS as follows:
  • e e eee fixes both s 1 s 1 s_(1)s_1s1 and s 2 s 2 s_(2)s_2s2 (as it must, being the identity).
  • g g ggg swaps s 1 s 1 s_(1)s_1s1 and s 2 s 2 s_(2)s_2s2, i.e., g s 1 = s 2 g s 1 = s 2 g*s_(1)=s_(2)g \cdot s_1 = s_2gs1=s2 and g s 2 = s 1 g s 2 = s 1 g*s_(2)=s_(1)g \cdot s_2 = s_1gs2=s1.
For this action, we have:
  • G s 1 = { e } G s 1 = { e } G_(s_(1))={e}G_{s_1} = \{e\}Gs1={e} because only the identity element fixes s 1 s 1 s_(1)s_1s1.
  • G s 2 = { e } G s 2 = { e } G_(s_(2))={e}G_{s_2} = \{e\}Gs2={e} because only the identity element fixes s 2 s 2 s_(2)s_2s2.
In this specific example, G s 1 G s 1 G_(s_(1))G_{s_1}Gs1 and G s 2 G s 2 G_(s_(2))G_{s_2}Gs2 happen to be equal because the non-identity element of G G GGG does not fix either of the elements of S S SSS but rather swaps them. However, the action of g g ggg clearly differentiates s 1 s 1 s_(1)s_1s1 and s 2 s 2 s_(2)s_2s2, showing that the nature of the action can lead to different behaviors for different elements of S S SSS, and it’s easy to construct examples where G s 1 G s 2 G s 1 G s 2 G_(s_(1))!=G_(s_(2))G_{s_1} \neq G_{s_2}Gs1Gs2 by adjusting the action.
For a more direct counterexample to the statement, consider a group G G GGG acting on a set S = { s 1 , s 2 , s 3 } S = { s 1 , s 2 , s 3 } S={s_(1),s_(2),s_(3)}S = \{s_1, s_2, s_3\}S={s1,s2,s3} where G G GGG fixes s 1 s 1 s_(1)s_1s1 and s 2 s 2 s_(2)s_2s2 but not s 3 s 3 s_(3)s_3s3, or any scenario where the action is not uniform across all elements of S S SSS. 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_4S4.
Answer:
To determine the truth of the statement "There are exactly 8 elements of order 3 in S 4 S 4 S_(4)S_4S4," we need to understand the structure of the symmetric group S 4 S 4 S_(4)S_4S4 and the concept of an element’s order.
Background:
  • The symmetric group S 4 S 4 S_(4)S_4S4 consists of all permutations of four elements, with a total of 4 ! = 24 4 ! = 24 4!=244! = 244!=24 elements.
  • The order of an element in a group is the smallest positive integer n n nn