2011

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UPSC Algebra

\(tan\:\theta =\frac{sin\:\theta }{cos\:\theta }\)

1(a) Show that the set \[ G=\left\{f_1, f_2, f_3, f_4, f_5, f_6\right\} \] of six transformations on the set of Complex numbers defined by \[ \begin{aligned} &f_1(z)=z, f_2(z)=1-z \\ &f_3(z)=\frac{z}{(z-1)}, f_4(z)=\frac{1}{z} \\ &f_5(z)=\frac{1}{(1-z)} \text { and } f_6(z)=\frac{(z-1)}{z} \end{aligned} \] is a non-abelian group of order 6 w.r.t. composition of mappings.

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To show that the set G = { f 1 , f 2 , f 3 , f 4 , f 5 , f 6 } G = { f 1 , f 2 , f 3 , f 4 , f 5 , f 6 } G={f_(1),f_(2),f_(3),f_(4),f_(5),f_(6)}G = \{ f_1, f_2, f_3, f_4, f_5, f_6 \}G={f1,f2,f3,f4,f5,f6} forms a group under composition of mappings, we need to verify the following group axioms:
  1. Closure
We need to show that the composition of any two functions in G G GGG is also in G G GGG.
For example, let’s consider f 3 f 4 f 3 f 4 f_(3)@f_(4)f_3 \circ f_4f3f4. The composition f 3 ( f 4 ( z ) ) f 3 ( f 4 ( z ) ) f_(3)(f_(4)(z))f_3(f_4(z))f3(f4(z)) is given by:
f 3 ( f 4 ( z ) ) = f 3 ( 1 z ) = 1 z 1 z 1 = 1 1 z f 3 ( f 4 ( z ) ) = f 3 1 z = 1 z 1 z 1 = 1 1 z f_(3)(f_(4)(z))=f_(3)((1)/(z))=((1)/(z))/((1)/(z)-1)=(1)/(1-z)f_3(f_4(z)) = f_3\left(\frac{1}{z}\right) = \frac{\frac{1}{z}}{\frac{1}{z} – 1} = \frac{1}{1 – z}f3(f4(z))=f3(1z)=1z1z1=11z
After calculating, we find that f 3 f 4 = f 5 f 3 f 4 = f 5 f_(3)@f_(4)=f_(5)f_3 \circ f_4 = f_5f3f4=f5, which is in G G GGG.
Similarly, we can show that the composition of any two functions in G G GGG will result in another function in G G GGG. Therefore, G G GGG is closed under composition.
  1. Associativity
Composition of functions is always associative. That is, for any f , g , h G f , g , h G f,g,h in Gf, g, h \in Gf,g,hG, ( f g ) h = f ( g h ) ( f g ) h = f ( g h ) (f@g)@h=f@(g@h)(f \circ g) \circ h = f \circ (g \circ h)(fg)h=f(gh).
  1. Identity Element
The function f 1 ( z ) = z f 1 ( z ) = z f_(1)(z)=zf_1(z) = zf1(z)=z serves as the identity element for G G GGG under composition. For any f G f G f in Gf \in GfG, f 1 f = f f 1 f = f f_(1)@f=ff_1 \circ f = ff1f=f and f f 1 = f f f 1 = f f@f_(1)=ff \circ f_1 = fff1=f.
  1. Inverse Element
Each function in G G GGG must have an inverse that is also in G G GGG.
  • f 1 f 1 f_(1)f_1f1 is its own inverse.
  • f 2 f 2 f_(2)f_2f2 is its own inverse.
  • f 3 f 3 f_(3)f_3f3 and f 6 f 6 f_(6)f_6f6 are inverses of each other.
  • f 4 f 4 f_(4)f_4f4 and f 5 f 5 f_(5)f_5f5 are inverses of each other.
  1. Non-Abelian
To show that G G GGG is non-abelian, we need to find f , g G f , g G f,g in Gf, g \in Gf,gG such that f g g f f g g f f@g!=g@ff \circ g \neq g \circ ffggf.
Let’s consider f 3 f 3 f_(3)f_3f3 and f 4 f 4 f_(4)f_4f4:
  • f 3 f 4 = f 5 f 3 f 4 = f 5 f_(3)@f_(4)=f_(5)f_3 \circ f_4 = f_5f3f4=f5
  • f 4 f 3 = f 6 f 4 f 3 = f 6 f_(4)@f_(3)=f_(6)f_4 \circ f_3 = f_6f4f3=f6
Since f 5 f 6 f 5 f 6 f_(5)!=f_(6)f_5 \neq f_6f5f6, G G GGG is non-abelian.
Conclusion
Since G G GGG satisfies all the group axioms and is non-abelian, it is a non-abelian group of order 6 with respect to composition of mappings.
Verified Answer
5/5
\(tan\:\theta =\frac{sin\:\theta }{cos\:\theta }\)
\(a^2=b^2+c^2-2bc\:Cos\left(A\right)\)

