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In the group G={0,1,2,3,4,5} under addition modulo 6, (2+3−1+4)−1=
- a) one
- b)two
- c)five
- d)three
Correct answer is option 'D'. Can you explain this answer?
In the group G={0,1,2,3,4,5} under addition modulo 6, (2+3−1+4)−1=
a)
one
b)
two
c)
five
d)
three
|
|
Never Smo answered |
This group is Z5. In Z5 possible order of elements are only 1 and 5 because o(a) /o(G) for all a in G.
As this is additive group so : 2+2+2+2+2 =10(mod 5) = 0 which is identity.
So order of 2 in given Group is 5 not 4.
As this is additive group so : 2+2+2+2+2 =10(mod 5) = 0 which is identity.
So order of 2 in given Group is 5 not 4.
Let Z be a set of integers, then under ordinary multiplication (Z, ·) is- a)monoid
- b) semi-group
- c)quasi-group
- d) group
Correct answer is option 'A'. Can you explain this answer?
Let Z be a set of integers, then under ordinary multiplication (Z, ·) is
a)
monoid
b)
semi-group
c)
quasi-group
d)
group
|
Ira Malhotra answered |
The question seems to be incomplete. Can you please provide the complete statement?
Let * be the binary operation on the rational number given by a*b=a+b+ab. Which of the following property does not exist for the group?
- a)Closure property
- b)Identity property
- c)Symmetric property
- d)Associative property
Correct answer is option 'B'. Can you explain this answer?
Let * be the binary operation on the rational number given by a*b=a+b+ab. Which of the following property does not exist for the group?
a)
Closure property
b)
Identity property
c)
Symmetric property
d)
Associative property
|
Chirag Verma answered |
Explanation: For identity e, a+e=e+a=e, a*e = a+e+ae = a => e=0 and e+a = e+a+ea = a => e=0. So e=0 will be identity, for e to be identity, a*e = a ⇒ a+e+ae = a ⇒ e+ae = 0 and e(1+a) = 0 which gives e=0 or a=-1. So, when a = -1, no identity element exist as e can be any value in that case.
If H ⊂ K are two subgroups of G and if [G : H] = 8 and [G : K] = 4, then [K : H] is- a)2
- b)3
- c)5
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
If H ⊂ K are two subgroups of G and if [G : H] = 8 and [G : K] = 4, then [K : H] is
a)
2
b)
3
c)
5
d)
None of these
|
Pranavi Kapoor answered |
If H is a constant, it means that the value of H does not change. This can be used in mathematical equations or formulas to indicate that a certain variable or parameter remains constant throughout the calculations.
Let f(x) = x3 +x2 - x + 15 and g(x) = x3 + 2x2 - x + 15. Then, over Q- a)f is irreducible and g is reducible
- b)f is reducible and g is irreducible
- c)Both f and g are reducible
- d)Both f and g are irreducible
Correct answer is option 'B'. Can you explain this answer?
Let f(x) = x3 +x2 - x + 15 and g(x) = x3 + 2x2 - x + 15. Then, over Q
a)
f is irreducible and g is reducible
b)
f is reducible and g is irreducible
c)
Both f and g are reducible
d)
Both f and g are irreducible
|
|
Chirag Verma answered |
Let f (x) = x3 + x2 - x + 15
Put x = -3
f(x) = -27 + 9 + 3 + 15 = 0
So, f(x) is reducible,
and g(x) = x3 + 2x2 - x + 15 is not reducible
Thus f is reducible and g is irreducible.
Put x = -3
f(x) = -27 + 9 + 3 + 15 = 0
So, f(x) is reducible,
and g(x) = x3 + 2x2 - x + 15 is not reducible
Thus f is reducible and g is irreducible.
Statement A : Every isomorphic image of a cyclic group is cyclic.
Statement B : Every homomorphic image of a cyclic group is cyclic
- a)B is true only
- b)Both A and B are false
- c)A is true only
- d)Both A and B are true
Correct answer is option 'D'. Can you explain this answer?
Statement A : Every isomorphic image of a cyclic group is cyclic.
Statement B : Every homomorphic image of a cyclic group is cyclic
Statement B : Every homomorphic image of a cyclic group is cyclic
a)
B is true only
b)
Both A and B are false
c)
A is true only
d)
Both A and B are true
|
|
Chirag Verma answered |
Correct option is D. Both the statements are true.
In a group G, we have ab = a or ba = a then- a)a = e
- b) a2 = e
- c)b = e
- d)b2 - e
Correct answer is option 'C'. Can you explain this answer?
In a group G, we have ab = a or ba = a then
a)
a = e
b)
a2 = e
c)
b = e
d)
b2 - e
|
|
Pranavi Kapoor answered |
Explanation:
The given conditions in the group G are:
1. ab = a
2. ba = a
We need to determine the correct statement among the given options.
Let's consider the first condition, ab = a. Multiplying both sides of this equation by the inverse of 'b' (denoted as b⁻¹), we get:
ab * b⁻¹ = a * b⁻¹
This simplifies to:
a * (b * b⁻¹) = a * b⁻¹
Since b * b⁻¹ is the identity element (denoted as e) in the group G, the equation becomes:
a * e = a * b⁻¹
And since a * e = a, we have:
a = a * b⁻¹
This result shows that the inverse of 'b' in the group G is equal to 'a'. Hence, option 'a' is incorrect.
