UPSC Optional (Maths) Paper-02 Algebra Solution

2020

Estimated reading: 36 minutes 152 views

UPSC Algebra

1. (a) Let $S_3$ and $Z_3$ be permutation group on 3 symbols and group of residue classes module 3 respectively. Show that there is no homomorphism of $S_3$ in $Z_3$ except the trivial homomorphism.

Expert Answer

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

To show that there is no homomorphism of

S_{3}

into

Z_{3}

except the trivial homomorphism, we can use the following properties of homomorphisms:

A homomorphism $ϕ : G \to H$ preserves the identity element, i.e., $ϕ (e_{G}) = e_{H}$ .
A homomorphism $ϕ : G \to H$ preserves the group operation, i.e., $ϕ (a * b) = ϕ (a) * ϕ (b)$ .
A homomorphism $ϕ : G \to H$ preserves the order of elements, i.e., if $a$ has order $n$ in $G$ , then $ϕ (a)$ has order dividing $n$ in $H$ .

Properties of

S_{3}

and

Z_{3}

$S_{3}$ is the permutation group on 3 symbols, and it has $3! = 6$ elements.
$Z_{3}$ is the group of residue classes modulo 3, and it has 3 elements: $[0], [1], [2]$ .

Steps to Show No Non-Trivial Homomorphism Exists

Identity Element: Any homomorphism $ϕ : S_{3} \to Z_{3}$ must map the identity element of $S_{3}$ (the identity permutation $e$ ) to the identity element of $Z_{3}$ ( $[0]$ ).

$ϕ (e) = [0]$
Order of Elements: The order of any element in $Z_{3}$ divides 3. In $S_{3}$ , we have elements of order 2 (e.g., transpositions) and elements of order 3 (e.g., 3-cycles). If there exists a non-trivial homomorphism $ϕ$ , then it must map elements of $S_{3}$ to elements of $Z_{3}$ in such a way that the order of the image divides the order of the original element.

However, $Z_{3}$ only has elements of order 1 ( $[0]$ ) and order 3 ( $[1], [2]$ ). There are no elements of order 2 in $Z_{3}$ .
Contradiction: $S_{3}$ contains elements of order 2 (transpositions). Any homomorphism $ϕ$ would have to map these elements to an element in $Z_{3}$ whose order divides 2. Since $Z_{3}$ contains no such elements (other than the identity), we reach a contradiction.

Therefore, the only homomorphism that can exist from

S_{3}

Z_{3}

is the trivial homomorphism that maps all elements of

S_{3}

to the identity element

[0]

Z_{3}

Verified Answer

 5/5

2. (a) Let $G$ be a finite cyclic group of order $n$. Then prove that $G$ has $\phi(n)$ generators (where $\phi$ is Euler's $\phi$-function).

Expert Answer

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

Introduction

The problem asks us to prove that a finite cyclic group

G

of order

n

has

ϕ (n)

generators, where

ϕ

is Euler’s

ϕ

-function. Euler’s

ϕ

-function

ϕ (n)

is defined as the number of positive integers less than

n

that are relatively prime to

n

Preliminaries

Let

G

be a finite cyclic group of order

n

generated by

a

. That is,

G = {a^{0}, a^{1}, a^{2}, \dots, a^{n - 1}}

Generators of

G

A generator

g

G

is an element such that every element in

G

can be written as a power of

g

. In other words,

G = {g^{0}, g^{1}, g^{2}, \dots, g^{n - 1}}

Euler’s

ϕ

-Function

Euler’s

ϕ

-function

ϕ (n)

counts the number of positive integers less than

n

that are relatively prime to

n

Proof

Claim: An element $a^{k}$ generates $G$ if and only if $gcd (k, n) = 1$ .

