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

Which of the following statements are true? Give reasons for your answers.

i) If a group

G

is isomorphic to one of its proper subgroups, then

G = Z

ii) If

x

and

y

are elements of a non-abelian group (

G, *

) such that

x * y = y * x

, then

x = e

y = e

, where

e

is the identity of

G

with respect to *.

iii) There exists a unique non-abelian group of prime order.

iv) If

(a, b) \in A \times A

, where

A

is a group, then

o ((a, b)) = o (a) o (b)

v) If

H

and

K

are normal subgroups of a group

G

, then

h k = k h \forall h \in H, k \in K

a) Prove that every non-trivial subgroup of a cyclic group has finite index. Hence prove that $(Q, +)$ is not cyclic.

b) Let

G

be an infinite group such that for any non-trivial subgroup

H

G, | G : H | < \infty

. Then prove that
i)

H \leq G \Rightarrow H = {e}

H

is infinite;
ii) If

g \in G, g \neq e

, then

o (g)

is infinite.

c) Prove that a cyclic group with only one generator can have at most 2 elements.

a) Using Cayley’s theorem, find the permutation group to which a cyclic group of order 12 is isomorphic.

b) Let

τ

be a fixed odd permutation in

S_{10}

. Show that every odd permutation in

S_{10}

a product of

τ

and some permutation in

A_{10}

c) List two distinct cosets of

⟨ r ⟩

D_{10}

, where

r

is a reflection in

D_{10}

d) Give the smallest

n \in N

for which

A_{n}

is non-abelian. Justify your answer.

Use the Fundamental Theorem of Homomorphism for Groups to prove the following theorem, which is called the Zassenhaus (Butterfly) Lemma:
Let $H$ and $K$ be subgroups of a group $G$ and $H^{'}$ and $K^{'}$ be normal subgroups of $H$ and $K$ , respectively. Then
i) $H^{'} (H \cap K^{'}) ◃ H^{'} (H \cap K)$
ii) $K^{'} (H^{'} \cap K) ◃ K^{'} (H \cap K)$

iii)

\frac{H^{'} (H \cap K)}{H^{'} (H \cap K^{'})} ≃ \frac{K^{'} (H \cap K)}{K^{'} (H^{'} \cap K)} ≃ \frac{(H \cap K)}{(H^{'} \cap K) (H \cap K^{'})}

The situation can be represented by the subgroup diagram below, which explains the name ‘butterfly’.

PART-B (MM: 30 Marks)
(Based on Block 3.)

Which of the following statements are true, and which are false? Give reasons for your answers.

i) For any ring

R

and

a, b \in R, (a + b)^{2} = a^{2} + 2 a b + b^{2}

ii) Every ring has at least two elements.

iii) If

R

is a ring with identity and

I

is an ideal of

R

, then the identity of

R / I

is the same as the identity of

R

iv) If

f : R \to S

is a ring homomorphism, then it is a group homomorphism from

(R, +)

(S, +)

v) If

R

is a ring, then any ring homomorphism from

R \times R

into

R

is surjective.

a) For an ideal $I$ of a commutative ring $R$ , define $\sqrt{I} = {x \in R ∣ x^{n} \in I$ for some $n \in N}$ . Show that
i) $\sqrt{I}$ is an ideal of $R$ .
ii) $I \subseteq \sqrt{I}$ .
iii) $I \neq \sqrt{I}$ in some cases.

b) Is

\frac{R}{I} \times \frac{R}{J} ≃ \frac{R \times R}{I \times J}

, for any two ideals

I

and

J

of a ring

R

? Give reasons for your answer.

Let $S$ be a set, $R$ a ring and $f$ be a 1-1 mapping of $S$ onto $R$ . Define + and $\cdot$ on $S$ by:

\begin{aligned} x + y = f^{- 1} (f (x)) + f (y)) \\ x \cdot y = f^{- 1} (f (x) \cdot f (y)) \\ \forall x, y \in S . \end{aligned}

Show that

(S, +, \cdot)

is a ring isomorphic to

R

PART-C (MM: 20 Marks)

(Based on Block 4.)

Which of the following statements are true, and which are false? Give reasons for your answers.
i) If $k$ is a field, then so is $k \times k$ .

ii) If

R

is an integral domain and

I

is an ideal of

R

, then Char (R) = Char (R/I).

iii) In a domain, every prime ideal is a maximal ideal.

iv) If

R

is a ring with zero divisors, and

S

is a subring of

R

, then

S

has zero divisors.

v) If

R

is a ring and

f (x) \in R [x]

is of degree

n \in N

, then

f (x)

has exactly

n

roots in

R

a) Find all the units of $Z [\sqrt{- 7}]$ .

b) Check whether or not

Q [x] / < 8 x^{3} + 6 x^{2} - 9 x + 24 >

is a field.

c) Construct a field with 125 elements.