Now, let's consider the second condition, ba = a. Multiplying both sides of this equation by the inverse of 'a' (denoted as a⁻¹), we get:
b * a * a⁻¹ = a * a⁻¹
This simplifies to:
b * (a * a⁻¹) = a * a⁻¹
Since a * a⁻¹ is the identity element (denoted as e) in the group G, the equation becomes:
b * e = a * a⁻¹
And since b * e = b, we have:
b = a * a⁻¹
This result shows that the inverse of 'a' in the group G is equal to 'b'. Hence, option 'b' is incorrect.
Now, let's consider the third condition, b = ed. Multiplying both sides of this equation by the inverse of 'd' (denoted as d⁻¹), we get:
b * d⁻¹ = ed * d⁻¹
This simplifies to:
b * (d * d⁻¹) = e * d⁻¹
Since d * d⁻¹ is the identity element (denoted as e) in the group G, the equation becomes:
b * e = e * d⁻¹
And since b * e = b, we have:
b = e * d⁻¹
This result shows that the inverse of 'd' in the group G is equal to 'b'. Hence, option 'd' is incorrect.
Therefore, the correct statement among the given options is option 'c': a² = e.
The given conditions in the group G are:
1. ab = a
2. ba = a
We need to determine the correct statement among the given options.
Let's consider the first condition, ab = a. Multiplying both sides of this equation by the inverse of 'b' (denoted as b⁻¹), we get:
ab * b⁻¹ = a * b⁻¹
This simplifies to:
a * (b * b⁻¹) = a * b⁻¹
Since b * b⁻¹ is the identity element (denoted as e) in the group G, the equation becomes:
a * e = a * b⁻¹
And since a * e = a, we have:
a = a * b⁻¹
This result shows that the inverse of 'b' in the group G is equal to 'a'. Hence, option 'a' is incorrect.
Now, let's consider the second condition, ba = a. Multiplying both sides of this equation by the inverse of 'a' (denoted as a⁻¹), we get:
b * a * a⁻¹ = a * a⁻¹
This simplifies to:
b * (a * a⁻¹) = a * a⁻¹
Since a * a⁻¹ is the identity element (denoted as e) in the group G, the equation becomes:
b * e = a * a⁻¹
And since b * e = b, we have:
b = a * a⁻¹
This result shows that the inverse of 'a' in the group G is equal to 'b'. Hence, option 'b' is incorrect.
Now, let's consider the third condition, b = ed. Multiplying both sides of this equation by the inverse of 'd' (denoted as d⁻¹), we get:
b * d⁻¹ = ed * d⁻¹
This simplifies to:
b * (d * d⁻¹) = e * d⁻¹
Since d * d⁻¹ is the identity element (denoted as e) in the group G, the equation becomes:
b * e = e * d⁻¹
And since b * e = b, we have:
b = e * d⁻¹
This result shows that the inverse of 'd' in the group G is equal to 'b'. Hence, option 'd' is incorrect.
Therefore, the correct statement among the given options is option 'c': a² = e.
Let f(x) = x2 + 1, g(x) = x3 + x2 + 1 and h(x) = x4 + x2 + 1. Then- a)f(x) and g(x) are reducible over Z2
- b)g(x) and h(x) are reducible over Z2
- c)f(x) and h(x) are reducible over Z2
- d)f (x), g(x) and h(x) are reducible over Z2
Correct answer is option 'B'. Can you explain this answer?
Let f(x) = x2 + 1, g(x) = x3 + x2 + 1 and h(x) = x4 + x2 + 1. Then
a)
f(x) and g(x) are reducible over Z2
b)
g(x) and h(x) are reducible over Z2
c)
f(x) and h(x) are reducible over Z2
d)
f (x), g(x) and h(x) are reducible over Z2
|
|
Chirag Verma answered |
Here f(x) = x2 + 1, g(x) = x3 +x2 + 1 and h(x) =x4 + x2 + 1
In Z2 ={0, 1},
f(x) =x2 + 1
g(x) = x3 + x2 + 1
f(0) =0 + 1 = 1 g(0) = 0 + 0 = 1
f( l ) =1 + 1 = 2(mod 2) = 0
g(l) -1 + 1 + 1 = 3(mod 2) = 1
So, f(x) is reducible
hence, g(x) is irreducible.
In Z2 = { 0 , 1}
h(x) = x4 +x2 + l
h(0) = 0 + 0 + 1 = 1
h(l) =1 + 1 + 1 = 3(mod 2) = 1
.’. h(x) is irreducible.
Hence, g(x) and h(x) are irreducible over Z2.
In Z2 ={0, 1},
f(x) =x2 + 1
g(x) = x3 + x2 + 1
f(0) =0 + 1 = 1 g(0) = 0 + 0 = 1
f( l ) =1 + 1 = 2(mod 2) = 0
g(l) -1 + 1 + 1 = 3(mod 2) = 1
So, f(x) is reducible
hence, g(x) is irreducible.
In Z2 = { 0 , 1}
h(x) = x4 +x2 + l
h(0) = 0 + 0 + 1 = 1
h(l) =1 + 1 + 1 = 3(mod 2) = 1
.’. h(x) is irreducible.
Hence, g(x) and h(x) are irreducible over Z2.
Consider the group G of non-zero real numbers under multiplication. Let H = {x ε G I x = 1 or x is irrational} and K = {x e G | x ≥ 1}. Then- a)H is a subgroup of G but K is not a subgroup of G.
- b)H is not a subgroup of G but K is a subgroup of G.
- c)Both H and K are subgroups of G.
