2017

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UPSC Algebra

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

1(b) Let \(G\) be a group of order \(n\). Show that \(G\) is isomorphic to a subgroup of the permutation group \(S_n\).

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To show that a group G G GGG of order n n nnn is isomorphic to a subgroup of the symmetric group S n S n S_(n)S_nSn, we can use Cayley’s Theorem. Cayley’s Theorem states that every group G G GGG is isomorphic to a subgroup of the symmetric group acting on G G GGG.
Proof:
  1. Define the Action: Consider the action of G G GGG on itself by left multiplication. For each g G g G g in Gg \in GgG, define a function f g : G G f g : G G f_(g):G rarr Gf_g: G \rightarrow Gfg:GG by f g ( x ) = g x f g ( x ) = g x f_(g)(x)=gxf_g(x) = gxfg(x)=gx for all x G x G x in Gx \in GxG.
  2. Show that f g f g f_(g)f_gfg is a Permutation:
    • Injective: Assume f g ( x ) = f g ( y ) f g ( x ) = f g ( y ) f_(g)(x)=f_(g)(y)f_g(x) = f_g(y)fg(x)=fg(y) for some x , y G x , y G x,y in Gx, y \in Gx,yG. Then g x = g y g x = g y gx=gygx = gygx=gy. Since G G GGG is a group, every element has an inverse, so we can multiply both sides on the left by g 1 g 1 g^(-1)g^{-1}g1 to get x = y x = y x=yx = yx=y. This shows that f g f g f_(g)f_gfg is injective.
    • Surjective: For any y G y G y in Gy \in GyG, let x = g 1 y x = g 1 y x=g^(-1)yx = g^{-1}yx=g1y. Then f g ( x ) = g x = g ( g 1 y ) = y f g ( x ) = g x = g ( g 1 y ) = y f_(g)(x)=gx=g(g^(-1)y)=yf_g(x) = gx = g(g^{-1}y) = yfg(x)=gx=g(g1y)=y. This shows that f g f g f_(g)f_gfg is surjective.
    Since f g f g f_(g)f_gfg is both injective and surjective, it is a permutation of G G GGG.
  3. Define a Homomorphism: Define a function Φ : G S n Φ : G S n Phi:G rarrS_(n)\Phi: G \rightarrow S_nΦ:GSn by Φ ( g ) = f g Φ ( g ) = f g Phi(g)=f_(g)\Phi(g) = f_gΦ(g)=fg. We need to show that Φ Φ Phi\PhiΦ is a homomorphism, i.e., Φ ( g h ) = Φ ( g ) Φ ( h ) Φ ( g h ) = Φ ( g ) Φ ( h ) Phi(gh)=Phi(g)Phi(h)\Phi(gh) = \Phi(g)\Phi(h)Φ(gh)=Φ(g)Φ(h) for all g , h G g , h G g,h in Gg, h \in Gg,hG.
    For any x G x G x in Gx \in GxG,
    Φ ( g h ) ( x ) = f g h ( x ) = ( g h ) x = g ( h x ) = f g ( f h ( x ) ) = Φ ( g ) ( Φ ( h ) ( x ) ) Φ ( g h ) ( x ) = f g h ( x ) = ( g h ) x = g ( h x ) = f g ( f h ( x ) ) = Φ ( g ) ( Φ ( h ) ( x ) ) Phi(gh)(x)=f_(gh)(x)=(gh)x=g(hx)=f_(g)(f_(h)(x))=Phi(g)(Phi(h)(x))\Phi(gh)(x) = f_{gh}(x) = (gh)x = g(hx) = f_g(f_h(x)) = \Phi(g)(\Phi(h)(x))Φ(gh)(x)=fgh(x)=(gh)x=g(hx)=fg(fh(x))=Φ(g)(Φ(h)(x))
    This shows that Φ ( g h ) = Φ ( g ) Φ ( h ) Φ ( g h ) = Φ ( g ) Φ ( h ) Phi(gh)=Phi(g)Phi(h)\Phi(gh) = \Phi(g)\Phi(h)Φ(gh)=Φ(g)Φ(h), so Φ Φ Phi\PhiΦ is a homomorphism.