BMTC-134 Sample Solution 2024

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

BMTC-134 Solved Assignment 2024

Which of the following statements are true? Give reasons for your answers.

i) If a group

G

is isomorphic to one of its proper subgroups, then

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

and the integers

Z

Statement: "If a group

G

is isomorphic to one of its proper subgroups, then

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

is a subgroup

H

G

such that

H \neq G

and

H \neq {e}

, where

e

is the identity element of

G

Now, let’s analyze the statement:

If $G$ is isomorphic to one of its proper subgroups: This means there exists a proper subgroup $H \subset G$ such that there is an isomorphism $f : G \to H$ . Since $f$ is bijective, every element in $G$ corresponds to a unique element in $H$ , and vice versa.
Then $G = Z$ : 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, $Z$ .

To justify or refute this statement, we need to consider whether there are any groups other than

Z

that can be isomorphic to one of their proper subgroups.

Counterexample:

Consider the group

Q

, the group of rational numbers under addition. Let’s take a proper subgroup of

Q

, say

H = {\frac{m}{2^{n}} : m \in Z, n \in N}

, which is the group of dyadic rationals.

We can define an isomorphism

f : Q \to H

f (q) = 2 q

. This function is bijective and preserves the group operation (addition in this case). Therefore,

Q

is isomorphic to its proper subgroup

H

, but

Q \neq Z

This counterexample shows that the statement "If a group

G

is isomorphic to one of its proper subgroups, then

G = Z

" is false. There exist other groups, like

Q

, that are also isomorphic to one of their proper subgroups.

ii) If

x

and

y

are elements of a non-abelian group (

G, *

) such that

x * y = y * x

, then

x = e

y = e

, where

e

is the identity of

G

with respect to *.

Answer:

The statement to be analyzed is: "If

x

and

y

are elements of a non-abelian group

(G, *)

such that

x * y = y * x

, then

x = e

y = e

, where

e

is the identity of

G

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

Non-Abelian Group: A group

G

is non-abelian if there exist elements in

G

such that the group operation is not commutative, i.e., there exist

a, b \in G

such that

a * b \neq b * a

Identity Element: The identity element

e

in a group

G

is the element that satisfies

e * a = a * e = a

for all

a \in G

Now, let’s analyze the statement:

Given: $x * y = y * x$ for some $x, y \in G$ , a non-abelian group.
To Prove: $x = e$ or $y = e$ .

Counterexample:

To refute the statement, we need to find a non-abelian group

G

and elements

x, y \in G

such that

x * y = y * x

but neither

x

nor

y

is the identity element.

Consider the group of

2 \times 2

invertible matrices under matrix multiplication, which is a non-abelian group. Let’s choose two specific matrices:

x = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}), y = (\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix})

These matrices represent

90^{\circ}

and

- 90^{\circ}

rotations in the plane, respectively. It’s easy to verify that

x * y = y * x

, as both products represent a

180^{\circ}

rotation (or a reflection). However, neither

x

nor

y

is the identity matrix, which in this case would be:

e = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})

Thus, we have found a counterexample where

x * y = y * x

in a non-abelian group, but neither

x

nor

y

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

is the number of elements in

G

. It is denoted as

| G |

Abelian Group: A group

G

is abelian if for every pair of elements

a, b

G

, the group operation is commutative; that is,

a * b = b * a

Non-Abelian Group: A group is non-abelian if it is not abelian, meaning there exists at least one pair of elements

a, b

G

such that

a * b \neq b * a

Prime Order: A group of prime order has a number of elements equal to a prime number.

Now, let’s analyze the statement:

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 $p$ , this means there are $p$ elements in the group.
Uniqueness and Structure of Groups of Prime Order: According to Lagrange’s Theorem in group theory, the order of every subgroup of a group $G$ must divide the order of $G$ . For a group of prime order $p$ , the only divisors are 1 and $p$ itself. This implies that the only subgroups of such a group are the trivial group (with just the identity element) and the group itself.
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 $p$ ). 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) \in A \times A

, where

A

is a group, then

o ((a, b)) = o (a) o (b)

Answer:

The statement to evaluate is: "If

(a, b) \in A \times A

, where

A

is a group, then

o ((a, b)) = o (a) o (b)

Here,

A \times A

denotes the direct product of the group

A

with itself, and

o (x)

denotes the order of the element

x

in its respective group. The order of an element

g

in a group is the smallest positive integer

n

such that

g^{n} = e

, where

e

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 \times A

, the elements are ordered pairs

(a, b)