- d)Neither K nor H is a subgroup of G.
Correct answer is option 'A'. Can you explain this answer?
Consider the group G of non-zero real numbers under multiplication. Let H = {x ε G I x = 1 or x is irrational} and K = {x e G | x ≥ 1}. Then
a)
H is a subgroup of G but K is not a subgroup of G.
b)
H is not a subgroup of G but K is a subgroup of G.
c)
Both H and K are subgroups of G.
d)
Neither K nor H is a subgroup of G.
|
|
Chirag Verma answered |
√2 ε H unit
Let G = {n ε Z : 1 ≤ n ≤ 55, gcd(n, 56) = 1} be a multiplicative group modulo 56. Consider the sets S1 - {1,9, 17,25,33,41} and S2 = { 1 ,1 5 ,2 9 ,43 }
Q. which one of the following is TRUE?- a)S1 is a subgroup of G but S2 is NOT a subgroup of G
- b)S1 is NOT a subgroup of G but S2 is a subgroup of G
- c)Both Sl and S2 are subgroups of G
- d)Neither S1 nor S2 is a subgroup of G
Correct answer is option 'C'. Can you explain this answer?
Let G = {n ε Z : 1 ≤ n ≤ 55, gcd(n, 56) = 1} be a multiplicative group modulo 56. Consider the sets S1 - {1,9, 17,25,33,41} and S2 = { 1 ,1 5 ,2 9 ,43 }
Q. which one of the following is TRUE?
Q. which one of the following is TRUE?
a)
S1 is a subgroup of G but S2 is NOT a subgroup of G
b)
S1 is NOT a subgroup of G but S2 is a subgroup of G
c)
Both Sl and S2 are subgroups of G
d)
Neither S1 nor S2 is a subgroup of G
|
|
Chirag Verma answered |
Given G = {n ε Z : 1 ≤ n ≤ 55, g.c.d(n, 56) = 1}
i.e. G = {1, 2, 5, 9,11, 13, 15, 17, 19,23,25,27, 2 9 ,3 3 ,3 7 ,3 9 ,4 1 ,4 3 ,4 5 ,4 7 ,5 1 ,5 3 ,5 5 }
i.e. G is finite abelian group.
Here, S1 = {1, 9 , 17 , 25, 33, 41}.
So, S1 is a subgroup of G because S, satisfies all properties of subgroup of G. i.e., closure , associative existence of id entity and existence of inverse.
S2 = {1,15, 29,43} is also subgroup of G.
Hence, Both S1 and S2 are subgroups of G.
i.e. G = {1, 2, 5, 9,11, 13, 15, 17, 19,23,25,27, 2 9 ,3 3 ,3 7 ,3 9 ,4 1 ,4 3 ,4 5 ,4 7 ,5 1 ,5 3 ,5 5 }
i.e. G is finite abelian group.
Here, S1 = {1, 9 , 17 , 25, 33, 41}.
So, S1 is a subgroup of G because S, satisfies all properties of subgroup of G. i.e., closure , associative existence of id entity and existence of inverse.
S2 = {1,15, 29,43} is also subgroup of G.
Hence, Both S1 and S2 are subgroups of G.
The number of all subgroups of the group (Z60, +) of integers modulo 60 is- a)2
- b)10
- c)12
- d)60
Correct answer is option 'C'. Can you explain this answer?
The number of all subgroups of the group (Z60, +) of integers modulo 60 is
a)
2
b)
10
c)
12
d)
60
|
|
Chirag Verma answered |
Group (Z60, +) of integer modulo 60.
Order of Subgroup will divide order of the group.
60 = 22 • 3 • 5
So, total number of divisor = 3 x 2 x 2 = 12
So, 12 subgroups are possible.
Since operation is addition modulo 60 thus each divisor will form a subgroup.
Thus, there are 12 subgroups.
Order of Subgroup will divide order of the group.
60 = 22 • 3 • 5
So, total number of divisor = 3 x 2 x 2 = 12
So, 12 subgroups are possible.
Since operation is addition modulo 60 thus each divisor will form a subgroup.
Thus, there are 12 subgroups.
The number of elements of order 5 in a non-abelian group of order 10 is- a)1
- b)5
- c)4
- d)8
Correct answer is option 'C'. Can you explain this answer?
The number of elements of order 5 in a non-abelian group of order 10 is
a)
1
b)
5
c)
4
d)
8
|
Priyo Adhikary answered |
Any group of order 5 is cyclic .
so number of element will be phi 5 = 4
so number of element will be phi 5 = 4
Condition for monoid is __________
- a)(a+e)=a
- b)(a*e)=(a+e)
- c)a= a*(a+e)
- d)(a*e)=(e*a)=a
Correct answer is option 'D'. Can you explain this answer?
Condition for monoid is __________
a)
(a+e)=a
b)
(a*e)=(a+e)
c)
a= a*(a+e)
d)
(a*e)=(e*a)=a
|
Leona Sheen answered |
B
2^m = 56+2^n
2^m-2^n = 56
m=6
n=3
2^m = 56+2^n
2^m-2^n = 56
m=6
n=3
The set of all non-singular square matrices of same order with respect to matrix multiplication is- a)quasi-group
- b)monoid
- c)group
- d)abelian group
Correct answer is option 'C'. Can you explain this answer?
