2019
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UPSC Algebra
\(cos\:2\theta =12\:sin^2\theta \)
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
Introduction:
We are given a finite group$G$ $G$ G G $G$ and two subgroups $H$ $H$ H H $H$ and $K$ $K$ K K $K$ such that $K\subset H$ $K\subset H$ K sub H K \subset H $K\subset H$ . 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 $K$ $K$ K K $K$ , $H$ $H$ H H $H$ , and $K$ $K$ K K $K$ in $G$ $G$ G G $G$ , $G$ $G$ G G $G$ , and $H$ $H$ H H $H$ respectively.
We are given a finite group
Work/Calculations:
Definition of Index:
The index$(A:B)$ $(A:B)$ (A:B) (A: B) $(A:B)$ of a subgroup $B$ $B$ B B $B$ in a group $A$ $A$ A A $A$ is defined as the number of distinct left cosets of $B$ $B$ B B $B$ in $A$ $A$ A A $A$ . 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.
The index
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 $K$ $K$ K K $K$ in $G$ $G$ G G $G$ . Let’s denote this set of cosets as $G/K$ $G/K$ G//K G/K $G/K$ .
The index
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 $H$ $H$ H H $H$ in $G$ $G$ G G $G$ , denoted as $G/H$ $G/H$ G//H G/H $G/H$ , and $(H:K)$ $(H:K)$ (H:K) (H: K) $(H:K)$ is the number of distinct left cosets of $K$ $K$ K K $K$ in $H$ $H$ H H $H$ , denoted as $H/K$ $H/K$ H//K H/K $H/K$ .
Similarly,
Step 3: Relate $G/K$ $G/K$ G//K G/K $G/K$ with $G/H$ $G/H$ G//H G/H $G/H$ and $H/K$ $H/K$ H//K H/K $H/K$
Each coset$gK$ $gK$ gK gK $gK$ in $G/K$ $G/K$ G//K G/K $G/K$ can be uniquely expressed as $hK$ $hK$ hK hK $hK$ where $h$ $h$ h h $h$ is in some coset $gH$ $gH$ gH gH $gH$ in $G/H$ $G/H$ G//H G/H $G/H$ . Furthermore, each coset $gH$ $gH$ gH gH $gH$ in $G/H$ $G/H$ G//H G/H $G/H$ contains exactly $(H:K)$ $(H:K)$ (H:K) (H: K) $(H:K)$ distinct cosets of the form $hK$ $hK$ hK hK $hK$ in $H/K$ $H/K$ H//K H/K $H/K$ .
Each coset
Therefore, the total number of distinct cosets $gK$ $gK$ gK gK $gK$ in $G/K$ $G/K$ G//K G/K $G/K$ can be obtained by multiplying the number of distinct cosets $gH$ $gH$ gH gH $gH$ in $G/H$ $G/H$ G//H G/H $G/H$ by the number of distinct cosets $hK$ $hK$ hK hK $hK$ in $H/K$ $H/K$ H//K H/K $H/K$ .
Step 4: Mathematical Expression
This relationship can be mathematically expressed as:
This relationship can be mathematically expressed as:
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 $K$ $K$ K K $K$ in $G$ $G$ G G $G$ with the number of distinct left cosets of $H$ $H$ H H $H$ in $G$ $G$ G G $G$ and $K$ $K$ K K $K$ in $H$ $H$ H H $H$ . This proves the statement for finite groups $G$ $G$ G G $G$ and subgroups $H$ $H$ H H $H$ and $K$ $K$ K K $K$ such that $K\subset H$ $K\subset H$ K sub H K \subset H $K\subset H$ .
We have successfully proven that
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
Introduction:
We are given two finite groups$G$ $G$ G G $G$ and $H$ $H$ H H $H$ whose orders are relatively prime. We are asked to prove that there is only one homomorphism from $G$ $G$ G G $G$ to $H$ $H$ H H $H$ , which is the trivial homomorphism.
We are given two finite groups
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 $G$ $G$ G G $G$ and $H$ $H$ H H $H$ are finite groups with relatively prime orders, $gcd(n,m)=1$ $gcd(n,m)=1$ gcd(n,m)=1 \gcd(n, m) = 1 $gcd(n,m)=1$ .
Let
Let $\varphi :G\to H$ $\varphi :G\to H$ phi:G rarr H \phi: G \to H $\varphi :G\to H$ be a homomorphism.
