2010

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UPSC Algebra

\(2\:sin\:\theta \:sin\:\phi =-cos\:\left(\theta +\phi \right)+cos\:\left(\theta -\phi \right)\)

1(a) Let \(G=\mathbb{I}-\{-1\}\) be the set of all real numbers omitting \(-1\). Define the binary relation \(*\) on \(G\) by \(a * b=a+b+a b\). Show \((G, *)\) is a group and it is abelian.

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To show that ( G , ) ( G , ) (G,**)(G, *)(G,) is a group, we need to verify the following group axioms:
  1. Closure under ***
For any a , b G a , b G a,b in Ga, b \in Ga,bG, a b = a + b + a b a b = a + b + a b a**b=a+b+aba * b = a + b + abab=a+b+ab is also a real number. However, we need to ensure that a b 1 a b 1 a**b!=-1a * b \neq -1ab1.
If a b = 1 a b = 1 a**b=-1a * b = -1ab=1, then a + b + a b = 1 a + b + a b = 1 a+b+ab=-1a + b + ab = -1a+b+ab=1, which simplifies to a b + a + b + 1 = 0 a b + a + b + 1 = 0 ab+a+b+1=0ab + a + b + 1 = 0ab+a+b+1=0 or ( a + 1 ) ( b + 1 ) = 0 ( a + 1 ) ( b + 1 ) = 0 (a+1)(b+1)=0(a + 1)(b + 1) = 0(a+1)(b+1)=0.
Since a , b G a , b G a,b in Ga, b \in Ga,bG, a 1 a 1 a!=-1a \neq -1a1 and b 1 b 1 b!=-1b \neq -1b1, so a + 1 0 a + 1 0 a+1!=0a + 1 \neq 0a+10 and b + 1 0 b + 1 0 b+1!=0b + 1 \neq 0b+10. Therefore, a b 1 a b 1 a**b!=-1a * b \neq -1ab1, and G G GGG is closed under ***.
  1. Associativity
For any a , b , c G a , b , c G a,b,c in Ga, b, c \in Ga,b,cG,
( a b ) c = ( a + b + a b ) c = a + b + a b + c + ( a + b + a b ) c = a + b + c + a b + a c + b c + a b c ( a b ) c = ( a + b + a b ) c = a + b + a b + c + ( a + b + a b ) c = a + b + c + a b + a c + b c + a b c (a**b)**c=(a+b+ab)**c=a+b+ab+c+(a+b+ab)c=a+b+c+ab+ac+bc+abc(a * b) * c = (a + b + ab) * c = a + b + ab + c + (a + b + ab)c = a + b + c + ab + ac + bc + abc(ab)c=(a+b+ab)c=a+b+ab+c+(a+b+ab)c=a+b+c+ab+ac+bc+abc