The set of all non-singular square matrices of same order with respect to matrix multiplication is
a)
quasi-group
b)
monoid
c)
group
d)
abelian group
|
Hetal Shah answered |
Explanation:
To determine whether the set of all non-singular square matrices of the same order form a group with respect to matrix multiplication, we need to check whether it satisfies the four group axioms:
1. Closure: The product of any two non-singular square matrices of the same order is also a non-singular square matrix of the same order. Therefore, the set is closed under matrix multiplication.
2. Associativity: Matrix multiplication is associative, which means that for any three matrices A, B, and C of the same order, (AB)C = A(BC). Since matrix multiplication is associative, the set satisfies the associativity property.
3. Identity element: The identity matrix I, which is a non-singular square matrix of the same order as any matrix in the set, serves as the identity element. For any matrix A in the set, AI = A and IA = A. Therefore, the set contains an identity element.
4. Inverse element: For every non-singular square matrix A in the set, there exists an inverse matrix A^(-1) such that AA^(-1) = A^(-1)A = I, where I is the identity matrix. The inverse of A is also a non-singular square matrix of the same order. Therefore, the set contains inverse elements for every matrix.
Since the set of all non-singular square matrices of the same order satisfies all four group axioms, it can be concluded that it forms a group with respect to matrix multiplication.
Hence, the correct answer is option 'C' - group.
To determine whether the set of all non-singular square matrices of the same order form a group with respect to matrix multiplication, we need to check whether it satisfies the four group axioms:
1. Closure: The product of any two non-singular square matrices of the same order is also a non-singular square matrix of the same order. Therefore, the set is closed under matrix multiplication.
2. Associativity: Matrix multiplication is associative, which means that for any three matrices A, B, and C of the same order, (AB)C = A(BC). Since matrix multiplication is associative, the set satisfies the associativity property.
3. Identity element: The identity matrix I, which is a non-singular square matrix of the same order as any matrix in the set, serves as the identity element. For any matrix A in the set, AI = A and IA = A. Therefore, the set contains an identity element.
4. Inverse element: For every non-singular square matrix A in the set, there exists an inverse matrix A^(-1) such that AA^(-1) = A^(-1)A = I, where I is the identity matrix. The inverse of A is also a non-singular square matrix of the same order. Therefore, the set contains inverse elements for every matrix.
Since the set of all non-singular square matrices of the same order satisfies all four group axioms, it can be concluded that it forms a group with respect to matrix multiplication.
Hence, the correct answer is option 'C' - group.
The number of positive divisiors of 50,000 is- a)20
- b)30
- c)40
- d)50
Correct answer is option 'B'. Can you explain this answer?
The number of positive divisiors of 50,000 is
a)
20
b)
30
c)
40
d)
50
|
Veda Institute answered |
An algebraic structure (P,*) is called a semigroup if a*(b*c) = (a*b)*c for all a,b,c belongs to S or the elements follow associative property under “*”. (Matrix,*) and (Set of integers,+) are examples of semigroup.
How many properties can be held by a group?
- a)2
- b)3
- c)4
- d)5
Correct answer is option 'C'. Can you explain this answer?
How many properties can be held by a group?
a)
2
b)
3
c)
4
d)
5
|
|
Chirag Verma answered |
A group holds five properties simultaneously –
i) Closure
ii) associative
iii) Identity element
iv) Inverse element.
i) Closure
ii) associative
iii) Identity element
iv) Inverse element.
The set M of square matrices ( of same order) with respect to matrix multiplication is
- a)quasi-group
- b)group
- c)monoid
- d)semi-group
Correct answer is option 'C'. Can you explain this answer?
The set M of square matrices ( of same order) with respect to matrix multiplication is
a)
quasi-group
b)
group
c)
monoid
d)
semi-group
|
Vedika Sharma answered |
Explanation:
To determine the set M of square matrices with respect to matrix multiplication, we need to analyze the properties of this set.
Definition of a Monoid:
A monoid is a set equipped with an associative binary operation and an identity element.
Associative Binary Operation:
Matrix multiplication is an associative binary operation, which means that for any three matrices A, B, and C of the same order, the following holds:
(A * B) * C = A * (B * C)
Identity Element:
The identity element for matrix multiplication is the identity matrix. The identity matrix I is a square matrix with ones on the main diagonal and zeros elsewhere, such that for any matrix A of the appropriate size, the following holds:
A * I = I * A = A
Analysis:
Considering the properties of matrix multiplication, we can conclude the following:
1. Associativity: Matrix multiplication is associative, satisfying the requirement of an associative binary operation.
2. Identity Element: The identity matrix serves as the identity element for matrix multiplication, satisfying the requirement of an identity element.
Therefore, the set M of square matrices with respect to matrix multiplication forms a monoid.
Conclusion:
The correct answer is option 'C' - monoid. The set M of square matrices with respect to matrix multiplication satisfies the properties of an associative binary operation and has an identity element, making it a monoid.
To determine the set M of square matrices with respect to matrix multiplication, we need to analyze the properties of this set.
Definition of a Monoid:
A monoid is a set equipped with an associative binary operation and an identity element.
Associative Binary Operation:
Matrix multiplication is an associative binary operation, which means that for any three matrices A, B, and C of the same order, the following holds:
(A * B) * C = A * (B * C)
Identity Element:
The identity element for matrix multiplication is the identity matrix. The identity matrix I is a square matrix with ones on the main diagonal and zeros elsewhere, such that for any matrix A of the appropriate size, the following holds:
A * I = I * A = A
Analysis:
Considering the properties of matrix multiplication, we can conclude the following:
1. Associativity: Matrix multiplication is associative, satisfying the requirement of an associative binary operation.
2. Identity Element: The identity matrix serves as the identity element for matrix multiplication, satisfying the requirement of an identity element.
