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

Which of the following statements are true and which are false? Give reasons for your answer.
(a) If a finite group $G$ acts on a finite set $S$ , then $G_{s_{1}} = G_{s_{2}}$ for all $s_{1}, s_{2} \in X$ .
(b) There are exactly 8 elements of order 3 in $S_{4}$ .
(c) If $F = Q (\sqrt[5]{2}, \sqrt[3]{5})$ , then $[F : Q] = 8$ .
(d) $F_{7} (\sqrt{3}) = F_{7} (\sqrt{5})$ .
(e) For any $α \in F_{2^{5}}^{*}, α \neq 1, F_{2^{5}} = F_{2} [α]$ .
(a) Consider the natural action of $G L_{2} (R)$ on $M_{2} (R)$ , the set of $2 \times 2$ real matrices, by left multiplication.
(i) Under this action, if $\det (x) \neq 0$ , show that the stabiliser of $x \in M_{2} (R)$ is ${I}$ , where $I$ is the $2 \times 2$ identity matrix.
(ii) Suppose that $\det (x) = 0$ in the remaining parts of this exercise. We will show that the stabiliser of $x$ is infinite. If $x = 0$ , the stabiliser of $x$ is $G L_{2} (R)$ . So suppose $x \neq 0$ . Let us write $x = [\begin{array}{ll} a & c \\ b & d \end{array}]$ . Then, $[\begin{array}{l} a \\ b \end{array}] = λ [\begin{array}{l} c \\ d \end{array}]$ for non-zero $λ \in R$ . Why ?
(iii) Let $[\begin{array}{l} a^{'} \\ b^{'} \end{array}]$ be a vector that is not a scalar multiple of $[\begin{array}{l} a \\ b \end{array}]$ . Show that there is a matrix $b =$ $[\begin{array}{ll} u & v \\ w & z \end{array}]$ such that $b [\begin{array}{l} a \\ b \end{array}] = 0$ and $b [\begin{array}{l} a^{'} \\ b^{'} \end{array}] = α [\begin{array}{l} a^{'} \\ b^{'} \end{array}]$ .(Hint: Set up two sets of simultaneous equations in two unknowns and argue why they have a solution.)
(iv) Check that $I - b$ is in the stabiliser of $x$ . Also, show that there are infinitely many choices of $α$ for which $I - b$ is invertible.
(b) Let $H$ be a finite group and, for some prime $p$ , let $P$ be a p-Sylow subgroup of $H$ which is normal in $H$ . Suppose $H$ is normal in $K$ , where $K$ is a finite group. Then, show that $P$ is normal in $K$ .
(c) Find the elementary divisors and invariant factors of $Z_{8} \times Z_{12} \times Z_{15}$ .
Describe the set of primes $p$ for which $x^{2} - 11$ splits into linear factors over $Z_{p}$ .
(a) Determine, up to isomorphism, all the finite groups with exactly 2 conjugacy classes.
(b) Is there a finite group with class equation $1 + 1 + 2 + 2 + 2 + 2 + 2 + 2$ ?
(c) Compute the following:
a) $(\frac{173}{211})$
b) $(\frac{167}{239})$ .
(a) Let $F (α)$ be a finite extension $F$ of odd degree(greater than 1). Show that $F (α^{2}) = F (α)$ .
(b) Let $F \subset K$ and let $α, β \in K$ be algebraic over $F$ of degree $m$ and $n$ , respectively. Show that $[F (α, β) : F] \leq m n$ . What can you say about $[F (α, β) : F]$ if $m$ and $n$ are coprime?
(c) Find $[Q (\sqrt[3]{2}, ω) : Q]$ where $ω^{3} = 1, ω \neq 1$ .
(a) If $char (F) \neq 2$ , show that a polynomial $a x^{2} + b x + c$ is irreducible iff $b^{2} - 4 a c \notin F^{* 2}$ where $F^{* 2}$ is the group of squares in $F^{*}$ .
(b) By looking at the factorisation of $x^{9} - x \in F_{3} [x]$ guess the number of irreducible polynomials of degree 2 over $F_{3}$ . Find all the irreducible polynomials of degree 2 over $F_{3}$ .
(c) If $F$ is a finite field show that there is always an irreducible polynomial of the form $x^{3} - x + a$ where $a \in F$ .(Hint: Show that $x \mapsto x^{3} - x$ is not a surjective map.)
(a) Suppose that $M = [\begin{array}{ll} A & B \\ C & D \end{array}]$ is $2 n \times 2 n$ matrix where $A, B, C$ and $D$ are $n \times n$ matrices. Show that $M$ is symplectic if and only if the following conditions are satisfied:

\begin{aligned} A^{t} D - C^{t} B = I \\ A^{t} C - C^{t} A = 0 \\ B^{t} D - D^{t} B = 0 \end{aligned}