a ( b c ) = a ( b + c + b c ) = a + b + c + b c + a ( b + c + b c ) = a + b + c + a b + a c + b c + a b c a ( b c ) = a ( b + c + b c ) = a + b + c + b c + a ( b + c + b c ) = a + b + c + a b + a c + b c + a b c a**(b**c)=a**(b+c+bc)=a+b+c+bc+a(b+c+bc)=a+b+c+ab+ac+bc+abca * (b * c) = a * (b + c + bc) = a + b + c + bc + a(b + c + bc) = a + b + c + ab + ac + bc + abca(bc)=a(b+c+bc)=a+b+c+bc+a(b+c+bc)=a+b+c+ab+ac+bc+abc
Since ( a b ) c = a ( b c ) ( a b ) c = a ( b c ) (a**b)**c=a**(b**c)(a * b) * c = a * (b * c)(ab)c=a(bc), the operation *** is associative.
  1. Existence of Identity Element
We need to find an element e G e G e in Ge \in GeG such that for any a G a G a in Ga \in GaG, a e = a a e = a a**e=aa * e = aae=a.
Let a e = a a e = a a**e=aa * e = aae=a, then a + e + a e = a a + e + a e = a a+e+ae=aa + e + ae = aa+e+ae=a, which simplifies to e + a e = 0 e + a e = 0 e+ae=0e + ae = 0e+ae=0 or e ( 1 + a ) = 0 e ( 1 + a ) = 0 e(1+a)=0e(1 + a) = 0e(1+a)=0.
Since a 1 a 1 a!=-1a \neq -1a1, 1 + a 0 1 + a 0 1+a!=01 + a \neq 01+a0, and therefore e = 0 e = 0 e=0e = 0e=0.
  1. Existence of Inverse Element
For each a G a G a in Ga \in GaG, we need to find an element a 1 G a 1 G a^(-1)in Ga^{-1} \in Ga1G such that a a 1 = e a a 1 = e a**a^(-1)=ea * a^{-1} = eaa1=e.
Let a a 1 = e a a 1 = e a**a^(-1)=ea * a^{-1} = eaa1=e, then a + a 1 + a a 1 = 0 a + a 1 + a a 1 = 0 a+a^(-1)+aa^(-1)=0a + a^{-1} + aa^{-1} = 0a+a1+aa1=0, which simplifies to a 1 ( 1 + a ) = a a 1 ( 1 + a ) = a a^(-1)(1+a)=-aa^{-1}(1 + a) = -aa1(1+a)=a.
Since a 1 a 1 a!=-1a \neq -1a1, 1 + a 0 1 + a 0 1+a!=01 + a \neq 01+a0, and therefore a 1 = a a + 1 a 1 = a a + 1 a^(-1)=-(a)/(a+1)a^{-1} = -\frac{a}{a + 1}a1=aa+1.
  1. Commutativity
For any a , b G a , b G a,b in Ga, b \in Ga,bG,
a b = a + b + a b = b + a + b a = b a a b = a + b + a b = b + a + b a = b a a**b=a+b+ab=b+a+ba=b**aa * b = a + b + ab = b + a + ba = b * aab=a+b+ab=b+a+ba=ba
Since a b = b a a b = b a a**b=b**aa * b = b * aab=ba, ( G , ) ( G , ) (G,**)(G, *)(G,) is abelian.
Conclusion
Since all these properties are satisfied, ( G , ) ( G , ) (G,**)(G, *)(G,) is an abelian group.
Verified Answer
5/5
\(cos\:2\theta =1-2\:sin^2\theta \)
\(cot\:\theta =\frac{cos\:\theta }{sin\:\theta }\)