Step 2: Properties of Homomorphism
We know that the order of$\varphi (g)$ $\varphi (g)$ phi(g) \phi(g) $\varphi (g)$ must divide the order of $g$ $g$ g g $g$ for any $g\in G$ $g\in G$ g in G g \in G $g\in G$ . This is because if ${g}^{k}={e}_{G}$ ${g}^{k}={e}_{G}$ g^(k)=e_(G) g^k = e_G ${g}^{k}={e}_{G}$ in $G$ $G$ G G $G$ , then $\varphi (g{)}^{k}=\varphi ({e}_{G})={e}_{H}$ $\varphi (g{)}^{k}=\varphi ({e}_{G})={e}_{H}$ phi(g)^(k)=phi(e_(G))=e_(H) \phi(g)^k = \phi(e_G) = e_H $\varphi (g{)}^{k}=\varphi ({e}_{G})={e}_{H}$ in $H$ $H$ H H $H$ .
We know that the order of
Step 3: Relatively Prime Orders
Since$gcd(n,m)=1$ $gcd(n,m)=1$ gcd(n,m)=1 \gcd(n, m) = 1 $gcd(n,m)=1$ , the only possible order for $\varphi (g)$ $\varphi (g)$ phi(g) \phi(g) $\varphi (g)$ that divides both $n$ $n$ n n $n$ and $m$ $m$ m m $m$ is 1. This means that $\varphi (g)$ $\varphi (g)$ phi(g) \phi(g) $\varphi (g)$ must be the identity element in $H$ $H$ H H $H$ for all $g\in G$ $g\in G$ g in G g \in G $g\in G$ .
Since
Step 4: The Trivial Homomorphism
The only homomorphism$\varphi :G\to H$ $\varphi :G\to H$ phi:G rarr H \phi: G \to H $\varphi :G\to H$ that satisfies these conditions is the trivial homomorphism, which maps every element of $G$ $G$ G G $G$ to the identity element of $H$ $H$ H H $H$ :
The only homomorphism
Conclusion:
We have shown that if$G$ $G$ G G $G$ and $H$ $H$ H H $H$ are finite groups with relatively prime orders, then the only homomorphism from $G$ $G$ G G $G$ to $H$ $H$ H H $H$ is the trivial one, which maps every element in $G$ $G$ G G $G$ to the identity element in $H$ $H$ H H $H$ .
We have shown that if
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
Introduction:
We are asked to find all the quotient groups of the group${\mathbb{Z}}_{12}$ ${\mathbb{Z}}_{12}$ Z_(12) \mathbb{Z}_{12} ${\mathbb{Z}}_{12}$ , which is the group of integers modulo 12. The group ${\mathbb{Z}}_{12}$ ${\mathbb{Z}}_{12}$ Z_(12) \mathbb{Z}_{12} ${\mathbb{Z}}_{12}$ consists of the elements $\{0,1,2,\dots ,11\}$ $\{0,1,2,\dots ,11\}$ {0,1,2,dots,11} \{0, 1, 2, \ldots, 11\} $\{0,1,2,\dots ,11\}$ under addition modulo 12.
We are asked to find all the quotient groups of the group
Work/Calculations:
Step 1: Identify Subgroups of ${\mathbb{Z}}_{12}$ ${\mathbb{Z}}_{12}$ Z_(12) \mathbb{Z}_{12} ${\mathbb{Z}}_{12}$
First, let’s identify the subgroups of${\mathbb{Z}}_{12}$ ${\mathbb{Z}}_{12}$ Z_(12) \mathbb{Z}_{12} ${\mathbb{Z}}_{12}$ . 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\} $\{1,2,3,4,6,12\}$ .
First, let’s identify the subgroups of
$\u27e81\u27e9={\mathbb{Z}}_{12}$ $\u27e81\u27e9={\mathbb{Z}}_{12}$ (:1:)=Z_(12) \langle 1 \rangle = \mathbb{Z}_{12} $\u27e81\u27e9={\mathbb{Z}}_{12}$ $\u27e82\u27e9=\{0,2,4,6,8,10\}$ $\u27e82\u27e9=\{0,2,4,6,8,10\}$ (:2:)={0,2,4,6,8,10} \langle 2 \rangle = \{0, 2, 4, 6, 8, 10\} $\u27e82\u27e9=\{0,2,4,6,8,10\}$ $\u27e83\u27e9=\{0,3,6,9\}$ $\u27e83\u27e9=\{0,3,6,9\}$ (:3:)={0,3,6,9} \langle 3 \rangle = \{0, 3, 6, 9\} $\u27e83\u27e9=\{0,3,6,9\}$ $\u27e84\u27e9=\{0,4,8\}$ $\u27e84\u27e9=\{0,4,8\}$ (:4:)={0,4,8} \langle 4 \rangle = \{0, 4, 8\} $\u27e84\u27e9=\{0,4,8\}$ $\u27e86\u27e9=\{0,6\}$ $\u27e86\u27e9=\{0,6\}$ (:6:)={0,6} \langle 6 \rangle = \{0, 6\} $\u27e86\u27e9=\{0,6\}$
Step 2: Find Quotient Groups
To find the quotient groups${\mathbb{Z}}_{12}/H$ ${\mathbb{Z}}_{12}/H$ Z_(12)//H \mathbb{Z}_{12} / H ${\mathbb{Z}}_{12}/H$ , where $H$ $H$ H H $H$ is a subgroup of ${\mathbb{Z}}_{12}$ ${\mathbb{Z}}_{12}$ Z_(12) \mathbb{Z}_{12} ${\mathbb{Z}}_{12}$ , we need to find the cosets of each subgroup $H$ $H$ H H $H$ .