Therefore, the set M of square matrices with respect to matrix multiplication forms a monoid.
Conclusion:
The correct answer is option 'C' - monoid. The set M of square matrices with respect to matrix multiplication satisfies the properties of an associative binary operation and has an identity element, making it a monoid.
Let G be a finite group of order 200, then the number of subgroup of G of order 25 is- a)1
- b)2
- c)5
- d)10
Correct answer is option 'A'. Can you explain this answer?
Let G be a finite group of order 200, then the number of subgroup of G of order 25 is
a)
1
b)
2
c)
5
d)
10
|
Sanika Mehta answered |
Let G be a finite group of order 200. We need to find the number of subgroups of G that have an order of 25.
Step 1: Prime Factorization of 200
To begin, let's find the prime factorization of 200:
200 = 2^3 * 5^2
Step 2: Sylow's Theorem
According to Sylow's theorem, for any prime factor p of the order of a finite group G, the number of subgroups of G of order p^k, denoted as n_p, satisfies the following conditions:
1. n_p ≡ 1 (mod p) - This means that n_p leaves a remainder of 1 when divided by p.
2. n_p | (order of G)/p^k - This means that n_p divides the order of G divided by p^k.
Step 3: Finding n_5
In this case, we are interested in finding the number of subgroups of G of order 25, which means we need to find n_5. According to Sylow's theorem, n_5 ≡ 1 (mod 5) and n_5 | 200/5^2.
Since 200/5^2 = 8, we need to find a value of n_5 that leaves a remainder of 1 when divided by 5 and divides 8.
Step 4: Possible Values of n_5
The possible values of n_5 are 1 and 6. However, since we are looking for the number of subgroups of order 25, which is a specific value, the correct answer is 1.
Step 5: Conclusion
Therefore, the number of subgroups of G of order 25 is 1.
Step 1: Prime Factorization of 200
To begin, let's find the prime factorization of 200:
200 = 2^3 * 5^2
Step 2: Sylow's Theorem
According to Sylow's theorem, for any prime factor p of the order of a finite group G, the number of subgroups of G of order p^k, denoted as n_p, satisfies the following conditions:
1. n_p ≡ 1 (mod p) - This means that n_p leaves a remainder of 1 when divided by p.
2. n_p | (order of G)/p^k - This means that n_p divides the order of G divided by p^k.
Step 3: Finding n_5
In this case, we are interested in finding the number of subgroups of G of order 25, which means we need to find n_5. According to Sylow's theorem, n_5 ≡ 1 (mod 5) and n_5 | 200/5^2.
Since 200/5^2 = 8, we need to find a value of n_5 that leaves a remainder of 1 when divided by 5 and divides 8.
Step 4: Possible Values of n_5
The possible values of n_5 are 1 and 6. However, since we are looking for the number of subgroups of order 25, which is a specific value, the correct answer is 1.
Step 5: Conclusion
Therefore, the number of subgroups of G of order 25 is 1.
Let G be a group with identity e such that for some a ε G, a2 ≠ e and a6 = e. Then which of the following is true?- a)a3 = e, a4 ≠ e
- b)a4 ≠ e, a5 ≠ e
- c)a4 ≠ e, a5 = e
- d)a3 = e, a4 = e
Correct answer is option 'B'. Can you explain this answer?
Let G be a group with identity e such that for some a ε G, a2 ≠ e and a6 = e. Then which of the following is true?
a)
a3 = e, a4 ≠ e
b)
a4 ≠ e, a5 ≠ e
c)
a4 ≠ e, a5 = e
d)
a3 = e, a4 = e
|
|
Chirag Verma answered |
Given G be a group with identity e such that for some a ε G, a2 ≠ e and a6 = e.
Hence a4 ≠ e , a5 ≠ e .
Hence a4 ≠ e , a5 ≠ e .
In the additive group of integers, the order of every elements a ≠ 0 is
- a)infinity
- b)one
- c)zero
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
In the additive group of integers, the order of every elements a ≠ 0 is
a)
infinity
b)
one
c)
zero
d)
None of these
|
|
Veda Institute answered |
The correct answer is:
1. infinity
In the additive group of integers, the order of an element aaa (where a≠0 ) refers to the smallest positive integer nnn such that n⋅a=0 However, for any non-zero integer aaa, there is no positive integer nnn that satisfies this equation because adding aaa to itself any number of times will never result in 0. Therefore, the order of any non-zero element in this group is infinite.
Let U(n) be the set of all positive integers less than n and relatively prime to n for n = 248, the number of elements in U(n) is- a)60
- b)120
- c)180
- d)240
Correct answer is option 'B'. Can you explain this answer?
Let U(n) be the set of all positive integers less than n and relatively prime to n for n = 248, the number of elements in U(n) is
a)
60
b)
120
c)
180
d)
240
|
|
Suvojit Nayek answered |
Basically by definition of U(n), it's order is phi(n) where n is the no of elements in group. so phi (248) =248×(1-1/2) ×(1-1/31) =120 [ U should know about a little bit of euler's phi function and it's properties to compute this]
Every finite group G is isomorphic to a permutation group; this statement is- a)Cayley’s theorem
- b)Lagrange’s theorem
- c)Liouville’s theorem
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Every finite group G is isomorphic to a permutation group; this statement is
a)
Cayley’s theorem
b)
Lagrange’s theorem
c)
Liouville’s theorem
d)
None of these
|
Gagan Bist answered |
Show that every finite group G is isomorphic to a permutation group.