  4. Show that Φ Φ Phi\PhiΦ is Injective:
    Assume Φ ( g ) = Φ ( h ) Φ ( g ) = Φ ( h ) Phi(g)=Phi(h)\Phi(g) = \Phi(h)Φ(g)=Φ(h) for some g , h G g , h G g,h in Gg, h \in Gg,hG. Then f g = f h f g = f h f_(g)=f_(h)f_g = f_hfg=fh, which means g x = h x g x = h x gx=hxgx = hxgx=hx for all x G x G x in Gx \in GxG. Letting x = g 1 x = g 1 x=g^(-1)x = g^{-1}x=g1, we get g ( g 1 ) = h ( g 1 ) g ( g 1 ) = h ( g 1 ) g(g^(-1))=h(g^(-1))g(g^{-1}) = h(g^{-1})g(g1)=h(g1), which simplifies to e = h g 1 e = h g 1 e=hg^(-1)e = hg^{-1}e=hg1, where e e eee is the identity element in G G GGG. Multiplying both sides on the right by g g ggg, we get g = h g = h g=hg = hg=h. This shows that Φ Φ Phi\PhiΦ is injective.
  5. Conclusion: Since Φ Φ Phi\PhiΦ is an injective homomorphism from G G GGG to S n S n S_(n)S_nSn, the image of Φ Φ Phi\PhiΦ is a subgroup of S n S n S_(n)S_nSn that is isomorphic to G G GGG. This completes the proof.
Verified Answer
5/5
\(c=a\:cos\:B+b\:cos\:A\)
\(cos\:2\theta =2\:cos^2\theta -1\)

2(c) Let \(F\) be a field and \(F[X]\) denote the ring of polynomials over \(F\) in a single variable \(X\). For \(f(X), g(X) \in F[X]\) with \(g(X) \neq 0\), show that there exist \(q(X), r(X) \in F[X]\) such that degree \((r(X))<\text{degree}(g(X))\) and \(f(X)=q(X) \cdot g(X)+r(X)\).

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To show that for any two polynomials f ( X ) , g ( X ) F [ X ] f ( X ) , g ( X ) F [ X ] f(X),g(X)in F[X]f(X), g(X) \in F[X]f(X),g(X)F[X] with g ( X ) 0 g ( X ) 0 g(X)!=0g(X) \neq 0g(X)0, there exist q ( X ) , r ( X ) F [ X ] q ( X ) , r ( X ) F [ X ] q(X),r(X)in F[X]q(X), r(X) \in F[X]q(X),r(X)F[X] such that degree ( r ( X ) ) < degree ( g ( X ) ) degree ( r ( X ) ) < degree ( g ( X ) ) “degree”(r(X)) < “degree”(g(X))\text{degree}(r(X)) < \text{degree}(g(X))degree(r(X))<degree(g(X)) and f ( X ) = q ( X ) g ( X ) + r ( X ) f ( X ) = q ( X ) g ( X ) + r ( X ) f(X)=q(X)*g(X)+r(X)f(X) = q(X) \cdot g(X) + r(X)f(X)=q(X)g(X)+r(X), we can use the division algorithm for polynomials. The proof is constructive and provides a method to find q ( X ) q ( X ) q(X)q(X)q(X) and r ( X ) r ( X ) r(X)r(X)r(X).
Proof:
  1. Base Case: If degree ( f ( X ) ) < degree ( g ( X ) ) degree ( f ( X ) ) < degree ( g ( X ) ) “degree”(f(X)) < “degree”(g(X))\text{degree}(f(X)) < \text{degree}(g(X))degree(f(X))<degree(g(X)), then we can directly take q ( X ) = 0 q ( X ) = 0 q(X)=0q(X) = 0q(X)=0 and r ( X ) = f ( X ) r ( X ) = f ( X ) r(X)=f(X)r(X) = f(X)r(X)=f(X). In this case, the condition degree ( r ( X ) ) < degree ( g ( X ) ) degree ( r ( X ) ) < degree ( g ( X ) ) “degree”(r(X)) < “degree”(g(X))\text{degree}(r(X)) < \text{degree}(g(X))degree(r(X))<degree(g(X)) is satisfied.
