IGNOU MMT-003 Solved Assignment 2023 | M.Sc. MACS

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

₹365.00

Get Instant Access

Access via our Android App Only

Please read the following points before ordering :

Assignment Question Paper
Sample Solution
Frequently Asked Questions (FAQs)
Terms and Conditions

Share with your Friends

IGNOU MMT-003 Assignment Question Paper 2023

mmt-003-2023-ext-aa2d2d76-613f-4e79-9c79-c97616952506

Which of the following statements are true? Give reasons for your answers. Marks will only be given for valid justification of your answers.

i) If

G

is a finite abelian group and

p

is a prime factor of

o (G)

, then the number of Sylow p-subgroups of

G

is a prime.

ii) The minimum polynomial of

5^{1 / 3}

over

Q

x^{1 / 3}

iii)

Z_{m n} ≃ Z_{m} \times Z_{h} \forall m, n \in Z

iv) If

G

is a finite group and

m ∣ o (G), m \in N

, then

G

has a subgroup of order

m

v) If

β_{1}

and

β_{2}

are two

15^{th}

roots of unity, then

Q (β_{1}) = Q (β_{2})

vi) There exists an extension field of

Z_{3}

of order 25 .

vii) Every group of order 18 has a normal subgroup of order 2.

viii) If

I

and

J

are ideals of a ring

R

, then

I J = I \cap J

ix) If

f : R \to S

is a ring homomorphism and

I

is an ideal of

R

, then

f (I)

is an ideal of S.

x

Every prime ideal of an integral domain is a maximal ideal.

a) Let $G$ be a group and let $H \leq G, K \leq G, o (H) = o (K) = p$ , a prime. Show that either $H \cap K = {e}$ or $H = K$ . Is this result still true if $p$ is not a prime? Justify your answer.

b) Let

G

be the group of all rigid motions of a plane and

S

be the set of all rectangles in the plane. Show that

G

acts on S . Also obtain the orbit and stabiliser of a square under this action.

c) Let

G

be a finite group and

H

be a normal subgroup of

G

. Prove that

H = \cup C_{x}

where the

C_{x}

are all the distinct conjugacy classes of

G

such that

H \cap C_{x} \neq \emptyset

a) Find the number of Sylow 5-subgroups, Sylow 7-subgroups and Sylow 2-subgroups $A_{5}$ has.

b) Let

G_{1}

and

G_{2}

be finite groups such that

p

divides

| G_{1} |

and

| G_{2} |

. Prove that the Sylow p-subgroups of

G_{1} \times G_{2}

are precisely of the form

P_{1} \times P_{2}

, where

P_{1}

and

P_{2}

are Sylow p-subgroups of

G_{1}

and

G_{2}

, respectively.

a) Write $[\begin{array}{ccc} 1 & - 1 & 0 \\ 2 & 0 & 3 \\ 5 & 5 & 1 \end{array}]$ as a product of $O (3)$ and an element of $B_{3} (R)$ .

b) Show that

S U (2)

and

S^{3}

are structurally the same.

a) Find all the possible abelian groups, up to isomorphism, of order 900 .

b) Construct the free group on the set

{α, β, γ}

. Further, check if it is a free abelian group or not.

a) Check whether or not the ring $R = Z_{3} [x] / < x^{6} - 1 >$
i) is finite;
ii) has zero divisors;
iii) has nilpotent elements.

b) Give two distinct rings whose quotient field is

{a + i b ∣ a, b \in R}

. Justify your answer.

c) Evaluate the following Legendre Symbols:
i)

(\frac{139}{431})

ii)

(\frac{149}{439})

d) Check whether 9782957210008 is a valid ISBN number.

a) Check whether or not $Z [\sqrt{- 7}]$ is a Euclidean domain.

b) Use the division algorithm to find the inverse of

\bar{18}

Z_{35}

i) Let $G$ be a group of automorphisms of a field $K$ . Is the fixed field $K^{G}$ a subfield of $K$ ? Why, or why not?

ii) Find

K^{G}

, where

K = Q (i, \sqrt{3}), G = G (K / Q)

MMT-003 Sample Solution 2023

untitled-document-16-9cbde385-ad0a-4c74-864f-362c1a2d74cc

MMT-003 Solved Assignment 2023

Question:-01

Which of the following statements are true? Give reasons for your answers. Marks will only be given for valid justification of your answers.
i) If $G$ is a finite abelian group and $p$ is a prime factor of $o (G)$ , then the number of Sylow p-subgroups of $G$ is a prime.

