UPSC Optional (Maths) Paper-02 Algebra Solution

2012

Estimated reading: 44 minutes 136 views

UPSC Algebra

1(a) How many elements of order 2 are there in the group of order 16 generated by $a$ and $b$ such that the order of $a$ is 8, the order of $b$ is 2, and $b a b^{-1}=a^{-1}$?

Expert Answer

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

In a group

G

generated by

a

and

b

with the given properties, we can describe the elements of

G

in terms of

a

and

b

. Specifically, any element

g

G

can be written as

g = a^{m} b^{n}

, where

0 \leq m < 8

and

n = 0, 1

Step 1: Identify Elements of Order 2

An element

g

has order 2 if

g^{2} = e

, where

e

is the identity element of the group.

Let’s consider the elements

a^{m} b^{n}

and find those that satisfy

(a^{m} b^{n})^{2} = e

Case 1:

n = 0

In this case,

(a^{m})^{2} = a^{2 m} = e

. Since

a

has order 8, this is only possible if

m = 4

. So,

a^{4}

is an element of order 2.

Case 2:

n = 1

In this case,

(a^{m} b)^{2} = a^{m} b a^{m} b

. Using the relation

b a b^{- 1} = a^{- 1}

, we get:

(a^{m} b)^{2} = a^{m} b a^{m} b = a^{m} b b^{- 1} a^{- m} b = a^{m} e a^{- m} b = a^{m} a^{- m} b = e b = b

So,

b

is also an element of order 2.

Case 3:

n = 1

and

m \neq 0

In this case,

(a^{m} b)^{2} = a^{m} b a^{m} b

. Using the relation

b a b^{- 1} = a^{- 1}

, we get:

(a^{m} b)^{2} = a^{m} b a^{m} b = a^{m} b b^{- 1} a^{- m} b = a^{m} e a^{- m} b = a^{m} a^{- m} b = e b = b

This equation simplifies to

b

, which means that

(a^{m} b)^{2} \neq e

unless

m = 0

Step 2: Count Elements of Order 2

From the above discussion, we find that there are only two elements of order 2:

a^{4}

and

b

So, there are 2 elements of order 2 in the group

G

Verified Answer

 5/5

2(a) How many conjugacy classes does the permutation group $S_5$ of permutations of 5 numbers have? Write down one element in each class (preferably in terms of cycles).

Expert Answer

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

In the symmetric group

S_{5}

, the conjugacy classes are determined by the cycle structure of the permutations. Two permutations are conjugate if and only if they have the same cycle structure.

Here are the different types of cycle structures that can appear in

S_{5}

Identity: $(1)$ (1 element)
Single transpositions: $(a b)$ (2-cycles) (10 elements)
Double transpositions: $(a b) (c d)$ (two 2-cycles) (15 elements)
3-cycles: $(a b c)$ (20 elements)
4-cycles: $(a b c d)$ (30 elements)
5-cycles: $(a b c d e)$ (24 elements)
3-cycle and a 2-cycle: $(a b c) (d e)$ (20 elements)

Each of these cycle structures represents a different conjugacy class. Therefore,

S_{5}

has 7 conjugacy classes.

Here is one element from each conjugacy class, written in terms of cycles:

Identity: $(1)$
Single transpositions: $(1 2)$
Double transpositions: $(1 2) (3 4)$
3-cycles: $(1 2 3)$
4-cycles: $(1 2 3 4)$
5-cycles: $(1 2 3 4 5)$
3-cycle and a 2-cycle: $(1 2 3) (4 5)$

These elements are representatives of their respective conjugacy classes.

Verified Answer

 5/5

3(a) Is the ideal generated by 2 and $X$ in the polynomial ring $Z[X]$ of polynomials in a single variable $X$ with coefficients in the ring of integers $Z$, a principal ideal? Justify your answer.

Expert Answer

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

In the polynomial ring

Z [X]

, consider the ideal

I

generated by 2 and

X

. We want to determine whether this ideal is a principal ideal, i.e., whether it can be generated by a single element.