Proof of Claim:
- Forward Direction: Suppose $a^{k}$ generates $G$ . Then $⟨ a^{k} ⟩ = G$ , which means $a^{k}$ has order $n$ . By Lagrange’s theorem, the order of $a^{k}$ must divide $n$ . Since $a^{k}$ has order $n$ , $k$ and $n$ must be relatively prime.
- Backward Direction: Suppose $gcd (k, n) = 1$ . We want to show that $a^{k}$ generates $G$ . To do this, we need to show that every element $a^{m}$ in $G$ can be written as $(a^{k})^{r}$ for some integer $r$ .
Since $gcd (k, n) = 1$ , there exist integers $p$ and $q$ such that $p k + q n = 1$ . For any $a^{m}$ in $G$ , we have:

$a^{m} = a^{m p k} = (a^{k})^{m p} = (a^{k})^{r}$

where $r = m p$ . This shows that $a^{k}$ generates $G$ .
Counting Generators: The number of generators of $G$ is the same as the number of integers $k$ such that $0 < k < n$ and $gcd (k, n) = 1$ . This is precisely $ϕ (n)$ .

Conclusion

We have proven that a finite cyclic group

G

of order

n

has

ϕ (n)

generators. The proof relies on the properties of Euler’s

ϕ

-function and the definition of a cyclic group.

Verified Answer

 5/5

3. (a) Let $R$ be a finite field of characteristic $p(>0)$. Show that the mapping $f: R \rightarrow R$ defined by $f(a)=a^p, \forall a \in R$ is an isomorphism.

Expert Answer

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

Introduction

The problem asks us to prove that the mapping

f : R \to R

defined by

f (a) = a^{p}

for all

a \in R

is an isomorphism, where

R

is a finite field of characteristic

p > 0

Definitions

Field: A set $R$ with two operations $+$ and $\times$ that satisfy the field axioms.
Characteristic: A field $R$ has characteristic $p$ if $p$ is the smallest positive integer such that $p \cdot a = 0$ for all $a \in R$ . If no such $p$ exists, the characteristic is zero.
Isomorphism: A bijective map $f : R \to R$ that preserves the field operations.

Properties Needed

Frobenius Endomorphism: In a field of characteristic $p$ , $(a + b)^{p} = a^{p} + b^{p}$ and $(a b)^{p} = a^{p} b^{p}$ .

Proof

To prove that

f

is an isomorphism, we need to show that

f

is a bijective map that preserves addition and multiplication.

$f$ is a Homomorphism
Preservation of Addition:

$f (a + b) = (a + b)^{p} = a^{p} + b^{p} = f (a) + f (b)$
Preservation of Multiplication:

$f (a b) = (a b)^{p} = a^{p} b^{p} = f (a) f (b)$
$f$ is Injective (One-to-One)

Suppose

f (a) = f (b)

. Then

a^{p} = b^{p}

which implies

a^{p} - b^{p} = 0

. Since

R

is a field, it has no zero divisors, and we can factor

a^{p} - b^{p}

(a - b)^{p}

. This means

a - b = 0

a = b

, proving that

f

is injective.

$f$ is Surjective (Onto)

Since

R

is finite and

f

is injective,

f

must also be surjective. Alternatively, for any

b \in R

b = b^{p^{2}}

(by Fermat’s Little Theorem or the fact that

R^{*}

, the multiplicative group of

R

, has order

p^{n} - 1

). Thus,

f (b^{p - 1}) = (b^{p - 1})^{p} = b^{p} = b

, showing that

f

is surjective.

Conclusion

We have shown that

f

preserves both addition and multiplication, and is both injective and surjective. Therefore,

f : R \to R

defined by

f (a) = a^{p}

is an isomorphism.

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

2020

1. (a) Let \(S_3\) and \(Z_3\) be permutation group on 3 symbols and group of residue classes module 3 respectively. Show that there is no homomorphism of \(S_3\) in \(Z_3\) except the trivial homomorphism.

Expert Answer

2. (a) Let \(G\) be a finite cyclic group of order \(n\). Then prove that \(G\) has \(\phi(n)\) generators (where \(\phi\) is Euler's \(\phi\)-function).

Expert Answer

3. (a) Let \(R\) be a finite field of characteristic \(p(>0)\). Show that the mapping \(f: R \rightarrow R\) defined by \(f(a)=a^p, \forall a \in R\) is an isomorphism.

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