1(b) Show that a cyclic group of order 6 is isomorphic to the product of a cyclic group of order 2 and a cyclic group of order 3. Can you generalize this? Justify.

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
Step 1: Define the Groups
Let G G GGG be a cyclic group of order 6, generated by an element a a aaa, i.e., G = a = { e , a , a 2 , a 3 , a 4 , a 5 } G = a = { e , a , a 2 , a 3 , a 4 , a 5 } G=(:a:)={e,a,a^(2),a^(3),a^(4),a^(5)}G = \langle a \rangle = \{ e, a, a^2, a^3, a^4, a^5 \}G=a={e,a,a2,a3,a4,a5}.
Let H H HHH be a cyclic group of order 2, generated by an element b b bbb, i.e., H = b = { e , b } H = b = { e , b } H=(:b:)={e,b}H = \langle b \rangle = \{ e, b \}H=b={e,b}.
Let K K KKK be a cyclic group of order 3, generated by an element c c ccc, i.e., K = c = { e , c , c 2 } K = c = { e , c , c 2 } K=(:c:)={e,c,c^(2)}K = \langle c \rangle = \{ e, c, c^2 \}K=c={e,c,c2}.
Step 2: Define the Product Group
Consider the product group H × K H × K H xx KH \times KH×K, consisting of elements ( b i , c j ) ( b i , c j ) (b^(i),c^(j))(b^i, c^j)(bi,cj) where 0 i < 2 0 i < 2 0 <= i < 20 \leq i < 20i<2 and 0 j < 3 0 j < 3 0 <= j < 30 \leq j < 30j<3.
Step 3: Define the Isomorphism
Define a function ϕ : G H × K ϕ : G H × K phi:G rarr H xx K\phi: G \to H \times Kϕ:GH×K by ϕ ( a n ) = ( b n mod 2 , c n mod 3 ) ϕ ( a n ) = ( b n mod 2 , c n mod 3 ) phi(a^(n))=(b^(nmod2),c^(nmod3))\phi(a^n) = (b^{n \mod 2}, c^{n \mod 3})ϕ(an)=(bnmod2,cnmod3).
Step 4: Show ϕ ϕ phi\phiϕ is an Isomorphism
  1. Well-defined: ϕ ϕ phi\phiϕ is well-defined because n mod 2 n mod 2 nmod2n \mod 2nmod2 and n mod 3 n mod 3 nmod3n \mod 3nmod3 are always in the correct ranges for H H HHH and K K KKK, respectively.
  2. Bijective: ϕ ϕ phi\phiϕ is bijective because each element a n a n a^(n)a^nan in G G GGG maps to a unique element ( b n mod 2 , c n mod 3 ) ( b n mod 2 , c n mod 3 ) (b^(nmod2),c^(nmod3))(b^{n \mod 2}, c^{n \mod 3})(bnmod2,cnmod3) in H × K H × K H xx KH \times KH×K, and vice versa.
  3. Homomorphism: ϕ ϕ phi\phiϕ is a homomorphism because for any a m , a n G a m , a n G a^(m),a^(n)in Ga^m, a^n \in Gam,anG,
    ϕ ( a m a n ) = ϕ ( a m + n ) = ( b ( m + n ) mod 2 , c ( m + n ) mod 3 ) = ( b m mod 2 b n mod 2 , c m mod 3 c n mod 3 ) = ϕ ( a m ) ϕ ( a n ) ϕ ( a m a n ) = ϕ ( a m + n ) = ( b ( m + n ) mod 2 , c ( m + n ) mod 3 ) = ( b m mod 2 b n mod 2 , c m mod 3 c n mod 3 ) = ϕ ( a m ) ϕ ( a n ) phi(a^(m)**a^(n))=phi(a^(m+n))=(b^((m+n)mod2),c^((m+n)mod3))=(b^(mmod2)**b^(nmod2),c^(mmod3)**c^(nmod3))=phi(a^(m))**phi(a^(n))\phi(a^m * a^n) = \phi(a^{m+n}) = (b^{(m+n) \mod 2}, c^{(m+n) \mod 3}) = (b^{m \mod 2} * b^{n \mod 2}, c^{m \mod 3} * c^{n \mod 3}) = \phi(a^m) * \phi(a^n)ϕ(aman)=ϕ(am+n)=(b(m+n)mod2,c(m+n)mod3)=(bmmod2bnmod2,cmmod3cnmod3)=ϕ(am)ϕ(an)
Generalization
This result can be generalized to say that a cyclic group of order m n m n mnmnmn where m m mmm and n n nnn are coprime is isomorphic to the product of a cyclic group of order m m mmm and a cyclic group of order n n nnn.
Justification for Generalization
  1. Chinese Remainder Theorem: If m m mmm and n n nnn are coprime, then the map ϕ : Z m n Z m × Z n ϕ : Z m n Z m × Z n phi:Z_(mn)rarrZ_(m)xxZ_(n)\phi: \mathbb{Z}_{mn} \to \mathbb{Z}_m \times \mathbb{Z}_nϕ:ZmnZm×Zn defined by ϕ ( a ) = ( a mod m , a mod n ) ϕ ( a ) = ( a mod m , a mod n ) phi(a)=(amodm,amodn)\phi(a) = (a \mod m, a \mod n)ϕ(a)=(amodm,amodn) is an isomorphism.
  2. Cyclic Groups: A cyclic group of order m n m n mnmnmn is isomorphic to Z m n Z m n Z_(mn)\mathbb{Z}_{mn}Zmn, and cyclic groups of orders m m mmm and n n nnn are isomorphic to Z m Z m Z_(m)\mathbb{Z}_mZm and Z n Z n Z_(n)\mathbb{Z}_nZn, respectively.
  3. Isomorphism: Combining these, we get that a cyclic group of order m n m n mnmnmn is isomorphic to the product of a cyclic group of order m m mmm and a cyclic group of order n n nnn.
Therefore, we can generalize that a cyclic group of order m n m n mnmnmn is isomorphic to the product of a cyclic group of order m m mmm and a cyclic group of order n n nnn when m m mmm and n n nnn are coprime.
Verified Answer
5/5
\(sin\left(2\theta \right)=2\:sin\:\theta \:cos\:\theta \)
\(a=b\:cos\:C+c\:cos\:B\)