To find the quotient groups
${\mathbb{Z}}_{12}/\u27e81\u27e9$ ${\mathbb{Z}}_{12}/\u27e81\u27e9$ Z_(12)//(:1:) \mathbb{Z}_{12} / \langle 1 \rangle has one coset,${\mathbb{Z}}_{12}/\u27e81\u27e9$ $\{0\}$ $\{0\}$ {0} \{0\} , so it is isomorphic to$\{0\}$ ${\mathbb{Z}}_{1}$ ${\mathbb{Z}}_{1}$ Z_(1) \mathbb{Z}_1 .${\mathbb{Z}}_{1}$ ${\mathbb{Z}}_{12}/\u27e82\u27e9$ ${\mathbb{Z}}_{12}/\u27e82\u27e9$ Z_(12)//(:2:) \mathbb{Z}_{12} / \langle 2 \rangle has two cosets,${\mathbb{Z}}_{12}/\u27e82\u27e9$ $\{0,2,4,6,8,10\}$ $\{0,2,4,6,8,10\}$ {0,2,4,6,8,10} \{0, 2, 4, 6, 8, 10\} and$\{0,2,4,6,8,10\}$ $\{1,3,5,7,9,11\}$ $\{1,3,5,7,9,11\}$ {1,3,5,7,9,11} \{1, 3, 5, 7, 9, 11\} , so it is isomorphic to$\{1,3,5,7,9,11\}$ ${\mathbb{Z}}_{2}$ ${\mathbb{Z}}_{2}$ Z_(2) \mathbb{Z}_2 .${\mathbb{Z}}_{2}$ ${\mathbb{Z}}_{12}/\u27e83\u27e9$ ${\mathbb{Z}}_{12}/\u27e83\u27e9$ Z_(12)//(:3:) \mathbb{Z}_{12} / \langle 3 \rangle has three cosets, so it is isomorphic to${\mathbb{Z}}_{12}/\u27e83\u27e9$ ${\mathbb{Z}}_{3}$ ${\mathbb{Z}}_{3}$ Z_(3) \mathbb{Z}_3 .${\mathbb{Z}}_{3}$ ${\mathbb{Z}}_{12}/\u27e84\u27e9$ ${\mathbb{Z}}_{12}/\u27e84\u27e9$ Z_(12)//(:4:) \mathbb{Z}_{12} / \langle 4 \rangle has four cosets, so it is isomorphic to${\mathbb{Z}}_{12}/\u27e84\u27e9$ ${\mathbb{Z}}_{4}$ ${\mathbb{Z}}_{4}$ Z_(4) \mathbb{Z}_4 .${\mathbb{Z}}_{4}$ ${\mathbb{Z}}_{12}/\u27e86\u27e9$ ${\mathbb{Z}}_{12}/\u27e86\u27e9$ Z_(12)//(:6:) \mathbb{Z}_{12} / \langle 6 \rangle has six cosets, so it is isomorphic to${\mathbb{Z}}_{12}/\u27e86\u27e9$ ${\mathbb{Z}}_{6}$ ${\mathbb{Z}}_{6}$ Z_(6) \mathbb{Z}_6 .${\mathbb{Z}}_{6}$
Step 3: List All Quotient Groups
The quotient groups are${\mathbb{Z}}_{1},{\mathbb{Z}}_{2},{\mathbb{Z}}_{3},{\mathbb{Z}}_{4},$ ${\mathbb{Z}}_{1},{\mathbb{Z}}_{2},{\mathbb{Z}}_{3},{\mathbb{Z}}_{4},$ Z_(1),Z_(2),Z_(3),Z_(4), \mathbb{Z}_1, \mathbb{Z}_2, \mathbb{Z}_3, \mathbb{Z}_4, ${\mathbb{Z}}_{1},{\mathbb{Z}}_{2},{\mathbb{Z}}_{3},{\mathbb{Z}}_{4},$ and ${\mathbb{Z}}_{6}$ ${\mathbb{Z}}_{6}$ Z_(6) \mathbb{Z}_6 ${\mathbb{Z}}_{6}$ .