, where

a, b \in A

. The group operation in

A \times A

is defined component-wise. That is, if

(a_{1}, b_{1}), (a_{2}, b_{2}) \in A \times A

, then their product is

(a_{1} * a_{2}, b_{1} * b_{2})

, where

*

is the group operation in

A

Order of an Element in $A \times A$ : The order of an element

(a, b)

A \times A

is the smallest positive integer

n

such that

(a, b)^{n} = (e, e)

, where

e

is the identity in

A

. This means

(a^{n}, b^{n}) = (e, e)

, which implies

a^{n} = e

and

b^{n} = e

Now, let’s analyze the statement:

Given: $(a, b) \in A \times A$ , where $A$ is a group.
To Prove: $o ((a, b)) = o (a) o (b)$ .

Counterexample:

Consider the case where

o (a)

and

o (b)

are not relatively prime. Let

A

be a group, and choose

a, b \in A

such that

o (a) = 2

and

o (b) = 4

. This means

a^{2} = e

and

b^{4} = e

A \times A

, consider the element

(a, b)

. We need to find the smallest positive integer

n

such that

(a, b)^{n} = (e, e)

For $n = 2$ , $(a, b)^{2} = (a^{2}, b^{2}) = (e, b^{2})$ . Since $b^{2} \neq e$ , $n = 2$ is not the order of $(a, b)$ .
For $n = 4$ , $(a, b)^{4} = (a^{4}, b^{4}) = (e, e)$ . Here, $n = 4$ is the smallest positive integer satisfying the condition.

So,

o ((a, b)) = 4

. However,

o (a) o (b) = 2 \times 4 = 8

This counterexample shows that

o ((a, b))

is not necessarily equal to

o (a) o (b)

. The statement is false. The order of an element in the direct product of groups is not always the product of the orders of its constituent elements, especially when the orders of these elements are not relatively prime.

v) If

H

and

K

are normal subgroups of a group

G

, then

h k = k h \forall h \in H, k \in K

Answer:

The statement to evaluate is: "If

H

and

K

are normal subgroups of a group

G

, then

h k = k h

for all

h \in H, k \in K

To analyze this statement, we need to understand the concept of normal subgroups and their properties in the context of group theory.

Normal Subgroup: A subgroup

N

of a group

G

is called a normal subgroup if it is invariant under conjugation by elements of

G

. This means for every

n \in N

and

g \in G

, the element

g n g^{- 1}

is also in

N

Given that

H

and

K

are normal subgroups of

G

, we need to determine whether the commutativity

h k = k h

holds for all

h \in H

and

k \in K

Analysis:

Given: $H$ and $K$ are normal subgroups of $G$ .
To Prove: $h k = k h$ for all $h \in H$ and $k \in K$ .

Since

H

and

K

are normal in

G

, for any

h \in H

and

k \in K

, and for any

g \in G

, we have

g h g^{- 1} \in H

and

g k g^{- 1} \in K

Now, consider the product

k h

. We want to show that this is equal to

h k

. Since

H

is normal,

k^{- 1} h k \in H

. Let’s denote

k^{- 1} h k

h^{'}

. So,

h^{'} = k^{- 1} h k

, which implies

h k = k h^{'}

Similarly, since

K

is normal,

h^{- 1} k h \in K

. Let’s denote

h^{- 1} k h

k^{'}

. So,

k^{'} = h^{- 1} k h

, which implies

k h = h k^{'}

Now, we need to show that

h^{'} = h

and

k^{'} = k

. Since

h^{'} \in H

and

h \in H

, and

H

is a subgroup,

h h^{' - 1} \in H

. But

h h^{' - 1} = h k^{- 1} h^{- 1} k

. Since

K

is normal,

h k^{- 1} h^{- 1} \in K

, and thus

h k^{- 1} h^{- 1} k \in K

. But this is in

H

as well, so it must be the identity element

e

, as

H \cap K = {e}

for normal subgroups

H

and

K

. Therefore,

h h^{' - 1} = e

, which implies

h^{'} = h

A similar argument shows that

k^{'} = k

. Therefore,

h k = k h

This proves that the statement "If

H

and

K

are normal subgroups of a group

G

, then

h k = k h

for all

h \in H, k \in K

" is true. The normality of the subgroups

H

and

K

G

ensures the commutativity of their elements.

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

IGNOU BMTC-134 Assignment Question Paper 2024

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

BMTC-134 Sample Solution 2024

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

IGNOU BMTC-134 Assignment Question Paper 2024

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