(Hint: Use block matrix multiplication.)
Also, check that the matrix

[\begin{array}{cc} 0 & - A \\ A & 0 \end{array}]

, where

A

is a

n \times n

orthogonal matrix, is a symplectic matrix.
(b) The aim of this exercise is to show that

S P_{2} (R)

acts transitively on

R^{2} ∖ {0}

.
(c) Show that
(i) Show that a matrix

[\begin{array}{ll} a & b \\ c & d \end{array}] \in G L_{2} (R)

is symplectic if and only if

a d - b c = 1

.
(ii) Show that, to prove that

{S P}_{2} (R)

acts transitively on

G L_{2} (R)

, it is enough to show that, for any vector

[\begin{array}{l} a \\ b \end{array}] \neq 0 \in R^{2}

, there is a

2 \times 2

symplectic matrix with

[\begin{array}{l} a \\ b \end{array}]

as the first column. (Hint: For any matrix

A

, what is

A [\begin{array}{l} 1 \\ 0 \end{array}]

?)
(iii) Complete the proof by showing that, given any non-zero vector

[\begin{array}{l} a \\ b \end{array}]

, there is always a non-zero vector

[\begin{array}{l} a^{'} \\ b^{'} \end{array}]

such that

[\begin{array}{ll} a & a^{'} \\ b & b^{'} \end{array}]

is symplectic.
8. In this exercise, we ask you to find the Sylow

p

-subgroups of the dihedral group

D_{n} = ⟨ x, y : x^{n}, y^{2}, y x y x ⟩, n \in N, n \geq 2 .

(a) Let

p

be an odd prime that divides

n, n = p^{r} l, p ∤ l

. Suppose

C = ⟨ x^{l} ⟩

. Show that

C

is the unique Sylow

p

-subgroup of

D_{n}

.
(b) Prove the relation

y^{i} x^{j} y^{k} x^{l} = {\begin{cases} y^{i} x^{j + l} & if k is even \\ y^{i + k} x^{l - j} & if k is odd. \end{cases}

Further, find all the elements of order 2 in

D_{n}

.
(c) Find all the Sylow 2-subgroups of

D_{n}

when

n

is odd. Describe them in terms of

x

and

y

.
(d) Suppose

n

is even,

n = 2^{k} m

, where

2 ∤ m, k \geq 2

. Let

N = ⟨ x^{m} ⟩

and

H = ⟨ y ⟩

. Show that

H N

is a subgroup of

D_{n}

. What is its order?
(e) Suppose

n

is as in the previous part. Find all the Sylow 2-supgroups of

D_{n}

. Describe them in terms of

x

and

y

.
9. (a) Let

G = ⟨ a, b ∣ a^{2}, b^{3}, a b a^{- 1} b^{- 1} ⟩

. Show that

G

is the cyclic group of order six.
(b) Solve the following set of congruences:

\begin{aligned} x & \equiv 2 & (mod 17) \\ 3 x & \equiv 4 & (mod 19) \\ x & \equiv 7 & (mod 23) \end{aligned}

Q (\sqrt{- 19})

is not a UFD by giving two different factorisations of 20 .

MMT-003 Sample Solution 2024

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

MMT-003 Solved Assignment 2024

Which of the following statements are true and which are false? Give reasons for your answer.
(a) If a finite group $G$ acts on a finite set $S$ , then $G_{s_{1}} = G_{s_{2}}$ for all $s_{1}, s_{2} \in X$ .

Answer:

The statement "If a finite group

G

acts on a finite set

S

, then

G_{s_{1}} = G_{s_{2}}

for all

s_{1}, s_{2} \in S

" is false. Here,

G_{s_{1}}

and

G_{s_{2}}

denote the stabilizer subgroups of

s_{1}

and

s_{2}

S

, respectively. The stabilizer subgroup

G_{s}

of an element

s \in S

is defined as the set of all elements in

G

that fix

s

, i.e.,

G_{s} = {g \in G ∣ g \cdot s = s}

Justification/Proof:

The statement claims that for any two elements

s_{1}

and

s_{2}

in the set

S

, their stabilizer subgroups in

G

are equal. This would mean that every element of

G

that fixes

s_{1}

also fixes

s_{2}

and vice versa. However, this is not necessarily true for all actions of

G

S

. The equality of stabilizers for all elements in

S

would imply a very rigid action, essentially saying that either the entire group fixes every element of

S

or that the action is trivial (where

G

acts on

S

in such a way that every element of

G

fixes every element of

S

Counterexample:

Consider the group

G = {e, g}

where

e

is the identity element and

g

is an element of order 2 (meaning

g^{2} = e

), acting on the set

S = {s_{1}, s_{2}}

with two elements. Define the action of

G

S

as follows:

$e$ fixes both $s_{1}$ and $s_{2}$ (as it must, being the identity).
$g$ swaps $s_{1}$ and $s_{2}$ , i.e., $g \cdot s_{1} = s_{2}$ and $g \cdot s_{2} = s_{1}$ .

For this action, we have:

$G_{s_{1}} = {e}$ because only the identity element fixes $s_{1}$ .
$G_{s_{2}} = {e}$ because only the identity element fixes $s_{2}$ .

In this specific example,

G_{s_{1}}

and

G_{s_{2}}

happen to be equal because the non-identity element of

G

does not fix either of the elements of

S

but rather swaps them. However, the action of

g

clearly differentiates

s_{1}

and

s_{2}

, showing that the nature of the action can lead to different behaviors for different elements of

S

, and it’s easy to construct examples where

G_{s_{1}} \neq G_{s_{2}}

by adjusting the action.

For a more direct counterexample to the statement, consider a group

G

acting on a set

S = {s_{1}, s_{2}, s_{3}}

where

G

fixes

s_{1}

and

s_{2}

but not

s_{3}

, or any scenario where the action is not uniform across all elements of

S

. This shows that the original statement is false.

(b) There are exactly 8 elements of order 3 in

S_{4}

Answer:

To determine the truth of the statement "There are exactly 8 elements of order 3 in

S_{4}

," we need to understand the structure of the symmetric group

S_{4}

and the concept of an element’s order.

Background:

The symmetric group $S_{4}$ consists of all permutations of four elements, with a total of $4! = 24$ elements.
The order of an element in a group is the smallest positive integer $n$ such that the element raised to the $n$ th power (under the group operation, which is composition in this case) equals the identity element. For permutations, this means repeating the permutation $n$ times returns the set to its original order.

Analysis:

Elements of order 3 in

S_{4}

are those permutations that cycle three elements while leaving the fourth unchanged. These can be represented as 3-cycles, such as

(123)

, which means that 1 goes to 2, 2 goes to 3, and 3 goes back to 1, with the fourth element (in this case, 4) remaining fixed.

To count the elements of order 3, we look for all possible 3-cycles within

S_{4}

. A 3-cycle can be chosen by selecting 3 elements out of 4 to participate in the cycle, and the order in which these three elements are arranged matters (since

(123)

is different from

(132)

, for example).

Counting 3-Cycles:

The number of ways to choose 3 elements out of 4 is given by the combination formula

(\binom{4}{3})

. For each selection of 3 elements, there are 2 ways to arrange them into a cycle (since for any three distinct elements

a

b

, and

c

, there are exactly two 3-cycles:

(a b c)

and

(a c b)

Thus, the total number of 3-cycles (elements of order 3) in

S_{4}

is:

(\binom{4}{3}) \times 2

Let’s calculate this value:

(\binom{4}{3}) \times 2 = 4 \times 2 = 8

After calculating, we find that there are indeed exactly 8 elements of order 3 in

S_{4}

, which means the statement is true. Each of these elements is a permutation that cycles three of the four elements back to their original positions after three applications, which is the definition of having order 3.

F = Q (\sqrt[5]{2}, \sqrt[3]{5})

, then

[F : Q] = 8

Answer:

To evaluate the statement "If

F = Q (\sqrt[5]{2}, \sqrt[3]{5})

, then

[F : Q] = 8

," we need to understand the concept of field extensions and the degree of an extension.

Background:

$F = Q (\sqrt[5]{2}, \sqrt[3]{5})$ is a field extension of $Q$ that includes both $\sqrt[5]{2}$ and $\sqrt[3]{5}$ .
The degree of the extension $[F : Q]$ is the dimension of $F$ as a vector space over $Q$ .
The degree of a composite extension can be determined by the multiplicativity of degrees in tower extensions: if $K \subseteq L \subseteq M$ are fields, then $[M : K] = [M : L] [L : K]$ .

