Comprehensive IGNOU BMTC-134 Solved Assignment 2024 for B.Sc (G) CBCS Students

IGNOU BMTC-134 Solved Assignment 2024 | B.Sc (G) CBCS

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

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IGNOU BMTC-134 Assignment Question Paper 2024

bmtc-134-solved-assignment-2024-10cd7422-4b11-4531-9171-a6445826d855

bmtc-134-solved-assignment-2024-10cd7422-4b11-4531-9171-a6445826d855

BMTC-134 Solved Assignment 2024
  1. Which of the following statements are true? Give reasons for your answers.
i) If a group G G GGG is isomorphic to one of its proper subgroups, then G = Z G = Z G=ZG=\mathbb{Z}G=Z.
ii) If x x xxx and y y yyy are elements of a non-abelian group ( G , G , G,**G, *G, ) such that x y = y x x y = y x x**y=y**xx * y=y * xxy=yx, then x = e x = e x=e\mathrm{x}=\mathrm{e}x=e or y = e y = e y=e\mathrm{y}=\mathrm{e}y=e, where e e e\mathrm{e}e is the identity of G G G\mathrm{G}G with respect to *.
iii) There exists a unique non-abelian group of prime order.
iv) If ( a , b ) A × A ( a , b ) A × A (a,b)inAxxA(\mathrm{a}, \mathrm{b}) \in \mathrm{A} \times \mathrm{A}(a,b)A×A, where A A A\mathrm{A}A is a group, then o ( ( a , b ) ) = o ( a ) o ( b ) o ( ( a , b ) ) = o ( a ) o ( b ) o((a,b))=o(a)o(b)\mathrm{o}((\mathrm{a}, \mathrm{b}))=\mathrm{o}(\mathrm{a}) \mathrm{o}(\mathrm{b})o((a,b))=o(a)o(b).
v) If H H HHH and K K KKK are normal subgroups of a group G G GGG, then h k = k h h H , k K h k = k h h H , k K hk=kh AA h in H,k in Kh k=k h \forall h \in H, k \in Khk=khhH,kK.
  1. a) Prove that every non-trivial subgroup of a cyclic group has finite index. Hence prove that ( Q , + ) ( Q , + ) (Q,+)(\mathbb{Q},+)(Q,+) is not cyclic.
b) Let G G G\mathrm{G}G be an infinite group such that for any non-trivial subgroup H H H\mathrm{H}H of G , | G : H | < G , | G : H | < G,|G:H| < oo\mathrm{G},|\mathrm{G}: \mathrm{H}|<\inftyG,|G:H|<. Then prove that
i) H G H = { e } H G H = { e } H <= G=>H={e}\mathrm{H} \leq \mathrm{G} \Rightarrow \mathrm{H}=\{\mathrm{e}\}HGH={e} or H H H\mathrm{H}H is infinite;
ii) If g G , g e g G , g e ginG,g!=e\mathrm{g} \in \mathrm{G}, \mathrm{g} \neq \mathrm{e}gG,ge, then o ( g ) o ( g ) o(g)\mathrm{o}(\mathrm{g})o(g) is infinite.
c) Prove that a cyclic group with only one generator can have at most 2 elements.
  1. a) Using Cayley’s theorem, find the permutation group to which a cyclic group of order 12 is isomorphic.
b) Let τ τ tau\tauτ be a fixed odd permutation in S 10 S 10 S_(10)\mathrm{S}_{10}S10. Show that every odd permutation in S 10 S 10 S_(10)\mathrm{S}_{10}S10 a product of τ τ tau\tauτ and some permutation in A 10 A 10 A_(10)\mathrm{A}_{10}A10.
c) List two distinct cosets of r r (:r:)\langle\mathrm{r}\rangler in D 10 D 10 D_(10)\mathrm{D}_{10}D10, where r r r\mathrm{r}r is a reflection in D 10 D 10 D_(10)\mathrm{D}_{10}D10.
d) Give the smallest n N n N ninN\mathrm{n} \in \mathbb{N}nN for which A n A n A_(n)\mathrm{A}_{\mathrm{n}}An is non-abelian. Justify your answer.