Answer:

The statement “If

G

is a finite abelian group and

p

is a prime factor of

o (G)

, then the number of Sylow

p

-subgroups of

G

is a prime” is false.

Justification:

In a finite abelian group

G

, every Sylow

p

-subgroup is normal. Therefore, all Sylow

p

-subgroups are conjugate to each other, and they are all isomorphic. In an abelian group, the number of Sylow

p

-subgroups is given by

n_{p}

, and according to Sylow’s Third Theorem,

n_{p}

divides

| G |

and

n_{p} \equiv 1 \mod p

However, there is no requirement that

n_{p}

must be a prime number.

Counterexample:

Consider the abelian group

G = Z_{4} \times Z_{2}

, where

Z_{n}

denotes the integers modulo

n

. The order of

G

o (G) = 4 \times 2 = 8

The prime factors of

o (G)

are

2

and

2

(i.e.,

8 = 2^{3}

The Sylow

2

-subgroups of

G

are generated by

(2, 0)

(0, 1)

, and

(2, 1)

. So, there are

3

Sylow

2

-subgroups, and

3

is not a prime divisor of

8

Thus, the statement is false, as demonstrated by this counterexample.

Page Break

ii) The minimum polynomial of

5^{1 / 3}

over

Q

x^{1 / 3}

Answer:

The statement “The minimum polynomial of

5^{1 / 3}

over

Q

x^{1 / 3}

” is false.

Justification:

The minimum polynomial of an algebraic number

α

over a field

F

is the unique monic polynomial

f (x)

of least degree such that

f (α) = 0

and

f (x)

has coefficients in

F

For

5^{1 / 3}

, the minimum polynomial over

Q

x^{3} - 5

, not

x^{1 / 3}

$x^{3} - 5$ is a polynomial with coefficients in $Q$ .
$(5^{1 / 3})^{3} - 5 = 5 - 5 = 0$ , so $5^{1 / 3}$ is a root of $x^{3} - 5$ .
$x^{3} - 5$ is irreducible over $Q$ , which means it has the least degree among all polynomials that have $5^{1 / 3}$ as a root and have coefficients in $Q$ .

Moreover,

x^{1 / 3}

is not even a polynomial over

Q

because its exponent

1 / 3

is not a non-negative integer.

Thus, the statement is false.

Page Break

iii)

Z_{m n} ≃ Z_{m} \times Z_{n} \forall m, n \in Z

Answer:

The statement “

Z_{m n} ≃ Z_{m} \times Z_{n} \forall m, n \in Z

” is false.

Justification:

The isomorphism

Z_{m n} ≃ Z_{m} \times Z_{n}

holds if and only if

m

and

n

are coprime (i.e.,

gcd (m, n) = 1

Counterexample:

Let

m = 4

and

n = 2

. Then

Z_{4 \times 2} = Z_{8}

However,

Z_{4} \times Z_{2}

is not isomorphic to

Z_{8}

$Z_{8}$ has elements $[0], [1], [2], [3], [4], [5], [6], [7]$ and is cyclic, generated by $[1]$ .
$Z_{4} \times Z_{2}$ has elements $(0, 0), (0, 1), (1, 0), (1, 1), (2, 0), (2, 1), (3, 0), (3, 1)$ and is not cyclic.

Since

Z_{8}

is cyclic and

Z_{4} \times Z_{2}

is not, they cannot be isomorphic.

Thus, the statement is false, as demonstrated by this counterexample.

Page Break

iv) If

G

is a finite group and

m ∣ o (G), m \in N

, then

G

has a subgroup of order

m

Answer:

The statement “If

G

is a finite group and

m ∣ o (G), m \in N

, then

G

has a subgroup of order

m

” is false.

Justification:

The statement would be true if

G

were a cyclic group, as every divisor of the order of a cyclic group corresponds to a unique subgroup. However, the statement is not generally true for all finite groups.

Counterexample:

Consider the symmetric group

S_{3}

with

o (S_{3}) = 6

. The divisors of 6 are

1, 2, 3, 6

While

S_{3}

does have subgroups of orders 1, 2, 3, and 6, it does not have a subgroup of order 4, even though 4 is a natural number (

m \in N

Thus, the statement is false, as demonstrated by this counterexample.

Page Break

v) If

β_{1}

and

β_{2}

are two

15^{th}

roots of unity, then

Q (β_{1}) = Q (β_{2})

Answer:

The statement “If

β_{1}

and

β_{2}

are two

15^{th}

roots of unity, then

Q (β_{1}) = Q (β_{2})

” is true.

