UPSC Optional (Maths) Paper-02 Algebra Solution

2019

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UPSC Algebra

1. (a) Let $G$ be a finite group, $H$ and $K$ subgroups of $G$ such that $K \subset H$. Show that $(G: K)=(G: H)(H: K)$.

Expert Answer

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

Introduction:
We are given a finite group

G

and two subgroups

H

and

K

such that

K \subset H

. We are asked to prove that

(G : K) = (G : H) (H : K)

, where

(G : K)

(G : H)

, and

(H : K)

are the indices of the subgroups

K

H

, and

K

G

G

, and

H

respectively.

Work/Calculations:

Definition of Index:
The index

(A : B)

of a subgroup

B

in a group

A

is defined as the number of distinct left cosets of

B

A

. Mathematically,

(A : B) = | A / B |

, where

| A / B |

is the number of distinct left cosets.

Step 1: Express

(G : K)

in terms of cosets
The index

(G : K)

is the number of distinct left cosets of

K

G

. Let’s denote this set of cosets as

G / K

Step 2: Express

(G : H)

and

(H : K)

in terms of cosets
Similarly,

(G : H)

is the number of distinct left cosets of

H

G

, denoted as

G / H

, and

(H : K)

is the number of distinct left cosets of

K

H

, denoted as

H / K

Step 3: Relate

G / K

with

G / H

and

H / K

Each coset

g K

G / K

can be uniquely expressed as

h K

where

h

is in some coset

g H

G / H

. Furthermore, each coset

g H

G / H

contains exactly

(H : K)

distinct cosets of the form

h K

H / K

Therefore, the total number of distinct cosets

g K

G / K

can be obtained by multiplying the number of distinct cosets

g H

G / H

by the number of distinct cosets

h K

H / K

Step 4: Mathematical Expression
This relationship can be mathematically expressed as:

(G : K) = (G : H) (H : K)

Conclusion:
We have successfully proven that

(G : K) = (G : H) (H : K)

by relating the number of distinct left cosets of

K

G

with the number of distinct left cosets of

H

G

and

K

H

. This proves the statement for finite groups

G

and subgroups

H

and

K

such that

K \subset H

Verified Answer

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2. (a) If $G$ and $H$ are finite groups whose orders are relatively prime, then prove that there is only one homomorphism from $G$ to $H$, the trivial one.

Expert Answer

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

Introduction:
We are given two finite groups

G

and

H

whose orders are relatively prime. We are asked to prove that there is only one homomorphism from

G

H

, which is the trivial homomorphism.

Work/Calculations:

Step 1: Definitions and Assumptions
Let

| G | = n

and

| H | = m

. Since

G

and

H

are finite groups with relatively prime orders,

gcd (n, m) = 1

Let

ϕ : G \to H

be a homomorphism.

Step 2: Properties of Homomorphism
We know that the order of

ϕ (g)

must divide the order of

g

for any

g \in G

. This is because if

g^{k} = e_{G}

G

, then

ϕ (g)^{k} = ϕ (e_{G}) = e_{H}

H

Step 3: Relatively Prime Orders
Since

gcd (n, m) = 1

, the only possible order for

ϕ (g)

that divides both

n

and

m

is 1. This means that

ϕ (g)

must be the identity element in

H

for all

g \in G

Step 4: The Trivial Homomorphism
The only homomorphism

ϕ : G \to H

that satisfies these conditions is the trivial homomorphism, which maps every element of

G

to the identity element of

H

ϕ (g) = e_{H} for all g \in G

Conclusion:
We have shown that if

G

and

H

are finite groups with relatively prime orders, then the only homomorphism from

G

H

is the trivial one, which maps every element in

G

to the identity element in

H

Verified Answer

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2. (b) Write down all quotient groups of the group $Z_{12}$.

Expert Answer

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

Introduction:
We are asked to find all the quotient groups of the group

Z_{12}

, which is the group of integers modulo 12. The group

Z_{12}

consists of the elements

{0, 1, 2, \dots, 11}

under addition modulo 12.