  1. Use the Fundamental Theorem of Homomorphism for Groups to prove the following theorem, which is called the Zassenhaus (Butterfly) Lemma:
    Let H H H\mathrm{H}H and K K K\mathrm{K}K be subgroups of a group G G G\mathrm{G}G and H H H^(‘)\mathrm{H}^{\prime}H and K K K^(‘)\mathrm{K}^{\prime}K be normal subgroups of H H H\mathrm{H}H and K K K\mathrm{K}K, respectively. Then
    i) H ( H K ) H ( H K ) H H K H ( H K ) H^(‘)(HnnK^(‘))◃H^(‘)(HnnK)\mathrm{H}^{\prime}\left(\mathrm{H} \cap \mathrm{K}^{\prime}\right) \triangleleft \mathrm{H}^{\prime}(\mathrm{H} \cap \mathrm{K})H(HK)H(HK)
    ii) K ( H K ) K ( H K ) K H K K ( H K ) K^(‘)(H^(‘)nnK)◃K^(‘)(HnnK)\mathrm{K}^{\prime}\left(\mathrm{H}^{\prime} \cap \mathrm{K}\right) \triangleleft \mathrm{K}^{\prime}(\mathrm{H} \cap \mathrm{K})K(HK)K(HK)
iii) H ( H K ) H ( H K ) K ( H K ) K ( H K ) ( H K ) ( H K ) ( H K ) H ( H K ) H H K K ( H K ) K H K ( H K ) H K H K (H^(‘)(HnnK))/(H^(‘)(HnnK^(‘)))≃(K^(‘)(HnnK))/(K^(‘)(H^(‘)nnK))≃((HnnK))/((H^(‘)nnK)(HnnK^(‘)))\frac{\mathrm{H}^{\prime}(\mathrm{H} \cap \mathrm{K})}{\mathrm{H}^{\prime}\left(\mathrm{H} \cap \mathrm{K}^{\prime}\right)} \simeq \frac{\mathrm{K}^{\prime}(\mathrm{H} \cap \mathrm{K})}{\mathrm{K}^{\prime}\left(\mathrm{H}^{\prime} \cap \mathrm{K}\right)} \simeq \frac{(\mathrm{H} \cap \mathrm{K})}{\left(\mathrm{H}^{\prime} \cap \mathrm{K}\right)\left(\mathrm{H} \cap \mathrm{K}^{\prime}\right)}H(HK)H(HK)K(HK)K(HK)(HK)(HK)(HK)
The situation can be represented by the subgroup diagram below, which explains the name ‘butterfly’.
original image
PART-B (MM: 30 Marks)
(Based on Block 3.)
  1. Which of the following statements are true, and which are false? Give reasons for your answers.
i) For any ring R R R\mathrm{R}R and a , b R , ( a + b ) 2 = a 2 + 2 a b + b 2 a , b R , ( a + b ) 2 = a 2 + 2 a b + b 2 a,binR,(a+b)^(2)=a^(2)+2ab+b^(2)\mathrm{a}, \mathrm{b} \in \mathrm{R},(\mathrm{a}+\mathrm{b})^2=\mathrm{a}^2+2 \mathrm{ab}+\mathrm{b}^2a,bR,(a+b)2=a2+2ab+b2.
ii) Every ring has at least two elements.
iii) If R R RRR is a ring with identity and I I III is an ideal of R R RRR, then the identity of R / I R / I R//IR / IR/I is the same as the identity of R R RRR.
iv) If f : R S f : R S f:RrarrS\mathrm{f}: \mathrm{R} \rightarrow \mathrm{S}f:RS is a ring homomorphism, then it is a group homomorphism from ( R , + ) ( R , + ) (R,+)(\mathrm{R},+)(R,+) to ( S , + ) ( S , + ) (S,+)(\mathrm{S},+)(S,+).
v) If R R R\mathrm{R}R is a ring, then any ring homomorphism from R × R R × R RxxR\mathrm{R} \times \mathrm{R}R×R into R R R\mathrm{R}R is surjective.