Justification:

The

15^{th}

roots of unity are the complex numbers of the form

β_{k} = \cos (\frac{2 π k}{15}) + i \sin (\frac{2 π k}{15}),

where

i

is the imaginary unit and

k = 0, 1, 2, \dots, 14

These roots of unity are all solutions to the equation

x^{15} - 1 = 0

, and they generate a cyclotomic field

Q (ζ_{15})

, where

ζ_{15} = e^{2 π i / 15}

Now, any

15^{th}

root of unity can be expressed as a power of

ζ_{15}

. Specifically,

β_{1} = ζ_{15}^{k_{1}}

and

β_{2} = ζ_{15}^{k_{2}}

for some

k_{1}, k_{2} \in {0, 1, 2, \dots, 14}

Since both

β_{1}

and

β_{2}

can be expressed using powers of

ζ_{15}

, it follows that

Q (β_{1})

and

Q (β_{2})

are both subfields of

Q (ζ_{15})

Moreover,

β_{1}

and

β_{2}

generate the same field as

ζ_{15}

because they are powers of

ζ_{15}

. Therefore,

Q (β_{1}) = Q (ζ_{15})

and

Q (β_{2}) = Q (ζ_{15})

, which implies

Q (β_{1}) = Q (β_{2})

Thus, the statement is true.

Page Break

vi) There exists an extension field of

Z_{3}

of order 25 .

Answer:

The statement “There exists an extension field of

Z_{3}

of order 25″ is false.

Justification:

An extension field

K

of a finite field

F

of order

p^{n}

(where

p

is a prime and

n

is a positive integer) must have order

p^{m}

for some integer

m > n

. This is because the extension field

K

can be thought of as a vector space over

F

, and the size of

K

must be

p^{m}

to be compatible with the vector space structure.

In the given statement,

Z_{3}

is a field of order

3

, and we are asked about an extension field of order

25

. The number

25 = 5^{2}

is not of the form

3^{m}

for any integer

m

Therefore, there cannot exist an extension field of

Z_{3}

with order

25

Thus, the statement is false.

Page Break

vii) Every group of order 18 has a normal subgroup of order 2 .

Answer:

The statement “Every group of order 18 has a normal subgroup of order 2” is false.

Justification:

A group of order 18 has order

2 \times 9

, which is divisible by the primes 2 and 3. According to the Sylow theorems, for a group

G

of order

18 = 2 \times 9

, the number

n_{2}

of Sylow 2-subgroups must divide 9 and be congruent to 1 modulo 2. The possible values for

n_{2}

are therefore 1, 3, or 9.

Counterexample:

Consider the dihedral group

D_{9}

, which is the group of symmetries of a regular 9-gon. The order of

D_{9}

is 18. In

D_{9}

, the number of Sylow 2-subgroups is 9, and none of them are normal in

D_{9}

Thus,

D_{9}

is a group of order 18 that does not have a normal subgroup of order 2, making the statement false.

Page Break

viii) If

I

and

J

are ideals of a ring

R

, then

I J = I \cap J

Answer:

The statement “If

I

and

J

are ideals of a ring

R

, then

I J = I \cap J

” is false.

Justification:

Product of Ideals $I J$ : The product $I J$ is defined as the set ${\sum_{i = 1}^{n} a_{i} b_{i} ∣ a_{i} \in I, b_{i} \in J, n \in N}$ . It is also an ideal of $R$ .
Intersection of Ideals $I \cap J$ : The intersection $I \cap J$ is the set of all elements that are both in $I$ and $J$ . It is also an ideal of $R$ .

Counterexample:

Consider the ring

R = Z

of integers. Let

I = 2 Z

and

J = 3 Z

$I J = {2 \times 3, 2 \times 6, 2 \times 9, \dots, 4 \times 3, 4 \times 6, \dots} = 6 Z$
$I \cap J = {n \in Z ∣ n is divisible by both 2 and 3} = 6 Z$

In this example,

I J = I \cap J

, but this is not generally true for all ideals

I

and

J

in all rings

R

For instance, consider the ring

R = Z [x]

of all polynomials with integer coefficients. Let

I = (2, x)

and

J = (3, x)

$I J = (6, 2 x, 3 x, x^{2})$
$I \cap J = (6, x)$

Here,

I J \neq I \cap J

, disproving the statement.