2(a) Let \(\left(\mathbb{R}^*, \cdot\right)\) be the multiplicative group of nonzero reals and \((G L(n, I R), X)\) be the multiplicative group of \(n \times n\) non-singular matrices. Show that the quotient group \(G L(n, \mathbb{R}) / S L(n, \mathbb{R})\) and \(\left(\mathbb{R}^* \cdot \cdot\right)\) are isomorphic where \(S L(n, \mathrm{IR})=\{A \in G L(n, \mathrm{IR}) / \operatorname{det} A=1\}\). What is the centre of \(G L(n\), IR ) ?

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
Step 1: Define the Groups and Quotient Group
  • R R R^(**)\mathbb{R}^*R is the set of all nonzero real numbers with multiplication as the operation.
  • G L ( n , R ) G L ( n , R ) GL(n,R)GL(n, \mathbb{R})GL(n,R) is the set of all n × n n × n n xx nn \times nn×n invertible matrices with matrix multiplication as the operation.
  • S L ( n , R ) S L ( n , R ) SL(n,R)SL(n, \mathbb{R})SL(n,R) is the set of all n × n n × n n xx nn \times nn×n matrices with determinant 1, i.e., S L ( n , R ) = { A G L ( n , R ) det ( A ) = 1 } S L ( n , R ) = { A G L ( n , R ) det ( A ) = 1 } SL(n,R)={A in GL(n,R)∣”det”(A)=1}SL(n, \mathbb{R}) = \{ A \in GL(n, \mathbb{R}) \mid \text{det}(A) = 1 \}SL(n,R)={AGL(n,R)det(A)=1}.
The quotient group G L ( n , R ) / S L ( n , R ) G L ( n , R ) / S L ( n , R ) GL(n,R)//SL(n,R)GL(n, \mathbb{R}) / SL(n, \mathbb{R})GL(n,R)/SL(n,R) consists of cosets A S L ( n , R ) A S L ( n , R ) A*SL(n,R)A \cdot SL(n, \mathbb{R})ASL(n,R) where A G L ( n , R ) A G L ( n , R ) A in GL(n,R)A \in GL(n, \mathbb{R})AGL(n,R).
Step 2: Define the Isomorphism
Define a function ϕ : G L ( n , R ) R ϕ : G L ( n , R ) R phi:GL(n,R)rarrR^(**)\phi: GL(n, \mathbb{R}) \to \mathbb{R}^*ϕ:GL(n,R)R by ϕ ( A ) = det ( A ) ϕ ( A ) = det ( A ) phi(A)=”det”(A)\phi(A) = \text{det}(A)ϕ(A)=det(A).
Step 3: Show ϕ ϕ phi\phiϕ Induces an Isomorphism between G L ( n , R ) / S L ( n , R ) G L ( n , R ) / S L ( n , R ) GL(n,R)//SL(n,R)GL(n, \mathbb{R}) / SL(n, \mathbb{R})GL(n,R)/SL(n,R) and R R R^(**)\mathbb{R}^*R
  1. Well-defined: ϕ ϕ phi\phiϕ is well-defined because the determinant of any n × n n × n n xx nn \times nn×n invertible matrix is a nonzero real number.
  2. Homomorphism: ϕ ϕ phi\phiϕ is a homomorphism because det ( A B ) = det ( A ) det ( B ) det ( A B ) = det ( A ) det ( B ) “det”(AB)=”det”(A)”det”(B)\text{det}(AB) = \text{det}(A) \text{det}(B)det(AB)=det(A)det(B).
  3. Kernel: The kernel of ϕ ϕ phi\phiϕ is S L ( n , R ) S L ( n , R ) SL(n,R)SL(n, \mathbb{R})SL(n,R) because ϕ ( A ) = 1 ϕ ( A ) = 1 phi(A)=1\phi(A) = 1ϕ(A)=1 if and only if det ( A ) = 1 det ( A ) = 1 “det”(A)=1\text{det}(A) = 1det(A)=1, i.e., A S L ( n , R ) A S L ( n , R ) A in SL(n,R)A \in SL(n, \mathbb{R})ASL(n,R).