  1. a) For an ideal I I III of a commutative ring R R RRR, define I = { x R x n I I = x R x n I sqrtI={xinR∣x^(n)inI:}\sqrt{\mathrm{I}}=\left\{\mathrm{x} \in \mathrm{R} \mid \mathrm{x}^{\mathrm{n}} \in \mathrm{I}\right.I={xRxnI for some n N } n N {:ninN}\left.\mathrm{n} \in \mathbb{N}\right\}nN}. Show that
    i) I I sqrtI\sqrt{I}I is an ideal of R R RRR.
    ii) I I I I I subesqrtII \subseteq \sqrt{I}II.
    iii) I I I I quad I!=sqrtI\quad I \neq \sqrt{I}II in some cases.
b) Is R I × R J R × R I × J R I × R J R × R I × J (R)/(I)xx(R)/(J)≃(R xx R)/(I xx J)\frac{R}{I} \times \frac{R}{J} \simeq \frac{R \times R}{I \times J}RI×RJR×RI×J, for any two ideals I I III and J J JJJ of a ring R R RRR ? Give reasons for your answer.
  1. Let S S SSS be a set, R R RRR a ring and f f fff be a 1-1 mapping of S S SSS onto R R RRR. Define + and *\cdot on S S SSS by:
x + y = f 1 ( f ( x ) ) + f ( y ) ) x y = f 1 ( f ( x ) f ( y ) ) x , y S . x + y = f 1 ( f ( x ) ) + f ( y ) x y = f 1 ( f ( x ) f ( y ) ) x , y S . {:[{:x+y=f^(-1)(f(x))+f(y))],[x*y=f^(-1)(f(x)*f(y))],[AAx”,”yinS.]:}\begin{aligned} & \left.\mathrm{x}+\mathrm{y}=\mathrm{f}^{-1}(\mathrm{f}(\mathrm{x}))+\mathrm{f}(\mathrm{y})\right) \\ & \mathrm{x} \cdot \mathrm{y}=\mathrm{f}^{-1}(\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y})) \\ & \forall \mathrm{x}, \mathrm{y} \in \mathrm{S} . \end{aligned}x+y=f1(f(x))+f(y))xy=f1(f(x)f(y))x,yS.
Show that ( S , + , ) ( S , + , ) (S,+,*)(\mathrm{S},+, \cdot)(S,+,) is a ring isomorphic to R R R\mathrm{R}R.
PART-C (MM: 20 Marks)
(Based on Block 4.)
  1. Which of the following statements are true, and which are false? Give reasons for your answers.
    i) If k k k\mathrm{k}k is a field, then so is k × k k × k kxxk\mathrm{k} \times \mathrm{k}k×k.
ii) If R R RRR is an integral domain and I I III is an ideal of R R RRR, then Char (R) = Char (R/I).
iii) In a domain, every prime ideal is a maximal ideal.
iv) If R R R\mathrm{R}R is a ring with zero divisors, and S S S\mathrm{S}S is a subring of R R R\mathrm{R}R, then S S S\mathrm{S}S has zero divisors.
v) If R R RRR is a ring and f ( x ) R [ x ] f ( x ) R [ x ] f(x)in R[x]f(x) \in R[x]f(x)R[x] is of degree n N n N n inNn \in \mathbb{N}nN, then f ( x ) f ( x ) f(x)f(x)f(x) has exactly n n nnn roots in R R R\mathrm{R}R.
  1. a) Find all the units of Z [ 7 ] Z [ 7 ] Z[sqrt(-7)]\mathbb{Z}[\sqrt{-7}]Z[7].
b) Check whether or not Q [ x ] / < 8 x 3 + 6 x 2 9 x + 24 > Q [ x ] / < 8 x 3 + 6 x 2 9 x + 24 > Q[x]// < 8x^(3)+6x^(2)-9x+24 >\mathbb{Q}[x] /<8 x^3+6 x^2-9 x+24>Q[x]/<8x3+6x29x+24> is a field.
c) Construct a field with 125 elements.
