UPSC Optional (Maths) Paper-02 Algebra Solution

2023

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UPSC Algebra

1(a) Let $ G $ be a group of order 10 and $ G' $ be a group of order 6. Examine whether there exists a homomorphism of $ G $ onto $ G' $.

Expert Answer

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let’s examine whether there exists a homomorphism of

G

onto

G^{'}

where

G

is a group of order 10 and

G^{'}

is a group of order 6.

Introduction

In group theory, a homomorphism is a map

f : G \to G^{'}

between two groups

G

and

G^{'}

that preserves the group operation. Specifically, for all

a, b \in G

f (a b) = f (a) f (b)

. A homomorphism is said to be “onto” if it is surjective, meaning that every element in

G^{'}

is the image of some element in

G

Step 1: Examine the Orders of the Groups

The first thing to consider is the orders of the groups

G

and

G^{'}

, which are 10 and 6, respectively. The order of a group is simply the number of elements it contains.

Let

| G | = 10

and

| G^{'} | = 6

Step 2: Check the Conditions for Existence of Homomorphism

For a homomorphism

f : G \to G^{'}

to be onto, the order of

G^{'}

must divide the order of

G

The formula to check this condition is:

Condition for onto homomorphism = (| G^{'} | divides | G |)

Let’s substitute the values:

6 divides 10

After Calculating, we find that 6 does not divide 10.

Summary

Since the order of

G^{'}

does not divide the order of

G

, there does not exist an onto homomorphism

f : G \to G^{'}

Verified Answer

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1(b) Determine all the Sylow $ p $-subgroups of the symmetric group $ S_3 $.

Expert Answer

untitled-document-18-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

Introduction

In group theory, a Sylow

p

-subgroup of a group

G

is a maximal

p

-subgroup, denoted

P

, such that the order of

P

p^{n}

for some

n

, and

p^{n}

divides the order of

G

. Here,

p

is a prime number. We are interested in finding all the Sylow

p

-subgroups of the symmetric group

S_{3}

Step 1: Determine the Order of

S_{3}

The symmetric group

S_{3}

consists of all the permutations of 3 elements. The order of

S_{3}

is the number of such permutations, which is

3!

| S_{3} | = 3! = 3 \times 2 \times 1

After Calculating, we find

| S_{3} | = 6

Step 2: Prime Factorization of the Order of

S_{3}

To find the Sylow

p

-subgroups, we first need to find the prime factors of the order of

S_{3}

Prime factors of 6 = 2 \times 3

After Calculating, we find that the prime factors are

2

and

3

Step 3: Determine Sylow

p

-subgroups for Each Prime Factor

For

p = 2

The Sylow

2

-subgroups will have order

2^{1} = 2

Let’s substitute the values:

Sylow 2-subgroups of S_{3} = {subgroups of order 2}

After Calculating, we find that the Sylow

2

-subgroups are

{(1, 2), (2, 3), (1, 3)}

For

p = 3

The Sylow

3

-subgroups will have order

3^{1} = 3

Let’s substitute the values:

Sylow 3-subgroups of S_{3} = {subgroups of order 3}

After Calculating, we find that the Sylow

3

-subgroups are

{(1, 2, 3), (1, 3, 2)}

Summary

The Sylow

p

-subgroups of

S_{3}

are as follows:

Sylow $2$ -subgroups: ${(1, 2), (2, 3), (1, 3)}$
Sylow $3$ -subgroups: ${(1, 2, 3), (1, 3, 2)}$

These are all the Sylow

p

-subgroups of

S_{3}

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2(a) Prove that a non-commutative group of order $ 2p $, where $ p $ is an odd prime, must have a subgroup of order $ p $.

Expert Answer

untitled-document-18-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

Introduction

We are tasked with proving that a non-commutative group

G

of order

2 p

, where

p

is an odd prime, must have a subgroup of order

p

Step 1: Establish the Order of the Group

G

Given that

G

is a non-commutative group of order

2 p

, where

p

is an odd prime, we can write:

| G | = 2 p

Step 2: Use Sylow’s Theorem

By Sylow’s First Theorem, for any prime

p

that divides the order of

G

, there exists a Sylow

p

-subgroup

P

G

such that

| P | = p^{n}

for some

n

, and

p^{n}

divides

| G |

The formula to find the number

n_{p}

of Sylow

p

-subgroups is:

n_{p} \equiv 1 \mod p and n_{p} divides \frac{| G |}{p^{n}}

Let’s substitute the values for

p

and

| G |

n_{p} \equiv 1 \mod p and n_{p} divides \frac{2 p}{p}

After Calculating, we find that

n_{p} \equiv 1 \mod p

and

n_{p} divides 2

Step 3: Examine the Possibilities for

n_{p}

Since

n_{p}

divides 2,

n_{p}

can only be 1 or 2. Also,

n_{p} \equiv 1 \mod p

If $n_{p} = 1$ , then there is a unique Sylow $p$ -subgroup of order $p$ , which must be normal. This contradicts the assumption that $G$ is non-commutative.
If $n_{p} = 2$ , then there are exactly two distinct Sylow $p$ -subgroups of $G$ of order $p$ .

