2020

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UPSC Algebra

\(b^2=c^2+a^2-2ac\:Cos\left(B\right)\)

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 S 3 S_(3)S_3S3 into Z 3 Z 3 Z_(3)Z_3Z3 except the trivial homomorphism, we can use the following properties of homomorphisms:
  1. A homomorphism ϕ : G H ϕ : G H phi:G rarr H\phi: G \to Hϕ:GH preserves the identity element, i.e., ϕ ( e G ) = e H ϕ ( e G ) = e H phi(e_(G))=e_(H)\phi(e_G) = e_Hϕ(eG)=eH.
  2. A homomorphism ϕ : G H ϕ : G H phi:G rarr H\phi: G \to Hϕ:GH preserves the group operation, i.e., ϕ ( a b ) = ϕ ( a ) ϕ ( b ) ϕ ( a b ) = ϕ ( a ) ϕ ( b ) phi(a**b)=phi(a)**phi(b)\phi(a \ast b) = \phi(a) \ast \phi(b)ϕ(ab)=ϕ(a)ϕ(b).
  3. A homomorphism ϕ : G H ϕ : G H phi:G rarr H\phi: G \to Hϕ:GH preserves the order of elements, i.e., if a a aaa has order n n nnn in G G GGG, then ϕ ( a ) ϕ ( a ) phi(a)\phi(a)ϕ(a) has order dividing n n nnn in H H HHH.
Properties of S 3 S 3 S_(3)S_3S3 and Z 3 Z 3 Z_(3)Z_3Z3
  1. S 3 S 3 S_(3)S_3S3 is the permutation group on 3 symbols, and it has 3 ! = 6 3 ! = 6 3!=63! = 63!=6 elements.
  2. Z 3 Z 3 Z_(3)Z_3Z3 is the group of residue classes modulo 3, and it has 3 elements: [ 0 ] , [ 1 ] , [ 2 ] [ 0 ] , [ 1 ] , [ 2 ] [0],[1],[2][0], [1], [2][0],[1],[2].
Steps to Show No Non-Trivial Homomorphism Exists
  1. Identity Element: Any homomorphism ϕ : S 3 Z 3 ϕ : S 3 Z 3 phi:S_(3)rarrZ_(3)\phi: S_3 \to Z_3ϕ:S3Z3 must map the identity element of S 3 S 3 S_(3)S_3S3 (the identity permutation e e eee) to the identity element of Z 3 Z 3 Z_(3)Z_3Z3 ( [ 0 ] [ 0 ] [0][0][0]).
    ϕ ( e ) = [ 0 ] ϕ ( e ) = [ 0 ] phi(e)=[0]\phi(e) = [0]ϕ(e)=[0]
  2. Order of Elements: The order of any element in Z 3 Z 3 Z_(3)Z_3Z3 divides 3. In S 3 S 3 S_(3)S_3S3, 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 ϕ phi\phiϕ, then it must map elements of S 3 S 3 S_(3)S_3S3 to elements of Z 3 Z 3 Z_(3)Z_3Z3 in such a way that the order of the image divides the order of the original element.
    However, Z 3 Z 3 Z_(3)Z_3Z3 only has elements of order 1 ( [ 0 ] [ 0 ] [0][0][0]) and order 3 ( [ 1 ] , [ 2 ] [ 1 ] , [ 2 ] [1],[2][1], [2][1],[2]). There are no elements of order 2 in Z 3 Z 3 Z_(3)Z_3Z3.
  3. Contradiction: S 3 S 3 S_(3)S_3S3 contains elements of order 2 (transpositions). Any homomorphism ϕ ϕ phi\phiϕ would have to map these elements to an element in Z 3 Z 3 Z_(3)Z_3Z3 whose order divides 2. Since Z 3 Z 3 Z_(3)Z_3Z3 contains no such elements (other than the identity), we reach a contradiction.
Therefore, the only homomorphism that can exist from S 3 S 3 S_(3)S_3S3 to Z 3 Z 3 Z_(3)Z_3Z3 is the trivial homomorphism that maps all elements of S 3 S 3 S_(3)S_3S3 to the identity element [ 0 ] [ 0 ] [0][0][0] in Z 3 Z 3 Z_(3)Z_3Z3.
Verified Answer
5/5
\(c^2=a^2+b^2-2ab\:Cos\left(C\right)\)
\(a=b\:cos\:C+c\:cos\:B\)

