2023

Estimated reading: 42 minutes 16 views

UPSC Algebra

\(cos\left(2\theta \right)=cos^2\theta -sin^2\theta \)

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
untitled-document-18-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
let’s examine whether there exists a homomorphism of G G GGG onto G G G^(‘)G’G where G G GGG is a group of order 10 and G G G^(‘)G’G is a group of order 6.
Introduction
In group theory, a homomorphism is a map f : G G f : G G f:G rarrG^(‘)f: G \to G’f:GG between two groups G G GGG and G G G^(‘)G’G that preserves the group operation. Specifically, for all a , b G a , b G a,b in Ga, b \in Ga,bG, f ( a b ) = f ( a ) f ( b ) f ( a b ) = f ( a ) f ( b ) f(ab)=f(a)f(b)f(ab) = f(a)f(b)f(ab)=f(a)f(b). A homomorphism is said to be “onto” if it is surjective, meaning that every element in G G G^(‘)G’G is the image of some element in G G GGG.
Step 1: Examine the Orders of the Groups
The first thing to consider is the orders of the groups G G GGG and G G G^(‘)G’G, which are 10 and 6, respectively. The order of a group is simply the number of elements it contains.
Let | G | = 10 | G | = 10 |G|=10|G| = 10|G|=10 and | G | = 6 | G | = 6 |G^(‘)|=6|G’| = 6|G|=6.
Step 2: Check the Conditions for Existence of Homomorphism
For a homomorphism f : G G f : G G f:G rarrG^(‘)f: G \to G’f:GG to be onto, the order of G G G^(‘)G’G must divide the order of G G GGG.
The formula to check this condition is:
Condition for onto homomorphism = ( | G | divides | G | ) Condition for onto homomorphism = ( | G | divides | G | ) “Condition for onto homomorphism”=(|G^(‘)|”divides”|G|)\text{Condition for onto homomorphism} = (|G’| \, \text{divides} \, |G|)Condition for onto homomorphism=(|G|divides|G|)
Let’s substitute the values:
6 divides 10 6 divides 10 6″divides”106 \, \text{divides} \, 106divides10
After Calculating, we find that 6 does not divide 10.
Summary
Since the order of G G G^(‘)G’G does not divide the order of G G GGG, there does not exist an onto homomorphism f : G G f : G G f:G rarrG^(‘)f: G \to G’f:GG.
Verified Answer
5/5
\(c=a\:cos\:B+b\:cos\:A\)
\(sin\left(2\theta \right)=2\:sin\:\theta \:cos\:\theta \)