\(sin\left(\theta -\phi \right)=sin\:\theta \:cos\:\phi -cos\:\theta \:sin\:\phi \)

BMTC-134 Sample Solution 2024

bmtc-134-solved-assignment-2024-10cd7422-4b11-4531-9171-a6445826d855

bmtc-134-solved-assignment-2024-10cd7422-4b11-4531-9171-a6445826d855

BMTC-134 Solved Assignment 2024
  1. Which of the following statements are true? Give reasons for your answers.
i) If a group G G GGG is isomorphic to one of its proper subgroups, then G = Z G = Z G=ZG=\mathbb{Z}G=Z.
Answer:
To address this statement, we need to understand the concepts of group isomorphism and proper subgroups, and then apply these concepts to the specific case of the group G G GGG and the integers Z Z Z\mathbb{Z}Z.
Statement: "If a group G G GGG is isomorphic to one of its proper subgroups, then G = Z G = Z G=ZG = \mathbb{Z}G=Z."
A group isomorphism is a bijective group homomorphism between two groups. If there exists an isomorphism between two groups, they are said to be isomorphic, meaning they have the same group structure, but not necessarily the same elements.
A proper subgroup of a group G G GGG is a subgroup H H HHH of G G GGG such that H G H G H!=GH \neq GHG and H { e } H { e } H!={e}H \neq \{e\}H{e}, where e e eee is the identity element of G G GGG.
Now, let’s analyze the statement:
  1. If G G GGG is isomorphic to one of its proper subgroups: This means there exists a proper subgroup H G H G H sub GH \subset GHG such that there is an isomorphism f : G H f : G H f:G rarr Hf: G \to Hf:GH. Since f f fff is bijective, every element in G G GGG corresponds to a unique element in H H HHH, and vice versa.
  2. Then G = Z G = Z G=ZG = \mathbb{Z}G=Z: This part of the statement claims that the only group for which the above condition can be true is the group of integers under addition, Z Z Z\mathbb{Z}Z.
To justify or refute this statement, we need to consider whether there are any groups other than Z Z Z\mathbb{Z}Z that can be isomorphic to one of their proper subgroups.
Counterexample:
Consider the group Q Q Q\mathbb{Q}Q, the group of rational numbers under addition. Let’s take a proper subgroup of Q Q Q\mathbb{Q}Q, say H = { m 2 n : m Z , n N } H = { m 2 n : m Z , n N } H={(m)/(2^(n)):m inZ,n inN}H = \{ \frac{m}{2^n} : m \in \mathbb{Z}, n \in \mathbb{N} \}H={m2n:mZ,nN}, which is the group of dyadic rationals.
We can define an isomorphism f : Q H f : Q H f:Qrarr Hf: \mathbb{Q} \to Hf:QH by f ( q ) = 2 q f ( q ) = 2 q f(q)=2qf(q) = 2qf(q)=2q. This function is bijective and preserves the group operation (addition in this case). Therefore, Q Q Q\mathbb{Q}Q is isomorphic to its proper subgroup H H HHH, but Q Z Q Z Q!=Z\mathbb{Q} \neq \mathbb{Z}QZ.
This counterexample shows that the statement "If a group G G GGG is isomorphic to one of its proper subgroups, then G = Z G = Z G=ZG = \mathbb{Z}G=Z" is false. There exist other groups, like Q Q Q\mathbb{Q}Q, that are also isomorphic to one of their proper subgroups.
ii) If x x xxx and y y yyy are elements of a non-abelian group ( G , G , G,**G, *G, ) such that x y = y x x y = y x x**y=y**xx * y=y * xxy=yx, then x = e x = e x=e\mathrm{x}=\mathrm{e}x=e or y = e y = e y=e\mathrm{y}=\mathrm{e}y=e, where e e e\mathrm{e}e is the identity of G G G\mathrm{G}G with respect to *.
Answer:
The statement to be analyzed is: "If x x xxx and y y yyy are elements of a non-abelian group ( G , ) ( G , ) (G,**)(G, *)(G,) such that x y = y x x y = y x x**y=y**xx * y = y * xxy=yx, then x = e x = e x=ex = ex=e or y = e y = e y=ey = ey=e, where e e eee is the identity of G G GGG with respect to ***."
To evaluate this statement, we need to understand the properties of non-abelian groups and the implications of the given condition x y = y x x y = y x x**y=y**xx * y = y * xxy=yx.