  2. Inductive Step: Assume that degree ( f ( X ) ) degree ( g ( X ) ) degree ( f ( X ) ) degree ( g ( X ) ) “degree”(f(X)) >= “degree”(g(X))\text{degree}(f(X)) \geq \text{degree}(g(X))degree(f(X))degree(g(X)). Let n = degree ( f ( X ) ) n = degree ( f ( X ) ) n=”degree”(f(X))n = \text{degree}(f(X))n=degree(f(X)) and m = degree ( g ( X ) ) m = degree ( g ( X ) ) m=”degree”(g(X))m = \text{degree}(g(X))m=degree(g(X)). Write the polynomials in the form:
    f ( X ) = a n X n + a n 1 X n 1 + + a 1 X + a 0 f ( X ) = a n X n + a n 1 X n 1 + + a 1 X + a 0 f(X)=a_(n)X^(n)+a_(n-1)X^(n-1)+dots+a_(1)X+a_(0)f(X) = a_nX^n + a_{n-1}X^{n-1} + \ldots + a_1X + a_0f(X)=anXn+an1Xn1++a1X+a0
    g ( X ) = b m X m + b m 1 X m 1 + + b 1 X + b 0 g ( X ) = b m X m + b m 1 X m 1 + + b 1 X + b 0 g(X)=b_(m)X^(m)+b_(m-1)X^(m-1)+dots+b_(1)X+b_(0)g(X) = b_mX^m + b_{m-1}X^{m-1} + \ldots + b_1X + b_0g(X)=bmXm+bm1Xm1++b1X+b0
    where a n , b m 0 a n , b m 0 a_(n),b_(m)!=0a_n, b_m \neq 0an,bm0.
  3. Construct the Quotient: Consider the polynomial h ( X ) = a n b m X n m g ( X ) h ( X ) = a n b m X n m g ( X ) h(X)=(a_(n))/(b_(m))X^(n-m)*g(X)h(X) = \frac{a_n}{b_m}X^{n-m} \cdot g(X)h(X)=anbmXnmg(X). The leading term of h ( X ) h ( X ) h(X)h(X)h(X) will cancel out the leading term of f ( X ) f ( X ) f(X)f(X)f(X) when subtracted. Define a new polynomial f 1 ( X ) = f ( X ) h ( X ) f 1 ( X ) = f ( X ) h ( X ) f_(1)(X)=f(X)-h(X)f_1(X) = f(X) – h(X)f1(X)=f(X)h(X).
  4. Recursive Application: Now, degree ( f 1 ( X ) ) < degree ( f ( X ) ) degree ( f 1 ( X ) ) < degree ( f ( X ) ) “degree”(f_(1)(X)) < “degree”(f(X))\text{degree}(f_1(X)) < \text{degree}(f(X))degree(f1(X))<degree(f(X)). Apply the division algorithm to f 1 ( X ) f 1 ( X ) f_(1)(X)f_1(X)f1(X) and g ( X ) g ( X ) g(X)g(X)g(X). By induction, there exist q 1 ( X ) , r ( X ) F [ X ] q 1 ( X ) , r ( X ) F [ X ] q_(1)(X),r(X)in F[X]q_1(X), r(X) \in F[X]q1(X),r(X)F[X] such that degree ( r ( X ) ) < degree ( g ( X ) ) degree ( r ( X ) ) < degree ( g ( X ) ) “degree”(r(X)) < “degree”(g(X))\text{degree}(r(X)) < \text{degree}(g(X))degree(r(X))<degree(g(X)) and f 1 ( X ) = q 1 ( X ) g ( X ) + r ( X ) f 1 ( X ) = q 1 ( X ) g ( X ) + r ( X ) f_(1)(X)=q_(1)(X)*g(X)+r(X)f_1(X) = q_1(X) \cdot g(X) + r(X)f1(X)=q1(X)g(X)+r(X).