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 p ppp-subgroup of a group G G GGG is a maximal p p ppp-subgroup, denoted P P PPP, such that the order of P P PPP is p n p n p^(n)p^npn for some n n nnn, and p n p n p^(n)p^npn divides the order of G G GGG. Here, p p ppp is a prime number. We are interested in finding all the Sylow p p ppp-subgroups of the symmetric group S 3 S 3 S_(3)S_3S3.
Step 1: Determine the Order of S 3 S 3 S_(3)S_3S3
The symmetric group S 3 S 3 S_(3)S_3S3 consists of all the permutations of 3 elements. The order of S 3 S 3 S_(3)S_3S3 is the number of such permutations, which is 3 ! 3 ! 3!3!3!.
| S 3 | = 3 ! = 3 × 2 × 1 | S 3 | = 3 ! = 3 × 2 × 1 |S_(3)|=3!=3xx2xx1|S_3| = 3! = 3 \times 2 \times 1|S3|=3!=3×2×1
After Calculating, we find | S 3 | = 6 | S 3 | = 6 |S_(3)|=6|S_3| = 6|S3|=6.
Step 2: Prime Factorization of the Order of S 3 S 3 S_(3)S_3S3
To find the Sylow p p ppp-subgroups, we first need to find the prime factors of the order of S 3 S 3 S_(3)S_3S3.
Prime factors of 6 = 2 × 3 Prime factors of  6 = 2 × 3 “Prime factors of “6=2xx3\text{Prime factors of } 6 = 2 \times 3Prime factors of 6=2×3
After Calculating, we find that the prime factors are 2 2 222 and 3 3 333.
Step 3: Determine Sylow p p ppp-subgroups for Each Prime Factor
For p = 2 p = 2 p=2p = 2p=2
The Sylow 2 2 222-subgroups will have order 2 1 = 2 2 1 = 2 2^(1)=22^1 = 221=2.
Let’s substitute the values:
Sylow 2-subgroups of S 3 = { subgroups of order 2 } Sylow 2-subgroups of  S 3 = { subgroups of order  2 } “Sylow 2-subgroups of “S_(3)={“subgroups of order “2}\text{Sylow 2-subgroups of } S_3 = \{ \text{subgroups of order } 2 \}Sylow 2-subgroups of S3={subgroups of order 2}
After Calculating, we find that the Sylow 2 2 222-subgroups are { ( 1 , 2 ) , ( 2 , 3 ) , ( 1 , 3 ) } { ( 1 , 2 ) , ( 2 , 3 ) , ( 1 , 3 ) } {(1,2),(2,3),(1,3)}\{(1,2), (2,3), (1,3)\}{(1,2),(2,3),(1,3)}.
For p = 3 p = 3 p=3p = 3p=3
The Sylow 3 3 333-subgroups will have order 3 1 = 3 3 1 = 3 3^(1)=33^1 = 331=3.
Let’s substitute the values:
Sylow 3-subgroups of S 3 = { subgroups of order 3 } Sylow 3-subgroups of  S 3 = { subgroups of order  3 } “Sylow 3-subgroups of “S_(3)={“subgroups of order “3}\text{Sylow 3-subgroups of } S_3 = \{ \text{subgroups of order } 3 \}Sylow 3-subgroups of S3={subgroups of order 3}
After Calculating, we find that the Sylow 3 3 333-subgroups are { ( 1 , 2 , 3 ) , ( 1 , 3 , 2 ) } { ( 1 , 2 , 3 ) , ( 1 , 3 , 2 ) } {(1,2,3),(1,3,2)}\{(1,2,3), (1,3,2)\}{(1,2,3),(1,3,2)}.
Summary
The Sylow p p ppp-subgroups of S 3 S 3 S_(3)S_3S3 are as follows:
  • Sylow 2 2 222-subgroups: { ( 1 , 2 ) , ( 2 , 3 ) , ( 1 , 3 ) } { ( 1 , 2 ) , ( 2 , 3 ) , ( 1 , 3 ) } {(1,2),(2,3),(1,3)}\{(1,2), (2,3), (1,3)\}{(1,2),(2,3),(1,3)}
  • Sylow 3 3 333-subgroups: { ( 1 , 2 , 3 ) , ( 1 , 3 , 2 ) } { ( 1 , 2 , 3 ) , ( 1 , 3 , 2 ) } {(1,2,3),(1,3,2)}\{(1,2,3), (1,3,2)\}{(1,2,3),(1,3,2)}
These are all the Sylow p p ppp-subgroups of S 3 S 3 S_(3)S_3S3.
Verified Answer
5/5
\(cos\left(\theta -\phi \right)=cos\:\theta \:cos\:\phi +sin\:\theta \:sin\:\phi \)
\(c^2=a^2+b^2-2ab\:Cos\left(C\right)\)

