2019

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UPSC Algebra

\(cos\:2\theta =1-2\:sin^2\theta \)

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 G GGG and two subgroups H H HHH and K K KKK such that K H K H K sub HK \subset HKH. We are asked to prove that ( G : K ) = ( G : H ) ( H : K ) ( G : K ) = ( G : H ) ( H : K ) (G:K)=(G:H)(H:K)(G: K) = (G: H)(H: K)(G:K)=(G:H)(H:K), where ( G : K ) ( G : K ) (G:K)(G: K)(G:K), ( G : H ) ( G : H ) (G:H)(G: H)(G:H), and ( H : K ) ( H : K ) (H:K)(H: K)(H:K) are the indices of the subgroups K K KKK, H H HHH, and K K KKK in G G GGG, G G GGG, and H H HHH respectively.
Work/Calculations:
Definition of Index:
The index ( A : B ) ( A : B ) (A:B)(A: B)(A:B) of a subgroup B B BBB in a group A A AAA is defined as the number of distinct left cosets of B B BBB in A A AAA. Mathematically, ( A : B ) = | A / B | ( A : B ) = | A / B | (A:B)=|A//B|(A: B) = |A/B|(A:B)=|A/B|, where | A / B | | A / B | |A//B||A/B||A/B| is the number of distinct left cosets.
Step 1: Express ( G : K ) ( G : K ) (G:K)(G: K)(G:K) in terms of cosets
The index ( G : K ) ( G : K ) (G:K)(G: K)(G:K) is the number of distinct left cosets of K K KKK in G G GGG. Let’s denote this set of cosets as G / K G / K G//KG/KG/K.
Step 2: Express ( G : H ) ( G : H ) (G:H)(G: H)(G:H) and ( H : K ) ( H : K ) (H:K)(H: K)(H:K) in terms of cosets
Similarly, ( G : H ) ( G : H ) (G:H)(G: H)(G:H) is the number of distinct left cosets of H H HHH in G G GGG, denoted as G / H G / H G//HG/HG/H, and ( H : K ) ( H : K ) (H:K)(H: K)(H:K) is the number of distinct left cosets of K K KKK in H H HHH, denoted as H / K H / K H//KH/KH/K.
Step 3: Relate G / K G / K G//KG/KG/K with G / H G / H G//HG/HG/H and H / K H / K H//KH/KH/K
Each coset g K g K gKgKgK in G / K G / K G//KG/KG/K can be uniquely expressed as h K h K hKhKhK where h h hhh is in some coset g H g H gHgHgH in G / H G / H G//HG/HG/H. Furthermore, each coset g H g H gHgHgH in G / H G / H G//HG/HG/H contains exactly ( H : K ) ( H : K ) (H:K)(H: K)(H:K) distinct cosets of the form h K h K hKhKhK in H / K H / K H//KH/KH/K.
Therefore, the total number of distinct cosets g K g K gKgKgK in G / K G / K G//KG/KG/K can be obtained by multiplying the number of distinct cosets g H g H gHgHgH in G / H G / H G//HG/HG/H by the number of distinct cosets h K h K hKhKhK in H / K H / K H//KH/KH/K.
Step 4: Mathematical Expression
This relationship can be mathematically expressed as:
( G : K ) = ( G : H ) ( H : K ) ( G : K ) = ( G : H ) ( H : K ) (G:K)=(G:H)(H:K)(G: K) = (G: H)(H: K)(G:K)=(G:H)(H:K)
Conclusion:
We have successfully proven that ( G : K ) = ( G : H ) ( H : K ) ( G : K ) = ( G : H ) ( H : K ) (G:K)=(G:H)(H:K)(G: K) = (G: H)(H: K)(G:K)=(G:H)(H:K) by relating the number of distinct left cosets of K K KKK in G G GGG with the number of distinct left cosets of H H HHH in G G GGG and K K KKK in H H HHH. This proves the statement for finite groups G G GGG and subgroups H H HHH and K K KKK such that K H K H K sub HK \subset HKH.
Verified Answer
5/5
\(cos^2\left(\frac{\theta }{2}\right)=\frac{1+cos\:\theta }{2}\)
\(2\:sin\:\theta \:sin\:\phi =-cos\:\left(\theta +\phi \right)+cos\:\left(\theta -\phi \right)\)