  5. Combine Results: Now, we can express f ( X ) f ( X ) f(X)f(X)f(X) as:
    f ( X ) = h ( X ) + f 1 ( X ) = h ( X ) + q 1 ( X ) g ( X ) + r ( X ) = ( h ( X ) + q 1 ( X ) ) g ( X ) + r ( X ) f ( X ) = h ( X ) + f 1 ( X ) = h ( X ) + q 1 ( X ) g ( X ) + r ( X ) = ( h ( X ) + q 1 ( X ) ) g ( X ) + r ( X ) f(X)=h(X)+f_(1)(X)=h(X)+q_(1)(X)*g(X)+r(X)=(h(X)+q_(1)(X))*g(X)+r(X)f(X) = h(X) + f_1(X) = h(X) + q_1(X) \cdot g(X) + r(X) = (h(X) + q_1(X)) \cdot g(X) + r(X)f(X)=h(X)+f1(X)=h(X)+q1(X)g(X)+r(X)=(h(X)+q1(X))g(X)+r(X)
    Let q ( X ) = h ( X ) + q 1 ( X ) q ( X ) = h ( X ) + q 1 ( X ) q(X)=h(X)+q_(1)(X)q(X) = h(X) + q_1(X)q(X)=h(X)+q1(X). Then f ( X ) = q ( X ) g ( X ) + r ( X ) f ( X ) = q ( X ) g ( X ) + r ( X ) f(X)=q(X)*g(X)+r(X)f(X) = q(X) \cdot g(X) + r(X)f(X)=q(X)g(X)+r(X) with degree ( r ( X ) ) < degree ( g ( X ) ) degree ( r ( X ) ) < degree ( g ( X ) ) “degree”(r(X)) < “degree”(g(X))\text{degree}(r(X)) < \text{degree}(g(X))degree(r(X))<degree(g(X)).
  6. Conclusion: We have shown that for any f ( X ) , g ( X ) F [ X ] f ( X ) , g ( X ) F [ X ] f(X),g(X)in F[X]f(X), g(X) \in F[X]f(X),g(X)F[X] with g ( X ) 0 g ( X ) 0 g(X)!=0g(X) \neq 0g(X)0, there exist q ( X ) , r ( X ) F [ X ] q ( X ) , r ( X ) F [ X ] q(X),r(X)in F[X]q(X), r(X) \in F[X]q(X),r(X)F[X] such that degree ( r ( X ) ) < degree ( g ( X ) ) degree ( r ( X ) ) < degree ( g ( X ) ) “degree”(r(X)) < “degree”(g(X))\text{degree}(r(X)) < \text{degree}(g(X))degree(r(X))<degree(g(X)) and f ( X ) = q ( X ) g ( X ) + r ( X ) f ( X ) = q ( X ) g ( X ) + r ( X ) f(X)=q(X)*g(X)+r(X)f(X) = q(X) \cdot g(X) + r(X)f(X)=q(X)g(X)+r(X). This completes the proof.
Verified Answer
5/5
\(a=b\:cos\:C+c\:cos\:B\)
\(cos\:2\theta =cos^2\theta -sin^2\theta\)

3(a) Show that the groups \(\mathbb{Z}_5 \times \mathbb{Z}_7\) and \(\mathbb{Z}_{35}\) are isomorphic.

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To show that the groups Z 5 × Z 7 Z 5 × Z 7 Z_(5)xxZ_(7)\mathbb{Z}_5 \times \mathbb{Z}_7Z5×Z7 and Z 35 Z 35 Z_(35)\mathbb{Z}_{35}Z35 are isomorphic, we need to find a bijective function f : Z 5 × Z 7 Z 35 f : Z 5 × Z 7 Z 35 f:Z_(5)xxZ_(7)rarrZ_(35)f: \mathbb{Z}_5 \times \mathbb{Z}_7 \rightarrow \mathbb{Z}_{35}f:Z5×Z7Z35 such that f ( a + b ) = f ( a ) + f ( b ) f ( a + b ) = f ( a ) + f ( b ) f(a+b)=f(a)+f(b)f(a+b) = f(a) + f(b)f(a+b)=f(a)+f(b) for all a , b Z 5 × Z 7 a , b Z 5 × Z 7 a,b inZ_(5)xxZ_(7)a, b \in \mathbb{Z}_5 \times \mathbb{Z}_7a,bZ5×Z7.