In the symmetric group Sn of degree n,n > 2 the set of all 3-cycles generates a subgroup of order
- a)3
- b)3n
- c)n!/2
- d)n!
Correct answer is option 'C'. Can you explain this answer?
In the symmetric group Sn of degree n,n > 2 the set of all 3-cycles generates a subgroup of order
a)
3
b)
3n
c)
n!/2
d)
n!
|
Ajay Mallettula answered |
Even permutations generate alternating group
Let G be an Abelian group o f order 10. Let S = {g ε G : g-1 = g}. Then the number of nonidentity elements in S is :- a)5
- b)2
- c)1
- d)0
Correct answer is option 'C'. Can you explain this answer?
Let G be an Abelian group o f order 10. Let S = {g ε G : g-1 = g}. Then the number of nonidentity elements in S is :
a)
5
b)
2
c)
1
d)
0
|
|
Veda Institute answered |
An idempotent element in a monoid is one that, when combined with itself, produces the same element. Mathematically, this is expressed as a² = a * a = a.
Let G be a group of order 49, then- a)G is abelian
- b)G is cyclic
- c)G is non abelian
- d)centre of G has order 7
Correct answer is option 'A'. Can you explain this answer?
Let G be a group of order 49, then
a)
G is abelian
b)
G is cyclic
c)
G is non abelian
d)
centre of G has order 7
|
|
Chirag Verma answered |
Since wk that every group of order p2 is abelian,
where p is a prime integer.
∴ Group of order p2 is abelian also.
∴ Group of order 72
where p is a prime integer.
∴ Group of order p2 is abelian also.
∴ Group of order 72
i.e. 49 is abelian and cyclic.
Set (1,2,3,4} is a finite abelian group of order... under multiplication modulo ... as composition.- a)3,4
- b)4, 5
- c)1 , 2
- d)2 , 3
Correct answer is option 'B'. Can you explain this answer?
Set (1,2,3,4} is a finite abelian group of order... under multiplication modulo ... as composition.
a)
3,4
b)
4, 5
c)
1 , 2
d)
2 , 3
|
|
Shlok Shah answered |
To determine the order of the group, we need to find the number of elements in the set.
Step 1: Counting the number of elements in the set
The given set is {1, 2, 3, 4}. Counting the number of elements, we find that there are 4 elements in the set.
Step 2: Checking if the set forms a group under multiplication modulo n
To determine if the set forms a group under multiplication modulo n, we need to check the following conditions:
1. Closure: For any two elements a and b in the set, a * b (mod n) should also be in the set.
2. Associativity: For any three elements a, b, and c in the set, (a * b) * c (mod n) should be equal to a * (b * c) (mod n).
3. Identity: There should exist an identity element e in the set such that for any element a in the set, a * e (mod n) = a.
4. Inverse: For any element a in the set, there should exist an inverse element b in the set such that a * b (mod n) = e.
Checking the closure property:
1 * 1 (mod 4) = 1
1 * 2 (mod 4) = 2
1 * 3 (mod 4) = 3
1 * 4 (mod 4) = 0
2 * 1 (mod 4) = 2
2 * 2 (mod 4) = 0
2 * 3 (mod 4) = 2
2 * 4 (mod 4) = 0
3 * 1 (mod 4) = 3
3 * 2 (mod 4) = 2
3 * 3 (mod 4) = 1
3 * 4 (mod 4) = 0
4 * 1 (mod 4) = 0
4 * 2 (mod 4) = 0
4 * 3 (mod 4) = 0
4 * 4 (mod 4) = 0
From the above calculations, we can see that all the results are in the set {1, 2, 3, 4}. Therefore, the set satisfies the closure property.
Checking the associativity property:
Since multiplication is associative, the set satisfies the associativity property.
Checking the identity property:
There is no element e in the set {1, 2, 3, 4} such that a * e (mod 4) = a for all elements a in the set. Therefore, the set does not have an identity element and does not satisfy the identity property.
Checking the inverse property:
For each element in the set, we need to find an inverse element such that the product of the element and its inverse is congruent to the identity element modulo 4. However, since the set does not have an identity element, it does not have inverse elements either.
Since the set does not satisfy all the conditions to be a group, it cannot be considered as an abelian group under multiplication modulo 4. Therefore, the correct answer is option B) 4, 5.
Step 1: Counting the number of elements in the set
The given set is {1, 2, 3, 4}. Counting the number of elements, we find that there are 4 elements in the set.
Step 2: Checking if the set forms a group under multiplication modulo n
To determine if the set forms a group under multiplication modulo n, we need to check the following conditions:
1. Closure: For any two elements a and b in the set, a * b (mod n) should also be in the set.
2. Associativity: For any three elements a, b, and c in the set, (a * b) * c (mod n) should be equal to a * (b * c) (mod n).
3. Identity: There should exist an identity element e in the set such that for any element a in the set, a * e (mod n) = a.
4. Inverse: For any element a in the set, there should exist an inverse element b in the set such that a * b (mod n) = e.
