Sample Solution

MMT-003 Solved Assignment 2023 Sample Solution

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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Next Lesson

MMT-003 Solved Assignment 2023

Sample Solution

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

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