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 G GGG of order 2 p 2 p 2p2p2p, where p p ppp is an odd prime, must have a subgroup of order p p ppp.
Step 1: Establish the Order of the Group G G GGG
Given that G G GGG is a non-commutative group of order 2 p 2 p 2p2p2p, where p p ppp is an odd prime, we can write:
| G | = 2 p | G | = 2 p |G|=2p|G| = 2p|G|=2p
Step 2: Use Sylow’s Theorem
By Sylow’s First Theorem, for any prime p p ppp that divides the order of G G GGG, there exists a Sylow p p ppp-subgroup P P PPP of G G GGG such that | P | = p n | P | = p n |P|=p^(n)|P| = p^n|P|=pn for some n n nnn, and p n p n p^(n)p^npn divides | G | | G | |G||G||G|.
The formula to find the number n p n p n_(p)n_pnp of Sylow p p ppp-subgroups is:
n p 1 mod p and n p divides | G | p n n p 1 mod p and n p divides | G | p n n_(p)-=1quadmodp quad”and”quadn_(p)”divides”(|G|)/(p^(n))n_p \equiv 1 \mod p \quad \text{and} \quad n_p \, \text{divides} \, \frac{|G|}{p^n}np1modpandnpdivides|G|pn
Let’s substitute the values for p p ppp and | G | | G | |G||G||G|:
n p 1 mod p and n p divides 2 p p n p 1 mod p and n p divides 2 p p n_(p)-=1quadmodp quad”and”quadn_(p)”divides”(2p)/(p)n_p \equiv 1 \mod p \quad \text{and} \quad n_p \, \text{divides} \, \frac{2p}{p}np1modpandnpdivides2pp
After Calculating, we find that n p 1 mod p n p 1 mod p n_(p)-=1modpn_p \equiv 1 \mod pnp1modp and n p divides 2 n p divides 2 n_(p)”divides”2n_p \, \text{divides} \, 2npdivides2.
Step 3: Examine the Possibilities for n p n p n_(p)n_pnp
Since n p n p n_(p)n_pnp divides 2, n p n p n_(p)n_pnp can only be 1 or 2. Also, n p 1 mod p n p 1 mod p n_(p)-=1modpn_p \equiv 1 \mod pnp1modp.
  1. If n p = 1 n p = 1 n_(p)=1n_p = 1np=1, then there is a unique Sylow p p ppp-subgroup of order p p ppp, which must be normal. This contradicts the assumption that G G GGG is non-commutative.
  2. If n p = 2 n p = 2 n_(p)=2n_p = 2np=2, then there are exactly two distinct Sylow p p ppp-subgroups of G G GGG of order p p ppp.
Summary
From the above discussion, we can conclude that if G G GGG is a non-commutative group of order 2 p 2 p 2p2p2p, where p p ppp is an odd prime, then G G GGG must have a subgroup of order p p ppp. Specifically, there will be exactly two such subgroups.
Thus, the statement is proven.
Verified Answer
5/5
\(cos\left(2\theta \right)=cos^2\theta -sin^2\theta \)
\(cot\:\theta =\frac{cos\:\theta }{sin\:\theta }\)