1(e) (i) Prove that a group of Prime order is abelian.

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To prove that a group G G GGG of prime order p p ppp is abelian, we need to show that for all a , b G a , b G a,b in Ga, b \in Ga,bG, a b = b a a b = b a a*b=b*aa \cdot b = b \cdot aab=ba.
Step 1: Group Order is Prime
Let p p ppp be a prime number, and let G G GGG be a group of order p p ppp.
Step 2: Elements in the Group
Since G G GGG has order p p ppp, it has p p ppp elements. One of these elements is the identity element e e eee.
Step 3: Properties of Subgroups
By Lagrange’s theorem, the order of any subgroup of G G GGG must divide the order of G G GGG. Since p p ppp is prime, the only divisors of p p ppp are 1 and p p ppp itself. Therefore, any subgroup of G G GGG must have order 1 or p p ppp.
Step 4: Subgroups are Trivial or the Whole Group
A subgroup of order 1 is trivial and contains only the identity element e e eee. A subgroup of order p p ppp must be G G GGG itself.
Step 5: Every Element Generates the Group
Take any element a G a G a in Ga \in GaG where a e a e a!=ea \neq eae. The subgroup generated by a a aaa, denoted a a (:a:)\langle a \ranglea, must have order greater than 1 (since a e a e a!=ea \neq eae). By Step 4, a a (:a:)\langle a \ranglea must therefore be the whole group G G GGG.
Step 6: Commutativity
Take any two elements a , b G a , b G a,b in Ga, b \in Ga,bG. Both a a aaa and b b bbb generate G G GGG (by Step 5). Therefore, b b bbb can be written as a n a n a^(n)a^nan for some integer n n nnn, and a a aaa can be written as b m b m b^(m)b^mbm for some integer m m mmm.
Now, a b = a a n = a n + 1 a b = a a n = a n + 1 a*b=a*a^(n)=a^(n+1)a \cdot b = a \cdot a^n = a^{n+1}ab=aan=an+1 and b a = a n a = a n + 1 b a = a n a = a n + 1 b*a=a^(n)*a=a^(n+1)b \cdot a = a^n \cdot a = a^{n+1}ba=ana=an+1.
Since a b = b a a b = b a a*b=b*aa \cdot b = b \cdot aab=ba, G G GGG is abelian.
Conclusion
We have shown that any group of prime order p p ppp must be abelian.
Verified Answer
5/5
\(c^2=a^2+b^2-2ab\:Cos\left(C\right)\)
\(2\:cos\:\theta \:sin\:\phi =sin\:\left(\theta +\phi \right)-sin\:\left(\theta -\phi \right)\)

1(e) (ii) How many generators are there of the cyclic group \( (G, .) \) of order 8?