Non-Abelian Group: A group G G GGG is non-abelian if there exist elements in G G GGG such that the group operation is not commutative, i.e., there exist a , b G a , b G a,b in Ga, b \in Ga,bG such that a b b a a b b a a**b!=b**aa * b \neq b * aabba.
Identity Element: The identity element e e eee in a group G G GGG is the element that satisfies e a = a e = a e a = a e = a e**a=a**e=ae * a = a * e = aea=ae=a for all a G a G a in Ga \in GaG.
Now, let’s analyze the statement:
  1. Given: x y = y x x y = y x x**y=y**xx * y = y * xxy=yx for some x , y G x , y G x,y in Gx, y \in Gx,yG, a non-abelian group.
  2. To Prove: x = e x = e x=ex = ex=e or y = e y = e y=ey = ey=e.
Counterexample:
To refute the statement, we need to find a non-abelian group G G GGG and elements x , y G x , y G x,y in Gx, y \in Gx,yG such that x y = y x x y = y x x**y=y**xx * y = y * xxy=yx but neither x x xxx nor y y yyy is the identity element.
Consider the group of 2 × 2 2 × 2 2xx22 \times 22×2 invertible matrices under matrix multiplication, which is a non-abelian group. Let’s choose two specific matrices:
x = ( 0 1 1 0 ) , y = ( 0 1 1 0 ) x = 0 1 1 0 , y = 0 1 1 0 x=([0,-1],[1,0]),quad y=([0,1],[-1,0])x = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \quad y = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}x=(0110),y=(0110)
These matrices represent 90 90 90^(@)90^\circ90 and 90 90 -90^(@)-90^\circ90 rotations in the plane, respectively. It’s easy to verify that x y = y x x y = y x x**y=y**xx * y = y * xxy=yx, as both products represent a 180 180 180^(@)180^\circ180 rotation (or a reflection). However, neither x x xxx nor y y yyy is the identity matrix, which in this case would be:
e = ( 1 0 0 1 ) e = 1 0 0 1 e=([1,0],[0,1])e = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}e=(1001)
Thus, we have found a counterexample where x y = y x x y = y x x**y=y**xx * y = y * xxy=yx in a non-abelian group, but neither x x xxx nor y y yyy is the identity element. This disproves the statement, showing that it is false. There can be elements in a non-abelian group that commute with each other without being the identity element.
iii) There exists a unique non-abelian group of prime order.
Answer:
The statement to evaluate is: "There exists a unique non-abelian group of prime order."
To analyze this statement, we need to understand the concepts of group order, abelian groups, and non-abelian groups, particularly in the context of groups of prime order.
Group Order: The order of a group G G GGG is the number of elements in G G GGG. It is denoted as | G | | G | |G||G||G|.
Abelian Group: A group G G GGG is abelian if for every pair of elements a , b a , b a,ba, ba,b in G G GGG, the group operation is commutative; that is, a b = b a a b = b a a**b=b**aa * b = b * aab=ba.
Non-Abelian Group: A group is non-abelian if it is not abelian, meaning there exists at least one pair of elements a , b a , b a,ba, ba,b in G G GGG such that a b b a a b b a a**b!=b**aa * b \neq b * aabba.
Prime Order: A group of prime order has a number of elements equal to a prime number.
Now, let’s analyze the statement:
  1. Existence of a Non-Abelian Group of Prime Order: The key property of prime numbers is that they have exactly two distinct positive divisors: 1 and themselves. For a group of prime order p p ppp, this means there are p p ppp elements in the group.
  2. Uniqueness and Structure of Groups of Prime Order: According to Lagrange’s Theorem in group theory, the order of every subgroup of a group G G GGG must divide the order of G G GGG. For a group of prime order p p ppp, the only divisors are 1 and p p ppp itself. This implies that the only subgroups of such a group are the trivial group (with just the identity element) and the group itself.
  3. Implication for Abelian Property: In a group of prime order, every element except the identity must generate the entire group (since any subgroup must have order 1 or p p ppp). This means that every non-identity element is a generator of the group. In such a scenario, the group operation must be commutative, as there is only one group structure possible under which all elements (except the identity) are generators. This group structure is essentially cyclic, and all cyclic groups are abelian.