Checking the closure property:
1 * 1 (mod 4) = 1
1 * 2 (mod 4) = 2
1 * 3 (mod 4) = 3
1 * 4 (mod 4) = 0
2 * 1 (mod 4) = 2
2 * 2 (mod 4) = 0
2 * 3 (mod 4) = 2
2 * 4 (mod 4) = 0
3 * 1 (mod 4) = 3
3 * 2 (mod 4) = 2
3 * 3 (mod 4) = 1
3 * 4 (mod 4) = 0
4 * 1 (mod 4) = 0
4 * 2 (mod 4) = 0
4 * 3 (mod 4) = 0
4 * 4 (mod 4) = 0
From the above calculations, we can see that all the results are in the set {1, 2, 3, 4}. Therefore, the set satisfies the closure property.
Checking the associativity property:
Since multiplication is associative, the set satisfies the associativity property.
Checking the identity property:
There is no element e in the set {1, 2, 3, 4} such that a * e (mod 4) = a for all elements a in the set. Therefore, the set does not have an identity element and does not satisfy the identity property.
Checking the inverse property:
For each element in the set, we need to find an inverse element such that the product of the element and its inverse is congruent to the identity element modulo 4. However, since the set does not have an identity element, it does not have inverse elements either.
Since the set does not satisfy all the conditions to be a group, it cannot be considered as an abelian group under multiplication modulo 4. Therefore, the correct answer is option B) 4, 5.
Let A be a set contains n elements, then its power set P(A) contains- a)n elements
- b)2n elements
- c)n2 elements
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
Let A be a set contains n elements, then its power set P(A) contains
a)
n elements
b)
2n elements
c)
n2 elements
d)
none of these
|
Reeti Padhan answered |
Option B is correct
no of elements in p(A)= 2^no of elements in set a
no of elements in p(A)= 2^no of elements in set a
If two sets A and B are having 99 elements in common, then the number of elements common to each of the set A x B and B x A are- a)299
- b)992
- c)100
- d)18
Correct answer is option 'B'. Can you explain this answer?
If two sets A and B are having 99 elements in common, then the number of elements common to each of the set A x B and B x A are
a)
299
b)
992
c)
100
d)
18
|
|
Veda Institute answered |
A cyclic group is always an abelian group but every abelian group is not a cyclic group. For instance, the rational numbers under addition is an abelian group but is not a cyclic one.
Let n be a non-negative integer. Which of the following numbers can be the number of elements in a finite Boolean algebra?- a)2n
- b)2n+1 - 1
- c)2n + 1
- d)2n+1 + l
Correct answer is option 'A'. Can you explain this answer?
Let n be a non-negative integer. Which of the following numbers can be the number of elements in a finite Boolean algebra?
a)
2n
b)
2n+1 - 1
c)
2n + 1
d)
2n+1 + l
|
|
Chirag Verma answered |
The number of elements in a finite Boolean algebra = 2n; where n be a non-negative integer.
∴ option (a) is correct.
∴ option (a) is correct.
If the order of elements a, a-1 ∈ G are m and n respectively, then- a)m = n
- b)m ≠ n
- c)m = n = 0
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
If the order of elements a, a-1 ∈ G are m and n respectively, then
a)
m = n
b)
m ≠ n
c)
m = n = 0
d)
None of these
|
|
Pranavi Kapoor answered |
The order of the elements a, a-1, a-2, a-3, and a-4 depends on the specific values of 'a'. Without knowing the value of 'a', it is not possible to determine the order of these elements.
Choose the correct answer. If His a normal subgroup of G and K is a normal subgroup of H, then- a)K is a normal subgroup of G
- b)K is not a normal subgroup of G
- c)K is a subgroup of G.
- d)K is a subgroup of G
Correct answer is option 'B'. Can you explain this answer?
Choose the correct answer. If His a normal subgroup of G and K is a normal subgroup of H, then
a)
K is a normal subgroup of G
b)
K is not a normal subgroup of G
c)
K is a subgroup of G.
d)
K is a subgroup of G
|
|
Chirag Verma answered |
Let G = A4, H = {(1 2)(3 4),(1 3)(2 4),(1 4)(2 3), e},
K = {e, (1 2) (3 4)}
Then H is normal in G and K is normal in H. But K is not normal in G
Let x, y and z be Boolean variables. The number of possible values for the expression 
is- a)1
- b)2
- c)4
- d)8
Correct answer is option 'B'. Can you explain this answer?
Let x, y and z be Boolean variables. The number of possible values for the expression is
isa)
1
b)
2
c)
4
d)
8
|
Qamar Siddiqui answered |
Closure Property:
The closure property states that when two elements from a set are combined using a certain operation, the result will also be an element of the same set. In this case, the set of even natural numbers is closed under the addition operation.
Explanation:
To determine if a set is closed under addition, we need to check if the sum of any two elements in the set is also an element of the same set.
Let's take two even numbers from the given set: 6 and 8.
When we add these two numbers, we get 6 + 8 = 14. Now, we need to check if 14 is also an even number.
An even number is defined as an integer that is divisible by 2 without leaving a remainder. In other words, if we divide an even number by 2, the result will be an integer.
If we divide 14 by 2, we get 14/2 = 7, which is an integer. Therefore, 14 is also an even number.
Since the sum of any two even numbers from the given set is also an even number, we can conclude that the set of even natural numbers is closed under addition.
Conclusion:
The set of even natural numbers satisfies the closure property under the addition operation.
The closure property states that when two elements from a set are combined using a certain operation, the result will also be an element of the same set. In this case, the set of even natural numbers is closed under the addition operation.
Explanation:
To determine if a set is closed under addition, we need to check if the sum of any two elements in the set is also an element of the same set.