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
In a cyclic group G G GGG of order n n nnn, the number of generators is equal to the number of integers k k kkk that are relatively prime to n n nnn. These integers k k kkk are in the range 1 k < n 1 k < n 1 <= k < n1 \leq k < n1k<n.
Step 1: Identify the Order of the Group
The order of the group G G GGG is given as 8.
Step 2: Find Integers Relatively Prime to 8
We need to find the integers k k kkk that are relatively prime to 8 and lie in the range 1 k < 8 1 k < 8 1 <= k < 81 \leq k < 81k<8.
The integers in this range are { 1 , 2 , 3 , 4 , 5 , 6 , 7 } { 1 , 2 , 3 , 4 , 5 , 6 , 7 } {1,2,3,4,5,6,7}\{1, 2, 3, 4, 5, 6, 7\}{1,2,3,4,5,6,7}.
Out of these, the integers that are relatively prime to 8 are { 1 , 3 , 5 , 7 } { 1 , 3 , 5 , 7 } {1,3,5,7}\{1, 3, 5, 7\}{1,3,5,7}.
Step 3: Count the Number of Generators
The number of generators of G G GGG is equal to the number of integers that are relatively prime to 8, which is 4.
Conclusion
There are 4 generators of the cyclic group G G GGG of order 8.
Verified Answer
5/5
\(sin\left(2\theta \right)=2\:sin\:\theta \:cos\:\theta \)
\(2\:cos\:\theta \:cos\:\phi =cos\:\left(\theta +\phi \right)+cos\:\left(\theta -\phi \right)\)

2(a) Give an example of a group \( G \) in which every proper subgroup is cyclic but the group itself is not cyclic.

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
A classic example of a group G G GGG where every proper subgroup is cyclic but G G GGG itself is not cyclic is the Klein four-group V 4 V 4 V_(4)V_4V4.
Definition of V 4 V 4 V_(4)V_4V4
The Klein four-group V 4 V 4 V_(4)V_4V4 is defined as the set { e , a , b , c } { e , a , b , c } {e,a,b,c}\{ e, a, b, c \}{e,a,b,c} with the following multiplication table:
e a b c e e a b c a a e c b b b c e a c c b a e e a b c e e a b c a a e c b b b c e a c c b a e {:[*,e,a,b,c],[e,e,a,b,c],[a,a,e,c,b],[b,b,c,e,a],[c,c,b,a,e]:}\begin{array}{c|cccc} \cdot & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & e & a \\ c & c & b & a & e \end{array}eabceeabcaaecbbbceaccbae
Proper Subgroups are Cyclic
The proper subgroups of V 4 V 4 V_(4)V_4V4 are:
  1. { e , a } { e , a } {e,a}\{ e, a \}{e,a} generated by a a aaa
  2. { e , b } { e , b } {e,b}\{ e, b \}{e,b} generated by b b bbb
  3. { e , c } { e , c } {e,c}\{ e, c \}{e,c} generated by c c ccc
Each of these subgroups is cyclic, as they are generated by a single element.
V 4 V 4 V_(4)V_4V4 is Not Cyclic
The group V 4 V 4 V_(4)V_4V4 itself is not cyclic because there is no single element that generates the entire group. Each of a , b , c a , b , c a,b,ca, b, ca,b,c only generates a subgroup of order 2, and e e eee generates only the trivial subgroup { e } { e } {e}\{ e \}{e}.
Conclusion
The Klein four-group V 4 V 4 V_(4)V_4V4 is an example of a group where every proper subgroup is cyclic, but the group itself is not cyclic.
Verified Answer
5/5
\(sin\left(\theta +\phi \right)=sin\:\theta \:cos\:\phi +cos\:\theta \:sin\:\phi \)
\(\sec ^2 \theta=1+\tan ^2 \theta\)

3(a) Let \( F \) be the set of all real-valued continuous functions defined on the closed interval \([0,1]\). Prove that \( (F,+,.) \) is a Commutative Ring with unity with respect to addition and multiplication of functions defined pointwise.