Therefore, any group of prime order must be abelian. This directly contradicts the possibility of a non-abelian group of prime order. Hence, the statement "There exists a unique non-abelian group of prime order" is false. There cannot exist any non-abelian group of prime order, let alone a unique one.
iv) If ( a , b ) A × A ( a , b ) A × A (a,b)inAxxA(\mathrm{a}, \mathrm{b}) \in \mathrm{A} \times \mathrm{A}(a,b)A×A, where A A A\mathrm{A}A is a group, then o ( ( a , b ) ) = o ( a ) o ( b ) o ( ( a , b ) ) = o ( a ) o ( b ) o((a,b))=o(a)o(b)\mathrm{o}((\mathrm{a}, \mathrm{b}))=\mathrm{o}(\mathrm{a}) \mathrm{o}(\mathrm{b})o((a,b))=o(a)o(b).
Answer:
The statement to evaluate is: "If ( a , b ) A × A ( a , b ) A × A (a,b)in A xx A(a, b) \in A \times A(a,b)A×A, where A A AAA is a group, then o ( ( a , b ) ) = o ( a ) o ( b ) o ( ( a , b ) ) = o ( a ) o ( b ) o((a,b))=o(a)o(b)o((a, b)) = o(a) o(b)o((a,b))=o(a)o(b)."
Here, A × A A × A A xx AA \times AA×A denotes the direct product of the group A A AAA with itself, and o ( x ) o ( x ) o(x)o(x)o(x) denotes the order of the element x x xxx in its respective group. The order of an element g g ggg in a group is the smallest positive integer n n nnn such that g n = e g n = e g^(n)=eg^n = egn=e, where e e eee is the identity element of the group.
To analyze this statement, we need to understand the order of an element in the direct product of groups and how it relates to the orders of its constituent elements.
Direct Product of Groups: In the direct product A × A A × A A xx AA \times AA×A, the elements are ordered pairs ( a , b ) ( a , b ) (a,b)(a, b)(a,b), where a , b A a , b A a,b in Aa, b \in Aa,bA. The group operation in A × A A × A A xx AA \times AA×A is defined component-wise. That is, if ( a 1 , b 1 ) , ( a 2 , b 2 ) A × A ( a 1 , b 1 ) , ( a 2 , b 2 ) A × A (a_(1),b_(1)),(a_(2),b_(2))in A xx A(a_1, b_1), (a_2, b_2) \in A \times A(a1,b1),(a2,b2)A×A, then their product is ( a 1 a 2 , b 1 b 2 ) ( a 1 a 2 , b 1 b 2 ) (a_(1)**a_(2),b_(1)**b_(2))(a_1 * a_2, b_1 * b_2)(a1a2,b1b2), where *** is the group operation in A A AAA.
Order of an Element in A × A A × A A xx AA \times AA×A: The order of an element ( a , b ) ( a , b ) (a,b)(a, b)(a,b) in A × A A × A A xx AA \times AA×A is the smallest positive integer n n nnn such that ( a , b ) n = ( e , e ) ( a , b ) n = ( e , e ) (a,b)^(n)=(e,e)(a, b)^n = (e, e)(a,b)n=(e,e), where e e eee is the identity in A A AAA. This means ( a n , b n ) = ( e , e ) ( a n , b n ) = ( e , e ) (a^(n),b^(n))=(e,e)(a^n, b^n) = (e, e)(an,bn)=(e,e), which implies a n = e a n = e a^(n)=ea^n = ean=e and b n = e b n = e b^(n)=eb^n = ebn=e.
Now, let’s analyze the statement:
  1. Given: ( a , b ) A × A ( a , b ) A × A (a,b)in A xx A(a, b) \in A \times A(a,b)A×A, where A A AAA is a group.
  2. To Prove: o ( ( a , b ) ) = o ( a ) o ( b ) o ( ( a , b ) ) = o ( a ) o ( b ) o((a,b))=o(a)o(b)o((a, b)) = o(a) o(b)