The quotient groups are
Conclusion:
The quotient groups of${\mathbb{Z}}_{12}$ ${\mathbb{Z}}_{12}$ Z_(12) \mathbb{Z}_{12} ${\mathbb{Z}}_{12}$ are ${\mathbb{Z}}_{1},{\mathbb{Z}}_{2},{\mathbb{Z}}_{3},{\mathbb{Z}}_{4},$ ${\mathbb{Z}}_{1},{\mathbb{Z}}_{2},{\mathbb{Z}}_{3},{\mathbb{Z}}_{4},$ Z_(1),Z_(2),Z_(3),Z_(4), \mathbb{Z}_1, \mathbb{Z}_2, \mathbb{Z}_3, \mathbb{Z}_4, ${\mathbb{Z}}_{1},{\mathbb{Z}}_{2},{\mathbb{Z}}_{3},{\mathbb{Z}}_{4},$ and ${\mathbb{Z}}_{6}$ ${\mathbb{Z}}_{6}$ Z_(6) \mathbb{Z}_6 ${\mathbb{Z}}_{6}$ .
The quotient groups of
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
Introduction
In this problem, we are given that $a$ $a$ a a $a$ is an irreducible element in a Euclidean ring $R$ $R$ R R $R$ . 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
$R$ $R$ R R is an integral domain with a Euclidean function$R$ $d:R\to \mathbb{N}$ $d:R\to \mathbb{N}$ d:R rarrN d: R \to \mathbb{N} satisfying certain properties.$d:R\to \mathbb{N}$  An element
$a\in R$ $a\in R$ a in R a \in R is said to be irreducible if it is not a unit and cannot be expressed as a product of two nonunit elements.$a\in R$  A field is a commutative ring with unity where every nonzero element has a multiplicative inverse.
Step 2: $a$ $a$ a a $a$ is Irreducible $\Rightarrow a$ $\Rightarrow a$ =>a \Rightarrow a $\Rightarrow a$ is Prime
In a Euclidean domain, every irreducible element is prime. Therefore, $a$ $a$ a a $a$ is a prime element in $R$ $R$ R R $R$ .
Step 3: $a$ $a$ a a $a$ is Prime $\Rightarrow (a)$ $\Rightarrow (a)$ =>(a) \Rightarrow (a) $\Rightarrow (a)$ is a Prime Ideal
For a prime element $a$ $a$ a a $a$ , the ideal generated by $a$ $a$ a a $a$ , denoted $(a)$ $(a)$ (a) (a) $(a)$ , is a prime ideal. This means that if $ab\in (a)$ $ab\in (a)$ ab in(a) ab \in (a) $ab\in (a)$ , then either $a\in (a)$ $a\in (a)$ a in(a) a \in (a) $a\in (a)$ or $b\in (a)$ $b\in (a)$ b in(a) b \in (a) $b\in (a)$ .
Step 4: $(a)$ $(a)$ (a) (a) $(a)$ is a Prime Ideal $\Rightarrow R/(a)$ $\Rightarrow R/(a)$ =>R//(a) \Rightarrow R/(a) $\Rightarrow R/(a)$ 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 $\Rightarrow R/(a)$ $\Rightarrow R/(a)$ =>R//(a) \Rightarrow R/(a) $\Rightarrow R/(a)$ is a Field
In a Euclidean domain, every nonzero, nonunit element can be uniquely factored into irreducible elements. Since $a$ $a$ a a $a$ is irreducible, the only elements in $R/(a)$ $R/(a)$ R//(a) R/(a) $R/(a)$ are the cosets $0+(a),1+(a),\dots ,(a1)+(a)$ $0+(a),1+(a),\dots ,(a1)+(a)$ 0+(a),1+(a),dots,(a1)+(a) 0 + (a), 1 + (a), \ldots, (a1) + (a) $0+(a),1+(a),\dots ,(a1)+(a)$ .
Every nonzero 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 $a$ $a$ a a $a$ is an irreducible element in a Euclidean ring $R$ $R$ R R $R$ , 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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