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 G GGG and H H HHH whose orders are relatively prime. We are asked to prove that there is only one homomorphism from G G GGG to H H HHH, which is the trivial homomorphism.
Work/Calculations:
Step 1: Definitions and Assumptions
Let | G | = n | G | = n |G|=n|G| = n|G|=n and | H | = m | H | = m |H|=m|H| = m|H|=m. Since G G GGG and H H HHH are finite groups with relatively prime orders, gcd ( n , m ) = 1 gcd ( n , m ) = 1 gcd(n,m)=1\gcd(n, m) = 1gcd(n,m)=1.
Let ϕ : G H ϕ : G H phi:G rarr H\phi: G \to Hϕ:GH be a homomorphism.
Step 2: Properties of Homomorphism
We know that the order of ϕ ( g ) ϕ ( g ) phi(g)\phi(g)ϕ(g) must divide the order of g g ggg for any g G g G g in Gg \in GgG. This is because if g k = e G g k = e G g^(k)=e_(G)g^k = e_Ggk=eG in G G GGG, then ϕ ( g ) k = ϕ ( e G ) = e H ϕ ( g ) k = ϕ ( e G ) = e H phi(g)^(k)=phi(e_(G))=e_(H)\phi(g)^k = \phi(e_G) = e_Hϕ(g)k=ϕ(eG)=eH in H H HHH.
Step 3: Relatively Prime Orders
Since gcd ( n , m ) = 1 gcd ( n , m ) = 1 gcd(n,m)=1\gcd(n, m) = 1gcd(n,m)=1, the only possible order for ϕ ( g ) ϕ ( g ) phi(g)\phi(g)ϕ(g) that divides both n n nnn and m m mmm is 1. This means that ϕ ( g ) ϕ ( g ) phi(g)\phi(g)ϕ(g) must be the identity element in H H HHH for all g G g G g in Gg \in GgG.
Step 4: The Trivial Homomorphism
The only homomorphism ϕ : G H ϕ : G H phi:G rarr H\phi: G \to Hϕ:GH that satisfies these conditions is the trivial homomorphism, which maps every element of G G GGG to the identity element of H H HHH:
ϕ ( g ) = e H for all g G ϕ ( g ) = e H for all g G phi(g)=e_(H)quad”for all”quad g in G\phi(g) = e_H \quad \text{for all} \quad g \in Gϕ(g)=eHfor allgG
Conclusion:
We have shown that if G G GGG and H H HHH are finite groups with relatively prime orders, then the only homomorphism from G G GGG to H H HHH is the trivial one, which maps every element in G G GGG to the identity element in H H HHH.
Verified Answer
5/5
\(\sec ^2 \theta=1+\tan ^2 \theta\)
\(2\:cos\:\theta \:sin\:\phi =sin\:\left(\theta +\phi \right)-sin\:\left(\theta -\phi \right)\)