Thus, the statement is false, as demonstrated by this counterexample.

Page Break

ix) If

f : R \to S

is a ring homomorphism and

I

is an ideal of

R

, then

f (I)

is an ideal of

S

Answer:

The statement “If

f : R \to S

is a ring homomorphism and

I

is an ideal of

R

, then

f (I)

is an ideal of

S

” is false.

Justification:

An ideal

I

of a ring

R

is a subset of

R

that is closed under addition and multiplication by any element in

R

. A ring homomorphism

f : R \to S

preserves addition and multiplication but does not necessarily map ideals to ideals.

Counterexample:

Consider the ring

R = Z

and the ring

S = Z / 2 Z

(the integers modulo 2). Let

f : R \to S

be the natural projection map defined by

f (x) = x \mod 2

Let

I = 2 Z

be the ideal of even integers in

R

The image

f (I)

consists of the set

{0}

S

, since all even integers are mapped to 0 in

Z / 2 Z

However,

{0}

is not an ideal in

S

because it is not closed under multiplication by any element in

S

. Specifically,

0 \times 1 = 0

but

1

is not in

{0}

Thus,

f (I)

is not an ideal in

S

, making the statement false.

Page Break

x) Every prime ideal of an integral domain is a maximal ideal.

Answer:

The statement “Every prime ideal of an integral domain is a maximal ideal” is false.

Justification:

Prime Ideal: An ideal $P$ in a commutative ring $R$ is called a prime ideal if for any $a, b \in R$ , $a b \in P$ implies $a \in P$ or $b \in P$ .
Maximal Ideal: An ideal $M$ in a commutative ring $R$ is called a maximal ideal if $M \neq R$ and there is no ideal $N$ such that $M ⊊ N ⊊ R$ .

Counterexample:

Consider the integral domain

R = Z [x]

, the ring of all polynomials with integer coefficients.

Let

P = (x)

, the ideal generated by

x

. This is a prime ideal because

Z [x] / (x) ≅ Z

, which is an integral domain. However,

P

is not maximal because it is properly contained in other ideals, such as

(x, 2)

, which in turn is properly contained in

R

Thus,

P

is a prime ideal that is not maximal, disproving the statement.

Therefore, the statement is false, as demonstrated by this counterexample.

Page Break

Question:-02

a) Let $G$ be a group and let $H \leq G, K \leq G, o (H) = o (K) = p$ , a prime. Show that either $H \cap K = {e}$ or $H = K$ . Is this result still true if $p$ is not a prime? Justify your answer.

Answer:

Part 1: $o (H) = o (K) = p$ , a prime

Let

G

be a group and

H, K \leq G

with

o (H) = o (K) = p

, where

p

is a prime number.

Case 1: $H \cap K \neq {e}$
- If $H \cap K$ contains an element other than the identity $e$ , then $H \cap K$ must be a subgroup of both $H$ and $K$ (since the intersection of two subgroups is always a subgroup).
- The order of $H \cap K$ must divide the order of $H$ and $K$ by Lagrange’s theorem. Since $o (H) = o (K) = p$ and $p$ is a prime, the only divisors are 1 and $p$ .
- Since $H \cap K$ contains an element other than $e$ , its order must be $p$ , which means $H \cap K = H = K$ .
Case 2: $H \cap K = {e}$
- If $H \cap K$ contains only the identity $e$ , then $H \cap K = {e}$ .

So, either

H \cap K = {e}

H = K

Part 2: $o (H) = o (K)$ not necessarily a prime

p

is not a prime, the result may not hold. For example, consider

G = Z_{12}

H = {0, 4, 8}

, and

K = {0, 6, 12}

. Here,

o (H) = o (K) = 3

, which is not a prime. We have

H \cap K = {0}

, but

H \neq K

Thus, the result is not necessarily true if

p

is not a prime.

Page Break

b) Let

G

be the group of all rigid motions of a plane and

S

be the set of all rectangles in the plane. Show that

G

acts on

S

. Also obtain the orbit and stabiliser of a square under this action.

Answer:

To show that

G

acts on

S

, we need to satisfy two conditions:

For each $g \in G$ and $s \in S$ , $g \cdot s$ is also in $S$ .
For each $g_{1}, g_{2} \in G$ and $s \in S$ , $(g_{1} \cdot g_{2}) \cdot s = g_{1} \cdot (g_{2} \cdot s)$ .