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To prove that ( F , + , ) ( F , + , ) (F,+,*)(F, +, \cdot)(F,+,) is a commutative ring with unity, we need to verify the following properties:
  1. Closure under Addition and Multiplication
For any f , g F f , g F f,g in Ff, g \in Ff,gF, f + g f + g f+gf+gf+g and f g f g f*gf \cdot gfg are also real-valued continuous functions defined on [ 0 , 1 ] [ 0 , 1 ] [0,1][0, 1][0,1]. Therefore, F F FFF is closed under addition and multiplication.
  1. Associativity of Addition and Multiplication
For any f , g , h F f , g , h F f,g,h in Ff, g, h \in Ff,g,hF,
( f + g ) + h = f + ( g + h ) and ( f g ) h = f ( g h ) ( f + g ) + h = f + ( g + h ) and ( f g ) h = f ( g h ) (f+g)+h=f+(g+h)quad”and”quad(f*g)*h=f*(g*h)(f + g) + h = f + (g + h) \quad \text{and} \quad (f \cdot g) \cdot h = f \cdot (g \cdot h)(f+g)+h=f+(g+h)and(fg)h=f(gh)
  1. Commutativity of Addition and Multiplication
For any f , g F f , g F f,g in Ff, g \in Ff,gF,
f + g = g + f and f g = g f f + g = g + f and f g = g f f+g=g+f quad”and”quad f*g=g*ff + g = g + f \quad \text{and} \quad f \cdot g = g \cdot ff+g=g+fandfg=gf
  1. Existence of Additive and Multiplicative Identity (Unity)
The zero function f ( x ) = 0 f ( x ) = 0 f(x)=0f(x) = 0f(x)=0 for all x [ 0 , 1 ] x [ 0 , 1 ] x in[0,1]x \in [0, 1]x[0,1] serves as the additive identity, and the function f ( x ) = 1 f ( x ) = 1 f(x)=1f(x) = 1f(x)=1 for all x [ 0 , 1 ] x [ 0 , 1 ] x in[0,1]x \in [0, 1]x[0,1] serves as the multiplicative identity (unity).
For any f F f F f in Ff \in FfF,
f + 0 = f and f 1 = f f + 0 = f and f 1 = f f+0=f quad”and”quad f*1=ff + 0 = f \quad \text{and} \quad f \cdot 1 = ff+0=fandf1=f
  1. Existence of Additive Inverse
For any f F f F f in Ff \in FfF, the function f f -f-ff defined by f ( x ) = f ( x ) f ( x ) = f ( x ) -f(x)=-f(x)-f(x) = -f(x)f(x)=f(x) for all x [ 0 , 1 ] x [ 0 , 1 ] x in[0,1]x \in [0, 1]x[0,1] is the additive inverse of f f fff.
f + ( f ) = 0 f + ( f ) = 0 f+(-f)=0f + (-f) = 0f+(f)=0
  1. Distributive Law
For any f , g , h F f , g , h F f,g,h in Ff, g, h \in Ff,g,hF,
f ( g + h ) = ( f g ) + ( f h ) f ( g + h ) = ( f g ) + ( f h ) f*(g+h)=(f*g)+(f*h)f \cdot (g + h) = (f \cdot g) + (f \cdot h)f(g+h)=(fg)+(fh)
Conclusion
Since all these properties are satisfied, ( F , + , ) ( F , + , ) (F,+,*)(F, +, \cdot)(F,+,) is a commutative ring with unity with respect to addition and multiplication of functions defined pointwise.
Verified Answer
5/5
\(a^2=b^2+c^2-2bc\:Cos\left(A\right)\)
\(cos\:2\theta =2\:cos^2\theta -1\)

4(a) Let \( a \) and \( b \) be the elements of a group, with \( a^2=e \), \( b^6=e \) and \( a b = b^4 a \). Find the order of \( ab \), and express its inverse in each of the.