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 Z 12 Z_(12)\mathbb{Z}_{12}Z12, which is the group of integers modulo 12. The group Z 12 Z 12 Z_(12)\mathbb{Z}_{12}Z12 consists of the elements { 0 , 1 , 2 , , 11 } { 0 , 1 , 2 , , 11 } {0,1,2,dots,11}\{0, 1, 2, \ldots, 11\}{0,1,2,,11} under addition modulo 12.
Work/Calculations:
Step 1: Identify Subgroups of Z 12 Z 12 Z_(12)\mathbb{Z}_{12}Z12
First, let’s identify the subgroups of Z 12 Z 12 Z_(12)\mathbb{Z}_{12}Z12. The subgroups are generated by the divisors of 12. The divisors of 12 are { 1 , 2 , 3 , 4 , 6 , 12 } { 1 , 2 , 3 , 4 , 6 , 12 } {1,2,3,4,6,12}\{1, 2, 3, 4, 6, 12\}{1,2,3,4,6,12}.
  • 1 = Z 12 1 = Z 12 (:1:)=Z_(12)\langle 1 \rangle = \mathbb{Z}_{12}1=Z12
  • 2 = { 0 , 2 , 4 , 6 , 8 , 10 } 2 = { 0 , 2 , 4 , 6 , 8 , 10 } (:2:)={0,2,4,6,8,10}\langle 2 \rangle = \{0, 2, 4, 6, 8, 10\}2={0,2,4,6,8,10}
  • 3 = { 0 , 3 , 6 , 9 } 3 = { 0 , 3 , 6 , 9 } (:3:)={0,3,6,9}\langle 3 \rangle = \{0, 3, 6, 9\}3={0,3,6,9}
  • 4 = { 0 , 4 , 8 } 4 = { 0 , 4 , 8 } (:4:)={0,4,8}\langle 4 \rangle = \{0, 4, 8\}4={0,4,8}
  • 6 = { 0 , 6 } 6 = { 0 , 6 } (:6:)={0,6}\langle 6 \rangle = \{0, 6\}6={0,6}
Step 2: Find Quotient Groups
To find the quotient groups Z 12 / H Z 12 / H Z_(12)//H\mathbb{Z}_{12} / HZ12/H, where H H HHH is a subgroup of Z 12 Z 12 Z_(12)\mathbb{Z}_{12}Z12, we need to find the cosets of each subgroup H H HHH.
  1. Z 12 / 1 Z 12 / 1 Z_(12)//(:1:)\mathbb{Z}_{12} / \langle 1 \rangleZ12/1 has one coset, { 0 } { 0 } {0}\{0\}{0}, so it is isomorphic to Z 1 Z 1 Z_(1)\mathbb{Z}_1Z1.
  2. Z 12 / 2 Z 12 / 2 Z_(12)//(:2:)\mathbb{Z}_{12} / \langle 2 \rangleZ12/2 has two cosets, { 0 , 2 , 4 , 6 , 8 , 10 } { 0 , 2 , 4 , 6 , 8 , 10 } {0,2,4,6,8,10}\{0, 2, 4, 6, 8, 10\}{0,2,4,6,8,10} and { 1 , 3 , 5 , 7 , 9 , 11 } { 1 , 3 , 5 , 7 , 9 , 11 } {1,3,5,7,9,11}\{1, 3, 5, 7, 9, 11\}{1,3,5,7,9,11}, so it is isomorphic to Z 2 Z 2 Z_(2)\mathbb{Z}_2Z2.
  3. Z 12 / 3 Z 12 / 3 Z_(12)//(:3:)\mathbb{Z}_{12} / \langle 3 \rangleZ12/3 has three cosets, so it is isomorphic to Z 3 Z 3 Z_(3)\mathbb{Z}_3Z3.
  4. Z 12 / 4 Z 12 / 4 Z_(12)//(:4:)\mathbb{Z}_{12} / \langle 4 \rangleZ12/4 has four cosets, so it is isomorphic to Z 4 Z 4 Z_(4)\mathbb{Z}_4Z4.
  5. Z 12 / 6 Z 12 / 6 Z_(12)//(:6:)\mathbb{Z}_{12} / \langle 6 \rangleZ12/6 has six cosets, so it is isomorphic to Z 6 Z 6 Z_(6)\mathbb{Z}_6Z6.
Step 3: List All Quotient Groups
The quotient groups are Z 1 , Z 2 , Z 3 , Z 4 , Z 1 , Z 2 , Z 3 , Z 4 , Z_(1),Z_(2),Z_(3),Z_(4),\mathbb{Z}_1, \mathbb{Z}_2, \mathbb{Z}_3, \mathbb{Z}_4,Z1,Z2,Z3,Z4, and Z 6 Z 6 Z_(6)\mathbb{Z}_6Z6.
Conclusion:
The quotient groups of Z 12 Z 12 Z_(12)\mathbb{Z}_{12}Z12 are Z 1 , Z 2 , Z 3 , Z 4 , Z 1 , Z 2 , Z 3 , Z 4 , Z_(1),Z_(2),Z_(3),Z_(4),\mathbb{Z}_1, \mathbb{Z}_2, \mathbb{Z}_3, \mathbb{Z}_4,Z1,Z2,Z3,Z4, and Z 6 Z 6 Z_(6)\mathbb{Z}_6Z6.
Verified Answer
5/5
\(2\:cos\:\theta \:sin\:\phi =sin\:\left(\theta +\phi \right)-sin\:\left(\theta -\phi \right)\)
\(2\:sin\:\theta \:cos\:\phi =sin\:\left(\theta +\phi \right)+sin\:\left(\theta -\phi \right)\)