Summary

From the above discussion, we can conclude that if

G

is a non-commutative group of order

2 p

, where

p

is an odd prime, then

G

must have a subgroup of order

p

. Specifically, there will be exactly two such subgroups.

Thus, the statement is proven.

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3(a) Prove that $ x^2 + 1 $ is an irreducible polynomial in $ Z_3[x] $. Further show that the quotient ring $ \frac{Z_3[x]}{\langle x^2 + 1 \rangle} $ is a field of 9 elements.

Expert Answer

untitled-document-18-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

Introduction

We are tasked with proving two things:

The polynomial $x^{2} + 1$ is irreducible in $Z_{3} [x]$ .
The quotient ring $\frac{Z_{3} [x]}{⟨ x^{2} + 1 ⟩}$ is a field with 9 elements.

Step 1: Prove

x^{2} + 1

is Irreducible in

Z_{3} [x]

To prove that

x^{2} + 1

is irreducible in

Z_{3} [x]

, we need to show that it cannot be factored into two non-constant polynomials in

Z_{3} [x]

The polynomial

x^{2} + 1

is of degree 2. If it is reducible, then it can be factored into two polynomials of degree 1. In

Z_{3} [x]

, the possible factors would be of the form

(x - a) (x - b)

where

a, b \in Z_{3}

Let’s substitute the values:

x^{2} + 1 = (x - a) (x - b)

After Calculating, we find that for

a, b \in {0, 1, 2}

, no such factorization exists. Therefore,

x^{2} + 1

is irreducible in

Z_{3} [x]

Step 2: Prove

\frac{Z_{3} [x]}{⟨ x^{2} + 1 ⟩}

is a Field of 9 Elements

To show that

\frac{Z_{3} [x]}{⟨ x^{2} + 1 ⟩}

is a field, we need to prove two things:

It is a commutative ring with unity.
Every non-zero element has a multiplicative inverse.

Commutative Ring with Unity

The ring

Z_{3} [x]

is a commutative ring with unity, and

x^{2} + 1

is an irreducible polynomial. Therefore,

\frac{Z_{3} [x]}{⟨ x^{2} + 1 ⟩}

is also a commutative ring with unity.

Existence of Multiplicative Inverse

Since

x^{2} + 1

is irreducible, the ideal

⟨ x^{2} + 1 ⟩

is maximal. In a commutative ring, the quotient by a maximal ideal is a field.

Number of Elements

The elements of

\frac{Z_{3} [x]}{⟨ x^{2} + 1 ⟩}

can be represented as

a + b x

where

a, b \in Z_{3}

. There are 3 choices for

a

and 3 choices for

b

, making a total of

3 \times 3 = 9

elements.

After Calculating, we find that

\frac{Z_{3} [x]}{⟨ x^{2} + 1 ⟩}

indeed has 9 elements.

Summary

The polynomial $x^{2} + 1$ is irreducible in $Z_{3} [x]$ as it cannot be factored into two non-constant polynomials in $Z_{3} [x]$ .
The quotient ring $\frac{Z_{3} [x]}{⟨ x^{2} + 1 ⟩}$ is a field with 9 elements, satisfying the properties of a commutative ring with unity and containing multiplicative inverses for all non-zero elements.

Thus, both statements are proven.

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2023

1(a) Let \( G \) be a group of order 10 and \( G' \) be a group of order 6. Examine whether there exists a homomorphism of \( G \) onto \( G' \).

Expert Answer

1(b) Determine all the Sylow \( p \)-subgroups of the symmetric group \( S_3 \).

Expert Answer

2(a) Prove that a non-commutative group of order \( 2p \), where \( p \) is an odd prime, must have a subgroup of order \( p \).

Expert Answer

3(a) Prove that \( x^2 + 1 \) is an irreducible polynomial in \( Z_3[x] \). Further show that the quotient ring \( \frac{Z_3[x]}{\langle x^2 + 1 \rangle} \) is a field of 9 elements.

Expert Answer

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