Condition 1: Closure

A rigid motion in the plane consists of rotations, translations, and reflections. When we apply any of these to a rectangle, we still get a rectangle. Therefore,

G

is closed under the action on

S

Condition 2: Associativity

Let

g_{1}, g_{2}

be any two rigid motions and

s

be any rectangle. Then

(g_{1} \cdot g_{2}) \cdot s

is the rectangle obtained by first applying

g_{2}

s

and then applying

g_{1}

to the result. This is the same as first applying

g_{2}

s

to get

g_{2} \cdot s

and then applying

g_{1}

g_{2} \cdot s

to get

g_{1} \cdot (g_{2} \cdot s)

. Therefore,

(g_{1} \cdot g_{2}) \cdot s = g_{1} \cdot (g_{2} \cdot s)

So both conditions are satisfied, and

G

acts on

S

Orbit and Stabilizer for a Square

Orbit

The orbit of a square under this action is the set of all rectangles that can be obtained by applying a rigid motion to the square. Since a square is a special case of a rectangle, any rigid motion (rotation, translation, reflection) will still produce a square. Therefore, the orbit of a square under this action is the set of all squares in the plane.

Stabilizer

The stabilizer of a square is the set of all rigid motions that leave the square invariant. This includes:

The identity motion (doing nothing).
90-degree rotations about the center of the square.
180-degree rotations about the center of the square.
270-degree rotations about the center of the square.
Reflections about the axes of symmetry of the square.

So, the stabilizer consists of these 8 rigid motions.

Thus, we have shown that

G

acts on

S

and have obtained the orbit and stabilizer of a square under this action.

Page Break

c) Let

G

be a finite group and

H

be a normal subgroup of

G

. Prove that

H = \cup C_{x}

where the

C_{x}

are all the distinct conjugacy classes of

G

such that

H \cap C_{x} \neq \emptyset

Answer:

To prove that

H = \cup C_{x}

where

C_{x}

are all the distinct conjugacy classes of

G

such that

H \cap C_{x} \neq \emptyset

, we need to show two things:

$H \subseteq \cup C_{x}$
$\cup C_{x} \subseteq H$

1. $H \subseteq \cup C_{x}$

Take any element

h \in H

. Since

H

is a normal subgroup of

G

, it is invariant under conjugation by elements of

G

. That is, for any

g \in G

g^{- 1} h g \in H

The conjugacy class

C_{h}

h

G

is defined as

C_{h} = {g^{- 1} h g ∣ g \in G}

. Since

H

is normal,

C_{h} \subseteq H

Therefore,

h \in C_{h}

and

C_{h} \cap H \neq \emptyset

. So,

h \in \cup C_{x}

This shows that

H \subseteq \cup C_{x}

2. $\cup C_{x} \subseteq H$

Take any element

x

such that

C_{x} \cap H \neq \emptyset

. Let

h \in C_{x} \cap H

The conjugacy class

C_{x}

is defined as

C_{x} = {g^{- 1} x g ∣ g \in G}

Since

h \in C_{x}

, there exists some

g \in G

such that

h = g^{- 1} x g

Since

h \in H

and

H

is a normal subgroup,

g^{- 1} h g \in H

. But

g^{- 1} h g = x

, so

x \in H

This shows that

C_{x} \subseteq H

for every

x

such that

C_{x} \cap H \neq \emptyset

Therefore,

\cup C_{x} \subseteq H

Conclusion

Since

H \subseteq \cup C_{x}

and

\cup C_{x} \subseteq H

, we have

H = \cup C_{x}

where

C_{x}

are all the distinct conjugacy classes of

G

such that

H \cap C_{x} \neq \emptyset

This completes the proof.

Page Break

Get The AppClick Here

Frequently Asked Questions (FAQs)

How do I access the Complete Solution?

You can access the Complete Solution through our app, which can be downloaded using this link:

App Link

Simply click “Install” to download and install the app, and then follow the instructions to purchase the required assignment solution. Currently, the app is only available for Android devices. We are working on making the app available for iOS in the future, but it is not currently available for iOS devices.

Is this Complete Solution for IGNOU Assignments?

Yes, It is Complete Solution, a comprehensive solution to the assignments for IGNOU. Valid from January 1, 2023 to December 31, 2023.

Is the Complete Solution aligned with the IGNOU requirements?

Yes, the Complete Solution is aligned with the IGNOU requirements and has been solved accordingly.

Is the Complete Solution guaranteed error-free?

Yes, the Complete Solution is guaranteed to be error-free.The solutions are thoroughly researched and verified by subject matter experts to ensure their accuracy.