Analysis:

Consider the extension $Q (\sqrt[5]{2})$ over $Q$ . The minimal polynomial of $\sqrt[5]{2}$ over $Q$ is $x^{5} - 2$ , which is irreducible over $Q$ by Eisenstein’s criterion (with prime $p = 2$ ). Thus, $[Q (\sqrt[5]{2}) : Q] = 5$ .
Consider the extension $Q (\sqrt[3]{5})$ over $Q$ . The minimal polynomial of $\sqrt[3]{5}$ over $Q$ is $x^{3} - 5$ , which is also irreducible over $Q$ by Eisenstein’s criterion (with prime $p = 5$ ). Thus, $[Q (\sqrt[3]{5}) : Q] = 3$ .
To find $[F : Q]$ where $F = Q (\sqrt[5]{2}, \sqrt[3]{5})$ , we need to consider how $\sqrt[5]{2}$ and $\sqrt[3]{5}$ 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 of $F$ over $Q$ 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.

Given that the degrees of the individual extensions are 5 and 3, respectively, and assuming no prior knowledge about a direct interaction that would reduce the combined degree, one might naively expect the degree of the combined extension to be

5 \times 3 = 15

. This is because the extensions are generated by roots of polynomials that are irreducible over

Q

and do not immediately suggest a common field that would reduce the degree of the extension.

Conclusion:

The statement "If

F = Q (\sqrt[5]{2}, \sqrt[3]{5})

, then

[F : Q] = 8

" is false. Based on the analysis above, without specific interaction that would reduce the degree, the expected degree of the field extension

F

over

Q

would naively be calculated as

15

, not

8

, assuming independent extensions generated by

\sqrt[5]{2}

and

\sqrt[3]{5}

. However, the actual degree calculation requires detailed knowledge of how these extensions interact, and in general, the degree of such a combined extension is not simply the product of the degrees if there’s a nontrivial intersection. But in this specific case, without evidence of such an intersection that would lower the degree to

8

, the initial analysis suggests the statement is incorrect.

(d)

F_{7} (\sqrt{3}) = F_{7} (\sqrt{5})

Answer:

To evaluate the statement "

F_{7} (\sqrt{3}) = F_{7} (\sqrt{5})

," we need to understand the structure of finite fields, specifically the field extensions of

F_{7}

, the finite field with 7 elements.

Background:

$F_{7}$ is the finite field with 7 elements, typically represented as ${0, 1, 2, 3, 4, 5, 6}$ with arithmetic operations performed modulo 7.
$F_{7} (\sqrt{3})$ denotes the smallest field extension of $F_{7}$ containing $\sqrt{3}$ , and similarly, $F_{7} (\sqrt{5})$ denotes the smallest field extension of $F_{7}$ containing $\sqrt{5}$ .
In the context of finite fields, an element $a$ is said to have a square root in $F_{7}$ if there exists an element $b \in F_{7}$ such that $b^{2} \equiv a \mod 7$ .

Analysis:

To determine whether

F_{7} (\sqrt{3}) = F_{7} (\sqrt{5})

, we need to check if

\sqrt{3}

and

\sqrt{5}

can be expressed in terms of each other within the field extensions of

F_{7}

Square Root of 3 in $F_{7}$ : We need to find if there exists an element $x \in F_{7}$ such that $x^{2} \equiv 3 \mod 7$ .
Square Root of 5 in $F_{7}$ : Similarly, we need to find if there exists an element $y \in F_{7}$ such that $y^{2} \equiv 5 \mod 7$ .

Let’s perform explicit checks for the existence of

\sqrt{3}

and

\sqrt{5}

F_{7}

For $\sqrt{3}$ : We look for an element $x$ such that $x^{2} \equiv 3 \mod 7$ .
For $\sqrt{5}$ : We look for an element $y$ such that $y^{2} \equiv 5 \mod 7$ .

Let’s calculate these explicitly to see if such elements exist and to understand the relationship between

F_{7} (\sqrt{3})

and

F_{7} (\sqrt{5})

After calculating, we find that there are no integer solutions for

x^{2} \equiv 3 \mod 7

and

x^{2} \equiv 5 \mod 7

within

F_{7}

. This means that neither

\sqrt{3}

nor

\sqrt{5}

can be directly represented by any element in

F_{7}

through simple squaring and modular arithmetic.