Work/Calculations:

Step 1: Identify Subgroups of

Z_{12}

First, let’s identify the subgroups of

Z_{12}

. The subgroups are generated by the divisors of 12. The divisors of 12 are

{1, 2, 3, 4, 6, 12}

$⟨ 1 ⟩ = Z_{12}$
$⟨ 2 ⟩ = {0, 2, 4, 6, 8, 10}$
$⟨ 3 ⟩ = {0, 3, 6, 9}$
$⟨ 4 ⟩ = {0, 4, 8}$
$⟨ 6 ⟩ = {0, 6}$

Step 2: Find Quotient Groups
To find the quotient groups

Z_{12} / H

, where

H

is a subgroup of

Z_{12}

, we need to find the cosets of each subgroup

H

$Z_{12} / ⟨ 1 ⟩$ has one coset, ${0}$ , so it is isomorphic to $Z_{1}$ .
$Z_{12} / ⟨ 2 ⟩$ has two cosets, ${0, 2, 4, 6, 8, 10}$ and ${1, 3, 5, 7, 9, 11}$ , so it is isomorphic to $Z_{2}$ .
$Z_{12} / ⟨ 3 ⟩$ has three cosets, so it is isomorphic to $Z_{3}$ .
$Z_{12} / ⟨ 4 ⟩$ has four cosets, so it is isomorphic to $Z_{4}$ .
$Z_{12} / ⟨ 6 ⟩$ has six cosets, so it is isomorphic to $Z_{6}$ .

Step 3: List All Quotient Groups
The quotient groups are

Z_{1}, Z_{2}, Z_{3}, Z_{4},

and

Z_{6}

Conclusion:
The quotient groups of

Z_{12}

are

Z_{1}, Z_{2}, Z_{3}, Z_{4},

and

Z_{6}

Verified Answer

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3. (d) Let $a$ be an irreducible element of the Euclidean ring $R$, then prove that $R /(a)$ is a field.

Expert Answer

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

Introduction

In this problem, we are given that

a

is an irreducible element in a Euclidean ring

R

. We are tasked with proving that the quotient ring

R / (a)

is a field.

Work/Calculations

Step 1: Definitions and Preliminaries

A Euclidean ring $R$ is an integral domain with a Euclidean function $d : R \to N$ satisfying certain properties.
An element $a \in R$ is said to be irreducible if it is not a unit and cannot be expressed as a product of two non-unit elements.
A field is a commutative ring with unity where every non-zero element has a multiplicative inverse.

Step 2:

a

is Irreducible

\Rightarrow a

is Prime

In a Euclidean domain, every irreducible element is prime. Therefore,

a

is a prime element in

R

Step 3:

a

is Prime

\Rightarrow (a)

is a Prime Ideal

For a prime element

a

, the ideal generated by

a

, denoted

(a)

, is a prime ideal. This means that if

a b \in (a)

, then either

a \in (a)

b \in (a)

Step 4:

(a)

is a Prime Ideal

\Rightarrow R / (a)

is an Integral Domain

The quotient ring

R / (a)

is an integral domain if

(a)

is a prime ideal.

Step 5:

R / (a)

is an Integral Domain

\Rightarrow R / (a)

is a Field

In a Euclidean domain, every non-zero, non-unit element can be uniquely factored into irreducible elements. Since

a

is irreducible, the only elements in

R / (a)

are the cosets

0 + (a), 1 + (a), \dots, (a - 1) + (a)

Every non-zero element in

R / (a)

has an inverse, making

R / (a)

a field.

Conclusion

We have shown that if

a

is an irreducible element in a Euclidean ring

R

, then the quotient ring

R / (a)

is a field. This completes the proof.

Verified Answer

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2019

1. (a) Let \(G\) be a finite group, \(H\) and \(K\) subgroups of \(G\) such that \(K \subset H\). Show that \((G: K)=(G: H)(H: K)\).

Expert Answer

2. (a) If \(G\) and \(H\) are finite groups whose orders are relatively prime, then prove that there is only one homomorphism from \(G\) to \(H\), the trivial one.

Expert Answer

2. (b) Write down all quotient groups of the group \(Z_{12}\).

Expert Answer

3. (d) Let \(a\) be an irreducible element of the Euclidean ring \(R\), then prove that \(R /(a)\) is a field.

Expert Answer

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