Can I access the Complete Solution anytime?

As of now, you have access to the Complete Solution for a period of 6 months after the date of purchase, which is sufficient to complete the assignment. However, we can extend the access period upon request. You can access the solution anytime through our app.

What if I need help with a specific assignment question?

The app provides complete solutions for all assignment questions. If you still need help, you can contact the support team for assistance at Whatsapp +91-9958288900

Can I access the educational materials on multiple devices?

No, access to the educational materials is limited to one device only, where you have first logged in. Logging in on multiple devices is not allowed and may result in the revocation of access to the educational materials.

How do I make a payment for accessing the solutions?

Payments can be made through various secure online payment methods available in the app.Your payment information is protected with industry-standard security measures to ensure its confidentiality and safety. You will receive a receipt for your payment through email or within the app, depending on your preference.

What are the instructions for formatting my assignments?

The instructions for formatting your assignments are detailed in the Assignment Booklet, which includes details on paper size, margins, precision, and submission requirements. It is important to strictly follow these instructions to facilitate evaluation and avoid delays.

Terms and Conditions

The educational materials provided in the app are the sole property of the app owner and are protected by copyright laws.
Reproduction, distribution, or sale of the educational materials without prior written consent from the app owner is strictly prohibited and may result in legal consequences.
Any attempt to modify, alter, or use the educational materials for commercial purposes is strictly prohibited.
The app owner reserves the right to revoke access to the educational materials at any time without notice for any violation of these terms and conditions.
The app owner is not responsible for any damages or losses resulting from the use of the educational materials.
The app owner reserves the right to modify these terms and conditions at any time without notice.
By accessing and using the app, you agree to abide by these terms and conditions.
Access to the educational materials is limited to one device only. Logging in to the app on multiple devices is not allowed and may result in the revocation of access to the educational materials.

Our educational materials are solely available on our website and application only. Users and students can report the dealing or selling of the copied version of our educational materials by any third party at our email address (abstract4math@gmail.com) or mobile no. (+91-9958288900).

In return, such users/students can expect free our educational materials/assignments and other benefits as a bonafide gesture which will be completely dependent upon our discretion.

Categories IGNOU Assignment Solution, M.Sc. MACS, M.Sc. MACS 2023 Tags "IGNOU, 2023, Assignment Solution, IGNOU 2023, IGNOU Assignment, IGNOU M.Sc. MACS, M.Sc. MACS, MACS, MACS Assignment, MMT-003, MMT-003 2023, MMT-003 Assignment, Solved Assignment

IGNOU MMT-003 Solved Assignment 2023 | M.Sc. MACS

IGNOU MMT-003 Assignment Question Paper 2023

MMT-003 Sample Solution 2023

Question:-01

Justification:

Counterexample:

Justification:

Justification:

Counterexample:

Justification:

Counterexample:

Justification:

Justification:

Justification:

Counterexample:

Justification:

Counterexample:

Justification:

Counterexample:

Justification:

Counterexample:

Question:-02

Part 1: o ( H ) = o ( K ) = p o ( H ) = o ( K ) = p o(H)=o(K)=po(H) = o(K) = po(H)=o(K)=p, a prime

Part 2: o ( H ) = o ( K ) o ( H ) = o ( K ) o(H)=o(K)o(H) = o(K)o(H)=o(K) not necessarily a prime

Condition 1: Closure

Condition 2: Associativity

Orbit and Stabilizer for a Square

Orbit

Stabilizer

1. H ⊆ ∪ C x H ⊆ ∪ C x H sube uuC_(x)H \subseteq \cup C_xH⊆∪Cx

2. ∪ C x ⊆ H ∪ C x ⊆ H uuC_(x)sube H\cup C_x \subseteq H∪Cx⊆H

Conclusion

Frequently Asked Questions (FAQs)

Terms and Conditions

Related products

IGNOU MEG-16 Solved Assignment 2022-2023 | MEG | Indian Folk Literature

IGNOU MEG-11 Solved Assignment 2022-2023 | MEG | AMERICAN NOVEL

IGNOU MEG-03 Solved Assignment 2022-2023 | MEG | British Novel

IGNOU MST-004 Solved Assignment 2023 | PGDAST

All Policies

Account

Company

Contact Us

Part 1: $o (H) = o (K) = p$ , a prime

Part 2: $o (H) = o (K)$ not necessarily a prime

1. $H \subseteq \cup C_{x}$

2. $\cup C_{x} \subseteq H$