However, the statement "

F_{7} (\sqrt{3}) = F_{7} (\sqrt{5})

" refers to the equality of field extensions of

F_{7}

that include

\sqrt{3}

and

\sqrt{5}

, respectively. Given that neither

\sqrt{3}

nor

\sqrt{5}

has a direct representation in

F_{7}

, both

\sqrt{3}

and

\sqrt{5}

would necessitate extensions of

F_{7}

to be included.

The absence of integer solutions for these equations in

F_{7}

suggests that both

\sqrt{3}

and

\sqrt{5}

would lead to the creation of quadratic extensions of

F_{7}

, as both would be roots of irreducible quadratic polynomials over

F_{7}

. The key insight here is not the absence of solutions within

F_{7}

but the nature of the extensions they generate.

Given the structure of finite fields, any extension of

F_{7}

to include a square root not in

F_{7}

would have to be a quadratic extension, implying that both

F_{7} (\sqrt{3})

and

F_{7} (\sqrt{5})

are quadratic extensions of

F_{7}

. Since all quadratic extensions of a given finite field have the same size (in this case,

49

elements, or

F_{7^{2}}

), and since there’s only one field of each possible order in the realm of finite fields, it follows that

F_{7} (\sqrt{3})

and

F_{7} (\sqrt{5})

must indeed be the same field, despite the initial intuition suggested by the absence of direct square roots in

F_{7}

Therefore, the statement "

F_{7} (\sqrt{3}) = F_{7} (\sqrt{5})

" is true, as both

\sqrt{3}

and

\sqrt{5}

lead to the same quadratic extension of

F_{7}

, given the unique structure of finite fields and their extensions.

(e) For any

α \in F_{2^{5}}^{*}, α \neq 1, F_{2^{5}} = F_{2} [α]

Answer:

The statement "For any

α \in F_{2^{5}}^{*}, α \neq 1, F_{2^{5}} = F_{2} [α]

" concerns the structure of finite fields, specifically the field with

2^{5} = 32

elements, denoted

F_{2^{5}}

, and the concept of field extensions.

Background:

$F_{2^{5}}$ is a finite field with $32$ elements. It is the splitting field for the polynomial $x^{32} - x$ over $F_{2}$ , and it contains all roots of this polynomial.
$F_{2^{5}}^{*}$ denotes the multiplicative group of $F_{2^{5}}$ , which consists of all non-zero elements of $F_{2^{5}}$ . This group has $31$ elements, and it is cyclic.
$F_{2} [α]$ denotes the smallest field containing both $F_{2}$ and the element $α$ . If $α$ is a primitive element of $F_{2^{5}}$ (i.e., a generator of the multiplicative group $F_{2^{5}}^{*}$ ), then $F_{2} [α] = F_{2^{5}}$ .
A primitive element of a finite field is an element whose powers generate all non-zero elements of the field.

Analysis:

The statement asserts that for any non-identity element

α

F_{2^{5}}^{*}

, the field generated by

α

over

F_{2}

is the entire field

F_{2^{5}}

. This would mean that every non-identity element in

F_{2^{5}}

is a primitive element.

However, this is not true for every non-identity element of

F_{2^{5}}

. While it is true that a primitive element

α

(an element of order

31

) will generate the entire field

F_{2^{5}}

through its powers, not every non-identity element in

F_{2^{5}}

is a primitive element. Some elements will have smaller orders due to the structure of the cyclic group

F_{2^{5}}^{*}

. For example, an element of order

3

would only cycle through

3

distinct non-zero elements (plus the identity element when raised to the power of

3

) before repeating, and thus, it cannot generate the entire field.

Justification/Proof:

To disprove the statement, we can consider the property of the elements’ orders in

F_{2^{5}}^{*}

. The order of any element

α

in a finite field must divide the order of the field’s multiplicative group, which is

31

in this case (since

F_{2^{5}}^{*}

has

31

elements). By Lagrange’s theorem, the possible orders of elements in

F_{2^{5}}^{*}

are divisors of

31

, which are

1

and

31

itself, because

31

is a prime number.

Therefore, while an element

α

with order

31

(a primitive element) will indeed generate the entire field

F_{2^{5}}

through its powers, implying

F_{2^{5}} = F_{2} [α]

for such

α

, not every non-identity element has this property, as the statement incorrectly generalizes. Only the primitive elements of

F_{2^{5}}

can generate the entire field.

Conclusion:

The statement is false because not every non-identity element in

F_{2^{5}}

is a primitive element capable of generating the entire field. Only specific elements, namely the primitive elements, have this property.

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

IGNOU MMT-003 Assignment Question Paper 2024

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

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