Step 1: Define the Function f f fff
Let’s define the function f : Z 5 × Z 7 Z 35 f : Z 5 × Z 7 Z 35 f:Z_(5)xxZ_(7)rarrZ_(35)f: \mathbb{Z}_5 \times \mathbb{Z}_7 \rightarrow \mathbb{Z}_{35}f:Z5×Z7Z35 as follows:
f ( a , b ) = 7 a + 5 b f ( a , b ) = 7 a + 5 b f(a,b)=7a+5bf(a, b) = 7a + 5bf(a,b)=7a+5b
Step 2: Show that f f fff is Well-Defined
To show that f f fff is well-defined, we need to show that f ( a , b ) f ( a , b ) f(a,b)f(a, b)f(a,b) is in Z 35 Z 35 Z_(35)\mathbb{Z}_{35}Z35 for all ( a , b ) Z 5 × Z 7 ( a , b ) Z 5 × Z 7 (a,b)inZ_(5)xxZ_(7)(a, b) \in \mathbb{Z}_5 \times \mathbb{Z}_7(a,b)Z5×Z7.
Let’s substitute the values:
  • a a aaa can be 0 , 1 , 2 , 3 , 4 0 , 1 , 2 , 3 , 4 0,1,2,3,40, 1, 2, 3, 40,1,2,3,4 (from Z 5 Z 5 Z_(5)\mathbb{Z}_5Z5)
  • b b bbb can be 0 , 1 , 2 , 3 , 4 , 5 , 6 0 , 1 , 2 , 3 , 4 , 5 , 6 0,1,2,3,4,5,60, 1, 2, 3, 4, 5, 60,1,2,3,4,5,6 (from Z 7 Z 7 Z_(7)\mathbb{Z}_7Z7)
After substituting these values into f ( a , b ) = 7 a + 5 b f ( a , b ) = 7 a + 5 b f(a,b)=7a+5bf(a, b) = 7a + 5bf(a,b)=7a+5b, we find that f ( a , b ) f ( a , b ) f(a,b)f(a, b)f(a,b) can range from 0 0 000 to 34 34 343434, which is exactly the set Z 35 Z 35 Z_(35)\mathbb{Z}_{35}Z35.
Step 3: Show that f f fff is a Homomorphism
To show that f f fff is a homomorphism, we need to show that f ( a 1 + a 2 , b 1 + b 2 ) = f ( a 1 , b 1 ) + f ( a 2 , b 2 ) f ( a 1 + a 2 , b 1 + b 2 ) = f ( a 1 , b 1 ) + f ( a 2 , b 2 ) f(a_(1)+a_(2),b_(1)+b_(2))=f(a_(1),b_(1))+f(a_(2),b_(2))f(a_1 + a_2, b_1 + b_2) = f(a_1, b_1) + f(a_2, b_2)f(a1+a2,b1+b2)=f(a1,b1)+f(a2,b2).
Let’s substitute the values:
f ( a 1 + a 2 , b 1 + b 2 ) = 7 ( a 1 + a 2 ) + 5 ( b 1 + b 2 ) f ( a 1 + a 2 , b 1 + b 2 ) = 7 ( a 1 + a 2 ) + 5 ( b 1 + b 2 ) f(a_(1)+a_(2),b_(1)+b_(2))=7(a_(1)+a_(2))+5(b_(1)+b_(2))f(a_1 + a_2, b_1 + b_2) = 7(a_1 + a_2) + 5(b_1 + b_2)f(a1+a2,b1+b2)=7(a1+a2)+5(b1+b2)
After calculating, we get:
f ( a 1 + a 2 , b 1 + b 2 ) = 7 a 1 + 7 a 2 + 5 b 1 + 5 b 2 = f ( a 1 , b 1 ) + f ( a 2 , b 2 ) f ( a 1 + a 2 , b 1 + b 2 ) = 7 a 1 + 7 a 2 + 5 b 1 + 5 b 2 = f ( a 1 , b 1 ) + f ( a 2 , b 2 ) f(a_(1)+a_(2),b_(1)+b_(2))=7a_(1)+7a_(2)+5b_(1)+5b_(2)=f(a_(1),b_(1))+f(a_(2),b_(2))f(a_1 + a_2, b_1 + b_2) = 7a_1 + 7a_2 + 5b_1 + 5b_2 = f(a_1, b_1) + f(a_2, b_2)