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 a aaa is an irreducible element in a Euclidean ring R R RRR. We are tasked with proving that the quotient ring R / ( a ) R / ( a ) R//(a)R/(a)R/(a) is a field.
Work/Calculations
Step 1: Definitions and Preliminaries
  • A Euclidean ring R R RRR is an integral domain with a Euclidean function d : R N d : R N d:R rarrNd: R \to \mathbb{N}d:RN satisfying certain properties.
  • An element a R a R a in Ra \in RaR 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 a aaa is Irreducible a a =>a\Rightarrow aa is Prime
In a Euclidean domain, every irreducible element is prime. Therefore, a a aaa is a prime element in R R RRR.
Step 3: a a aaa is Prime ( a ) ( a ) =>(a)\Rightarrow (a)(a) is a Prime Ideal
For a prime element a a aaa, the ideal generated by a a aaa, denoted ( a ) ( a ) (a)(a)(a), is a prime ideal. This means that if a b ( a ) a b ( a ) ab in(a)ab \in (a)ab(a), then either a ( a ) a ( a ) a in(a)a \in (a)a(a) or b ( a ) b ( a ) b in(a)b \in (a)b(a).
Step 4: ( a ) ( a ) (a)(a)(a) is a Prime Ideal R / ( a ) R / ( a ) =>R//(a)\Rightarrow R/(a)R/(a) is an Integral Domain
The quotient ring R / ( a ) R / ( a ) R//(a)R/(a)R/(a) is an integral domain if ( a ) ( a ) (a)(a)(a) is a prime ideal.
Step 5: R / ( a ) R / ( a ) R//(a)R/(a)R/(a) is an Integral Domain R / ( a ) R / ( a ) =>R//(a)\Rightarrow R/(a)R/(a) is a Field
In a Euclidean domain, every non-zero, non-unit element can be uniquely factored into irreducible elements. Since a a aaa is irreducible, the only elements in R / ( a ) R / ( a ) R//(a)R/(a)R/(a) are the cosets 0 + ( a ) , 1 + ( a ) , , ( a 1 ) + ( a ) 0 + ( a ) , 1 + ( a ) , , ( a 1 ) + ( a ) 0+(a),1+(a),dots,(a-1)+(a)0 + (a), 1 + (a), \ldots, (a-1) + (a)0+(a),1+(a),,(a1)+(a).
Every non-zero element in R / ( a ) R / ( a ) R//(a)R/(a)R/(a) has an inverse, making R / ( a ) R / ( a ) R//(a)R/(a)R/(a) a field.
Conclusion
We have shown that if a a aaa is an irreducible element in a Euclidean ring R R RRR, then the quotient ring R / ( a ) R / ( a ) R//(a)R/(a)R/(a) is a field. This completes the proof.
Verified Answer
5/5
\(2\:sin\:\theta \:cos\:\phi =sin\:\left(\theta +\phi \right)+sin\:\left(\theta -\phi \right)\)

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