Expert Answer
upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To find the order of a b a b ababab and its inverse, we need to consider the given relations:
  1. a 2 = e a 2 = e a^(2)=ea^2 = ea2=e
  2. b 6 = e b 6 = e b^(6)=eb^6 = eb6=e
  3. a b = b 4 a a b = b 4 a ab=b^(4)aab = b^4aab=b4a
Step 1: Find the Order of a b a b ababab
To find the order of a b a b ababab, we need to find the smallest positive integer n n nnn such that ( a b ) n = e ( a b ) n = e (ab)^(n)=e(ab)^n = e(ab)n=e.
Let’s substitute the values into the formula and simplify:
( a b ) 2 = a b a b ( a b ) 2 = a b a b (ab)^(2)=abab(ab)^2 = abab(ab)2=abab
Using the relation a b = b 4 a a b = b 4 a ab=b^(4)aab = b^4aab=b4a, we get:
( a b ) 2 = b 4 a b 4 a = b 4 b 4 a a = b 8 a 2 = b 2 e = b 2 ( a b ) 2 = b 4 a b 4 a = b 4 b 4 a a = b 8 a 2 = b 2 e = b 2 (ab)^(2)=b^(4)a*b^(4)a=b^(4)*b^(4)*a*a=b^(8)*a^(2)=b^(2)*e=b^(2)(ab)^2 = b^4a \cdot b^4a = b^4 \cdot b^4 \cdot a \cdot a = b^8 \cdot a^2 = b^2 \cdot e = b^2(ab)2=b4ab4a=b4b4aa=b8a2=b2e=b2
Similarly, we can find ( a b ) 3 ( a b ) 3 (ab)^(3)(ab)^3(ab)3:
( a b ) 3 = ( a b ) 2 a b = b 2 b 4 a = b 6 a = e a = a ( a b ) 3 = ( a b ) 2 a b = b 2 b 4 a = b 6 a = e a = a (ab)^(3)=(ab)^(2)*ab=b^(2)*b^(4)a=b^(6)*a=e*a=a(ab)^3 = (ab)^2 \cdot ab = b^2 \cdot b^4a = b^6 \cdot a = e \cdot a = a(ab)3=(ab)2ab=b2b4a=b6a=ea=a
Finally, let’s find ( a b ) 6 ( a b ) 6 (ab)^(6)(ab)^6(ab)6:
( a b ) 6 = ( a b ) 3 ( a b ) 3 = a a = a 2 = e ( a b ) 6 = ( a b ) 3 ( a b ) 3 = a a = a 2 = e (ab)^(6)=(ab)^(3)*(ab)^(3)=a*a=a^(2)=e(ab)^6 = (ab)^3 \cdot (ab)^3 = a \cdot a = a^2 = e(ab)6=(ab)3(ab)3=aa=a2=e
So, the smallest positive integer n n nnn such that ( a b ) n = e ( a b ) n = e (ab)^(n)=e(ab)^n = e(ab)n=e is n = 6 n = 6 n=6n = 6n=6. Therefore, the order of a b a b ababab is 6.
Step 2: Find the Inverse of a b a b ababab
The inverse of an element x x xxx in a group is an element x 1 x 1 x^(-1)x^{-1}x1 such that x x 1 = e x x 1 = e x*x^(-1)=ex \cdot x^{-1} = exx1=e.
To find ( a b ) 1 ( a b ) 1 (ab)^(-1)(ab)^{-1}(ab)1, we need to find an element that, when multiplied by a b a b ababab, gives e e eee.
Since ( a b ) 6 = e ( a b ) 6 = e (ab)^(6)=e(ab)^6 = e(ab)6=e, we know that ( a b ) 1 = ( a b ) 5 ( a b ) 1 = ( a b ) 5 (ab)^(-1)=(ab)^(5)(ab)^{-1} = (ab)^5(ab)1=(ab)5.
Let’s substitute the values into the formula and simplify:
( a b ) 5 = ( a b ) 2 ( a b ) 2 a b = b 2 b 2 b 4 a = b 8 a = b 2 a ( a b ) 5 = ( a b ) 2 ( a b ) 2 a b = b 2 b 2 b 4 a = b 8 a = b 2 a (ab)^(5)=(ab)^(2)*(ab)^(2)*ab=b^(2)*b^(2)*b^(4)a=b^(8)*a=b^(2)*a(ab)^5 = (ab)^2 \cdot (ab)^2 \cdot ab = b^2 \cdot b^2 \cdot b^4a = b^8 \cdot a = b^2 \cdot a(ab)5=(ab)2(ab)2ab=b2b2b4a=b8a=b2a
After calculating, we find that ( a b ) 1 = b 2 a ( a b ) 1 = b 2 a (ab)^(-1)=b^(2)a(ab)^{-1} = b^2a(ab)1=b2a.
Conclusion
The order of a b a b ababab is 6, and its inverse is b 2 a b 2 a b^(2)ab^2ab2a.
Verified Answer
5/5
\(c^2=a^2+b^2-2ab\:Cos\left(C\right)\)

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