  4. Induced Isomorphism: By the First Isomorphism Theorem, G L ( n , R ) / Ker ( ϕ ) G L ( n , R ) / Ker ( ϕ ) GL(n,R)//”Ker”(phi)GL(n, \mathbb{R}) / \text{Ker}(\phi)GL(n,R)/Ker(ϕ) is isomorphic to the image of ϕ ϕ phi\phiϕ, which is R R R^(**)\mathbb{R}^*R. Therefore, G L ( n , R ) / S L ( n , R ) G L ( n , R ) / S L ( n , R ) GL(n,R)//SL(n,R)GL(n, \mathbb{R}) / SL(n, \mathbb{R})GL(n,R)/SL(n,R) is isomorphic to R R R^(**)\mathbb{R}^*R.
Step 4: Find the Center of G L ( n , R ) G L ( n , R ) GL(n,R)GL(n, \mathbb{R})GL(n,R)
The center Z ( G L ( n , R ) ) Z ( G L ( n , R ) ) Z(GL(n,R))Z(GL(n, \mathbb{R}))Z(GL(n,R)) of G L ( n , R ) G L ( n , R ) GL(n,R)GL(n, \mathbb{R})GL(n,R) consists of all matrices A A AAA such that A B = B A A B = B A AB=BAAB = BAAB=BA for all B G L ( n , R ) B G L ( n , R ) B in GL(n,R)B \in GL(n, \mathbb{R})BGL(n,R).
For n > 1 n > 1 n > 1n > 1n>1, the center consists only of scalar multiples of the identity matrix, i.e., Z ( G L ( n , R ) ) = { c I c R } Z ( G L ( n , R ) ) = { c I c R } Z(GL(n,R))={cI∣c inR^(**)}Z(GL(n, \mathbb{R})) = \{ cI \mid c \in \mathbb{R}^* \}Z(GL(n,R))={cIcR}.
For n = 1 n = 1 n=1n = 1n=1, G L ( 1 , R ) G L ( 1 , R ) GL(1,R)GL(1, \mathbb{R})GL(1,R) is essentially R R R^(**)\mathbb{R}^*R, and every element commutes with every other element, so the center is G L ( 1 , R ) G L ( 1 , R ) GL(1,R)GL(1, \mathbb{R})GL(1,R) itself.
Conclusion
The quotient group G L ( n , R ) / S L ( n , R ) G L ( n , R ) / S L ( n , R ) GL(n,R)//SL(n,R)GL(n, \mathbb{R}) / SL(n, \mathbb{R})GL(n,R)/SL(n,R) is isomorphic to R R R^(**)\mathbb{R}^*R, and the center of G L ( n , R ) G L ( n , R ) GL(n,R)GL(n, \mathbb{R})GL(n,R) is { c I c R } { c I c R } {cI∣c inR^(**)}\{ cI \mid c \in \mathbb{R}^* \}{cIcR} for n > 1 n > 1 n > 1n > 1n>1 and G L ( 1 , R ) G L ( 1 , R ) GL(1,R)GL(1, \mathbb{R})GL(1,R) for n = 1 n = 1 n=1n = 1n=1.
Verified Answer
5/5
\(2\:cos\:\theta \:cos\:\phi =cos\:\left(\theta +\phi \right)+cos\:\left(\theta -\phi \right)\)
\(Sin^2\left(\theta \:\right)+Cos^2\left(\theta \right)=1\)

3(a) Consider the polynomial ring \(Q[x]\). Show \(p(x)=x^3-2\) is irreducible over \(Q\). Let \(I\) be the ideal in \(Q[x]\) generated by \(p(x)\). Then show that \(Q[x] / I\) is a field and that each element of it is of the form \(a_0+a_1 t+a_2 t^2\) with \(a_0, a_1, a_2\) in \(Q\) and \(t=x+I\).

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