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 G GGG of order n n nnn has ϕ ( n ) ϕ ( n ) phi(n)\phi(n)ϕ(n) generators, where ϕ ϕ phi\phiϕ is Euler’s ϕ ϕ phi\phiϕ-function. Euler’s ϕ ϕ phi\phiϕ-function ϕ ( n ) ϕ ( n ) phi(n)\phi(n)ϕ(n) is defined as the number of positive integers less than n n nnn that are relatively prime to n n nnn.
Preliminaries
Let G G GGG be a finite cyclic group of order n n nnn generated by a a aaa. That is, G = { a 0 , a 1 , a 2 , , a n 1 } G = { a 0 , a 1 , a 2 , , a n 1 } G={a^(0),a^(1),a^(2),dots,a^(n-1)}G = \{ a^0, a^1, a^2, \ldots, a^{n-1} \}G={a0,a1,a2,,an1}.
Generators of G G GGG
A generator g g ggg of G G GGG is an element such that every element in G G GGG can be written as a power of g g ggg. In other words, G = { g 0 , g 1 , g 2 , , g n 1 } G = { g 0 , g 1 , g 2 , , g n 1 } G={g^(0),g^(1),g^(2),dots,g^(n-1)}G = \{ g^0, g^1, g^2, \ldots, g^{n-1} \}G={g0,g1,g2,,gn1}.
Euler’s ϕ ϕ phi\phiϕ-Function
Euler’s ϕ ϕ phi\phiϕ-function ϕ ( n ) ϕ ( n ) phi(n)\phi(n)ϕ(n) counts the number of positive integers less than n n nnn that are relatively prime to n n nnn.
Proof
  1. Claim: An element a k a k a^(k)a^kak generates G G GGG if and only if gcd ( k , n ) = 1 gcd ( k , n ) = 1 gcd(k,n)=1\gcd(k, n) = 1gcd(k,n)=1.
    Proof of Claim:
    • Forward Direction: Suppose a k a k a^(k)a^kak generates G G GGG. Then a k = G a k = G (:a^(k):)=G\langle a^k \rangle = Gak=G, which means a k a k a^(k)a^kak has order n n nnn. By Lagrange’s theorem, the order of a k a k a^(k)a^kak must divide n n nnn. Since a k a k a^(k)a^kak has order n n nnn, k k kkk and n n nnn must be relatively prime.
    • Backward Direction: Suppose gcd ( k , n ) = 1 gcd ( k , n ) = 1 gcd(k,n)=1\gcd(k, n) = 1gcd(k,n)=1. We want to show that a k a k a^(k)a^kak generates G G GGG. To do this, we need to show that every element a m a m a^(m)a^mam in G G GGG can be written as ( a k ) r ( a k ) r (a^(k))^(r)(a^k)^r(ak)r for some integer r r rrr.
    Since gcd ( k , n ) = 1 gcd ( k , n ) = 1 gcd(k,n)=1\gcd(k, n) = 1gcd(k,n)=1, there exist integers p p ppp and q q qqq such that p k + q n = 1 p k + q n = 1 pk+qn=1pk + qn = 1pk+qn=1. For any a m a m a^(m)a^mam in G G GGG, we have:
    a m = a m p k = ( a k ) m p = ( a k ) r a m = a m p k = ( a k ) m p = ( a k ) r a^(m)=a^(mpk)=(a^(k))^(mp)=(a^(k))^(r)a^m = a^{mpk} = (a^k)^{mp} = (a^k)^ram=ampk=(ak)mp=(ak)r
    where r = m p r = m p r=mpr = mpr=mp. This shows that a k a k a^(k)a^kak generates G G GGG.
  2. Counting Generators: The number of generators of G G GGG is the same as the number of integers k k kkk such that 0 < k < n 0 < k < n 0 < k < n0 < k < n0<k<n and gcd ( k , n ) = 1 gcd ( k , n ) = 1 gcd(k,n)=1\gcd(k, n) = 1gcd(k,n)=1. This is precisely ϕ ( n ) ϕ ( n ) phi(n)\phi(n)ϕ(n).
Conclusion
We have proven that a finite cyclic group G G GGG of order n n nnn has ϕ ( n ) ϕ ( n ) phi(n)\phi(n)ϕ(n) generators. The proof relies on the properties of Euler’s ϕ ϕ phi\phiϕ-function and the definition of a cyclic group.
Verified Answer
5/5
\(2\:cos\:\theta \:sin\:\phi =sin\:\left(\theta +\phi \right)-sin\:\left(\theta -\phi \right)\)
\(\operatorname{cosec}^2 \theta=1+\cot ^2 \theta\)