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:
  1. The polynomial x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1 is irreducible in Z 3 [ x ] Z 3 [ x ] Z_(3)[x]\mathbb{Z}_3[x]Z3[x].
  2. The quotient ring Z 3 [ x ] x 2 + 1 Z 3 [ x ] x 2 + 1 (Z_(3)[x])/((:x^(2)+1:))\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}Z3[x]x2+1 is a field with 9 elements.
Step 1: Prove x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1 is Irreducible in Z 3 [ x ] Z 3 [ x ] Z_(3)[x]\mathbb{Z}_3[x]Z3[x]
To prove that x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1 is irreducible in Z 3 [ x ] Z 3 [ x ] Z_(3)[x]\mathbb{Z}_3[x]Z3[x], we need to show that it cannot be factored into two non-constant polynomials in Z 3 [ x ] Z 3 [ x ] Z_(3)[x]\mathbb{Z}_3[x]Z3[x].
The polynomial x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1 is of degree 2. If it is reducible, then it can be factored into two polynomials of degree 1. In Z 3 [ x ] Z 3 [ x ] Z_(3)[x]\mathbb{Z}_3[x]Z3[x], the possible factors would be of the form ( x a ) ( x b ) ( x a ) ( x b ) (x-a)(x-b)(x – a)(x – b)(xa)(xb) where a , b Z 3 a , b Z 3 a,b inZ_(3)a, b \in \mathbb{Z}_3a,bZ3.
Let’s substitute the values:
x 2 + 1 = ( x a ) ( x b ) x 2 + 1 = ( x a ) ( x b ) x^(2)+1=(x-a)(x-b)x^2 + 1 = (x – a)(x – b)x2+1=(xa)(xb)
After Calculating, we find that for a , b { 0 , 1 , 2 } a , b { 0 , 1 , 2 } a,b in{0,1,2}a, b \in \{0, 1, 2\}a,b{0,1,2}, no such factorization exists. Therefore, x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1 is irreducible in Z 3 [ x ] Z 3 [ x ] Z_(3)[x]\mathbb{Z}_3[x]Z3[x].
Step 2: Prove Z 3 [ x ] x 2 + 1 Z 3 [ x ] x 2 + 1 (Z_(3)[x])/((:x^(2)+1:))\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}Z3[x]x2+1 is a Field of 9 Elements
To show that Z 3 [ x ] x 2 + 1 Z 3 [ x ] x 2 + 1 (Z_(3)[x])/((:x^(2)+1:))\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}Z3[x]x2+1 is a field, we need to prove two things:
  1. It is a commutative ring with unity.
  2. Every non-zero element has a multiplicative inverse.
Commutative Ring with Unity
The ring Z 3 [ x ] Z 3 [ x ] Z_(3)[x]\mathbb{Z}_3[x]Z3[x] is a commutative ring with unity, and x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1 is an irreducible polynomial. Therefore, Z 3 [ x ] x 2 + 1 Z 3 [ x ] x 2 + 1 (Z_(3)[x])/((:x^(2)+1:))\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}Z3[x]x2+1 is also a commutative ring with unity.
Existence of Multiplicative Inverse
Since x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1 is irreducible, the ideal x 2 + 1 x 2 + 1 (:x^(2)+1:)\langle x^2 + 1 \ranglex2+1 is maximal. In a commutative ring, the quotient by a maximal ideal is a field.
Number of Elements
The elements of Z 3 [ x ] x 2 + 1 Z 3 [ x ] x 2 + 1 (Z_(3)[x])/((:x^(2)+1:))\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}Z3[x]x2+1 can be represented as a + b x a + b x a+bxa + bxa+bx where a , b Z 3 a , b Z 3 a,b inZ_(3)a, b \in \mathbb{Z}_3a,bZ3. There are 3 choices for a a aaa and 3 choices for b b bbb, making a total of 3 × 3 = 9 3 × 3 = 9 3xx3=93 \times 3 = 93×3=9 elements.
After Calculating, we find that Z 3 [ x ] x 2 + 1 Z 3 [ x ] x 2 + 1 (Z_(3)[x])/((:x^(2)+1:))\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}Z3[x]x2+1 indeed has 9 elements.
Summary
  1. The polynomial x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1 is irreducible in Z 3 [ x ] Z 3 [ x ] Z_(3)[x]\mathbb{Z}_3[x]Z3[x] as it cannot be factored into two non-constant polynomials in Z 3 [ x ] Z 3 [ x ] Z_(3)[x]\mathbb{Z}_3[x]Z3[x].
  2. The quotient ring Z 3 [ x ] x 2 + 1 Z 3 [ x ] x 2 + 1 (Z_(3)[x])/((:x^(2)+1:))\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}Z3[x]x2+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.
Verified Answer
5/5
\(2\:cos\:\theta \:cos\:\phi =cos\:\left(\theta +\phi \right)+cos\:\left(\theta -\phi \right)\)

Get Complete Solution Mathematics Optional Paper | UPSC

CONTENTS
Insert math as
Block
Inline
Additional settings
Formula color
Text color
#333333
Type math using LaTeX
Preview
\({}\)
Nothing to preview
Insert
Scroll to Top
Scroll to Top