Step 1: Describe the Ideal

I

The ideal

I

consists of all polynomials that can be formed as

f (X) = 2 a (X) + X b (X)

, where

a (X)

and

b (X)

are arbitrary polynomials in

Z [X]

Step 2: Check for Principal Ideal

To be a principal ideal, there must exist a single polynomial

g (X)

Z [X]

such that every element in

I

can be written as

g (X) \cdot h (X)

, where

h (X)

is another polynomial in

Z [X]

Let’s assume for the sake of contradiction that such a

g (X)

exists. Then

g (X)

must divide both 2 and

X

Z [X]

$g (X)$ divides 2: The only polynomials that divide 2 are the constant polynomials $\pm 1$ and $\pm 2$ .
$g (X)$ divides $X$ : The polynomials that divide $X$ are $\pm 1$ and $\pm X$ .

The only polynomials that satisfy both conditions are

\pm 1

. If

g (X) = 1

g (X) = - 1

, then

I

would be the entire ring

Z [X]

, which it is not. Therefore, no such

g (X)

can exist.

Conclusion

The ideal

I

generated by 2 and

X

Z [X]

is not a principal ideal.

Verified Answer

 5/5

4(a) Describe the maximal ideals in the ring of Gaussian integers $\mathbb{Z}[i]=\{a+b i \mid a, b \in \mathbb{Z}\}$.

Expert Answer

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

In the ring of Gaussian integers

Z [i]

, a maximal ideal is an ideal

M

such that there are no other ideals

N

with

M ⊊ N ⊊ Z [i]

Properties of Maximal Ideals in

Z [i]

Nonzero Principal Ideals: In $Z [i]$ , every maximal ideal is a nonzero principal ideal, meaning it is generated by a single Gaussian integer $a + b i$ where $a + b i \neq 0$ .
Generated by Irreducible Elements: In $Z [i]$ , maximal ideals are generated by irreducible elements. An element $z$ in $Z [i]$ is irreducible if it is not a unit and cannot be expressed as a product of two non-units in $Z [i]$ .

Classification of Maximal Ideals

The maximal ideals in

Z [i]

can be classified as follows:

Ideals generated by primes congruent to 3 mod 4: For a prime $p$ in $Z$ that is congruent to 3 mod 4, $(p)$ is a maximal ideal in $Z [i]$ . This is because $p$ remains irreducible in $Z [i]$ .
Ideals generated by $1 + i$ : The ideal $(1 + i)$ is maximal. This is because $Z [i] / (1 + i)$ is isomorphic to $Z / (2)$ , which is a field.
Ideals generated by Gaussian primes not in $Z$ : A Gaussian integer $a + b i$ is a Gaussian prime if it is irreducible in $Z [i]$ . For example, $2 + 3 i$ is a Gaussian prime, and $(2 + 3 i)$ is a maximal ideal.

Summary

The maximal ideals in

Z [i]

are generated by:

Primes in $Z$ that are congruent to 3 mod 4.
The Gaussian integer $1 + i$ .
Gaussian primes that are not in $Z$ .

Each of these generates a unique maximal ideal in

Z [i]

Verified Answer

 5/5

Get Complete Solution Mathematics Optional Paper | UPSC

Mathematics Optional
UPSC Previous Years Maths Optional Papers with Solution | Paper-02
₹1,095.00 Add to cart
Mathematics Optional
UPSC Previous Years Maths Optional Papers with Solution | Paper-01
₹1,095.00 Add to cart

2012

1(a) How many elements of order 2 are there in the group of order 16 generated by \(a\) and \(b\) such that the order of \(a\) is 8, the order of \(b\) is 2, and \(b a b^{-1}=a^{-1}\)?

Expert Answer

2(a) How many conjugacy classes does the permutation group \(S_5\) of permutations of 5 numbers have? Write down one element in each class (preferably in terms of cycles).

Expert Answer

3(a) Is the ideal generated by 2 and \(X\) in the polynomial ring \(Z[X]\) of polynomials in a single variable \(X\) with coefficients in the ring of integers \(Z\), a principal ideal? Justify your answer.

Expert Answer

4(a) Describe the maximal ideals in the ring of Gaussian integers \(\mathbb{Z}[i]=\{a+b i \mid a, b \in \mathbb{Z}\}\).

Expert Answer

Get Complete Solution Mathematics Optional Paper | UPSC

UPSC Previous Years Maths Optional Papers with Solution | Paper-02

UPSC Previous Years Maths Optional Papers with Solution | Paper-01

CONTENTS

All Policies

Account

Company

Affiliate Program

Shipping Policy

Privacy Policy

Return Policy

Terms & Conditions

Cancellation Policy

Registration & Security

Refund Policy

Copyright & DMCA Notice

Privacy Policy

Content Copyright

Dispute Resolution

Terms of Services

Contact Us