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 R f : R R f:R rarr Rf: R \rightarrow Rf:RR defined by f ( a ) = a p f ( a ) = a p f(a)=a^(p)f(a) = a^pf(a)=ap for all a R a R a in Ra \in RaR is an isomorphism, where R R RRR is a finite field of characteristic p > 0 p > 0 p > 0p > 0p>0.
Definitions
  1. Field: A set R R RRR with two operations + + +++ and × × xx\times× that satisfy the field axioms.
  2. Characteristic: A field R R RRR has characteristic p p ppp if p p ppp is the smallest positive integer such that p a = 0 p a = 0 p*a=0p \cdot a = 0pa=0 for all a R a R a in Ra \in RaR. If no such p p ppp exists, the characteristic is zero.
  3. Isomorphism: A bijective map f : R R f : R R f:R rarr Rf: R \rightarrow Rf:RR that preserves the field operations.
Properties Needed
  1. Frobenius Endomorphism: In a field of characteristic p p ppp, ( a + b ) p = a p + b p ( a + b ) p = a p + b p (a+b)^(p)=a^(p)+b^(p)(a+b)^p = a^p + b^p(a+b)p=ap+bp and ( a b ) p = a p b p ( a b ) p = a p b p (ab)^(p)=a^(p)b^(p)(ab)^p = a^p b^p(ab)p=apbp.
Proof
To prove that f f fff is an isomorphism, we need to show that f f fff is a bijective map that preserves addition and multiplication.
  1. f f fff is a Homomorphism
  2. Preservation of Addition:
    f ( a + b ) = ( a + b ) p = a p + b p = f ( a ) + f ( b ) f ( a + b ) = ( a + b ) p = a p + b p = f ( a ) + f ( b ) f(a+b)=(a+b)^(p)=a^(p)+b^(p)=f(a)+f(b)f(a+b) = (a+b)^p = a^p + b^p = f(a) + f(b)f(a+b)=(a+b)p=ap+bp=f(a)+f(b)
  3. Preservation of Multiplication:
    f ( a b ) = ( a b ) p = a p b p = f ( a ) f ( b ) f ( a b ) = ( a b ) p = a p b p = f ( a ) f ( b ) f(ab)=(ab)^(p)=a^(p)b^(p)=f(a)f(b)f(ab) = (ab)^p = a^p b^p = f(a) f(b)f(ab)=(ab)p=apbp=f(a)f(b)
  4. f f fff is Injective (One-to-One)
Suppose f ( a ) = f ( b ) f ( a ) = f ( b ) f(a)=f(b)f(a) = f(b)f(a)=f(b). Then a p = b p a p = b p a^(p)=b^(p)a^p = b^pap=bp which implies a p b p = 0 a p b p = 0 a^(p)-b^(p)=0a^p – b^p = 0apbp=0. Since R R RRR is a field, it has no zero divisors, and we can factor a p b p a p b p a^(p)-b^(p)a^p – b^papbp as ( a b ) p ( a b ) p (a-b)^(p)(a-b)^p(ab)p. This means a b = 0 a b = 0 a-b=0a-b = 0ab=0 or a = b a = b a=ba = ba=b, proving that f f fff is injective.
  1. f f fff is Surjective (Onto)
Since R R RRR is finite and f f fff is injective, f f fff must also be surjective. Alternatively, for any b R b R b in Rb \in RbR, b = b p 2 b = b p 2 b=b^(p^(2))b = b^{p^2}b=bp2 (by Fermat’s Little Theorem or the fact that R R R^(**)R^*R, the multiplicative group of R R RRR, has order p n 1 p n 1 p^(n)-1p^n – 1pn1). Thus, f ( b p 1 ) = ( b p 1 ) p = b p = b f ( b p 1 ) = ( b p 1 ) p = b p = b f(b^(p-1))=(b^(p-1))^(p)=b^(p)=bf(b^{p-1}) = (b^{p-1})^p = b^p = bf(bp1)=(bp1)p=bp=b, showing that f f fff is surjective.
Conclusion
We have shown that f f fff preserves both addition and multiplication, and is both injective and surjective. Therefore, f : R R f : R R f:R rarr Rf: R \rightarrow Rf:RR defined by f ( a ) = a p f ( a ) = a p f(a)=a^(p)f(a) = a^pf(a)=ap is an isomorphism.
Verified Answer
5/5
\(cot\:\theta =\frac{cos\:\theta }{sin\:\theta }\)

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