Let's take two even numbers from the given set: 6 and 8.
When we add these two numbers, we get 6 + 8 = 14. Now, we need to check if 14 is also an even number.
An even number is defined as an integer that is divisible by 2 without leaving a remainder. In other words, if we divide an even number by 2, the result will be an integer.
If we divide 14 by 2, we get 14/2 = 7, which is an integer. Therefore, 14 is also an even number.
Since the sum of any two even numbers from the given set is also an even number, we can conclude that the set of even natural numbers is closed under addition.
Conclusion:
The set of even natural numbers satisfies the closure property under the addition operation.
The number of 2x2 matrices over Z3 (the field with three elements) with determinant 1 is- a)24
- b)60
- c)(a) 24 (b) 60
- d)(a) 24 (b) 60 (c) 20 (<7)30
Correct answer is option 'A'. Can you explain this answer?
The number of 2x2 matrices over Z3 (the field with three elements) with determinant 1 is
a)
24
b)
60
c)
(a) 24 (b) 60
d)
(a) 24 (b) 60 (c) 20 (<7)30
|
|
Veda Institute answered |
Consider the binary operation on X, a∗b=a+b+4, for a,b∈X. We are asked to determine the properties satisfied by this operation.
This operation satisfies the properties of an abelian group since it is closed under addition, associative, has an identity element (0), and every element has an inverse.
This operation satisfies the properties of an abelian group since it is closed under addition, associative, has an identity element (0), and every element has an inverse.
For n ε Z, Let nZ= {nk : k ε Z}. Then the number of units of Z/11Z and Z/12Z, respectively, are- a)11,12
- b)10, 11
- c)10, 4
- d)10, 8
Correct answer is option 'C'. Can you explain this answer?
For n ε Z, Let nZ= {nk : k ε Z}. Then the number of units of Z/11Z and Z/12Z, respectively, are
a)
11,12
b)
10, 11
c)
10, 4
d)
10, 8
|
|
Arghya answered |
Number of units of Z11 = φ(11) = 10.
Again
Number of units of Z12 = φ(12) = 4
The number of generators in group ({1,2,3,4,5,6} x7) are- a)4
- b)3
- c)2
- d)5
Correct answer is option 'C'. Can you explain this answer?
The number of generators in group ({1,2,3,4,5,6} x7) are
a)
4
b)
3
c)
2
d)
5
|
Aakash Singh answered |
The number of generators in group ({1,2,3,4,5,6} x7) area
To find the number of generators in the group ({1,2,3,4,5,6} x 7), we need to understand what a generator is in the context of group theory.
Group Theory and Generators
In group theory, a group is a set of elements along with an operation that combines any two elements to form a third element. The operation must satisfy certain properties such as closure, associativity, identity element, and inverse element.
A generator in a group is an element that, when combined with itself multiple times using the group operation, can generate all the elements of the group.
The Group ({1,2,3,4,5,6} x 7)
The group ({1,2,3,4,5,6} x 7) represents the Cartesian product of the set {1,2,3,4,5,6} with the group of integers modulo 7. This means that each element in the Cartesian product is a pair consisting of an element from the set {1,2,3,4,5,6} and an element from the group of integers modulo 7.
Calculating the Generators
To calculate the number of generators in the group ({1,2,3,4,5,6} x 7), we can use the following formula:
Number of generators = φ(n)
where φ(n) represents Euler's totient function, which gives the number of positive integers less than or equal to n that are relatively prime to n.
In this case, n = 7 since we are considering the group of integers modulo 7. Therefore, we need to calculate φ(7).
The Euler's totient function φ(n) for a prime number n is simply n-1. Since 7 is a prime number, φ(7) = 7-1 = 6.
Therefore, the number of generators in the group ({1,2,3,4,5,6} x 7) is 6.
Conclusion
The correct answer is option 'C', which states that there are 2 generators in the group ({1,2,3,4,5,6} x 7). This is determined by calculating the Euler's totient function φ(7), which is equal to 6. The number of generators in the group is equal to φ(7), so there are 6 generators.
To find the number of generators in the group ({1,2,3,4,5,6} x 7), we need to understand what a generator is in the context of group theory.
Group Theory and Generators
In group theory, a group is a set of elements along with an operation that combines any two elements to form a third element. The operation must satisfy certain properties such as closure, associativity, identity element, and inverse element.
A generator in a group is an element that, when combined with itself multiple times using the group operation, can generate all the elements of the group.
The Group ({1,2,3,4,5,6} x 7)
The group ({1,2,3,4,5,6} x 7) represents the Cartesian product of the set {1,2,3,4,5,6} with the group of integers modulo 7. This means that each element in the Cartesian product is a pair consisting of an element from the set {1,2,3,4,5,6} and an element from the group of integers modulo 7.
Calculating the Generators
To calculate the number of generators in the group ({1,2,3,4,5,6} x 7), we can use the following formula:
Number of generators = φ(n)
where φ(n) represents Euler's totient function, which gives the number of positive integers less than or equal to n that are relatively prime to n.
In this case, n = 7 since we are considering the group of integers modulo 7. Therefore, we need to calculate φ(7).
The Euler's totient function φ(n) for a prime number n is simply n-1. Since 7 is a prime number, φ(7) = 7-1 = 6.
Therefore, the number of generators in the group ({1,2,3,4,5,6} x 7) is 6.
Conclusion
The correct answer is option 'C', which states that there are 2 generators in the group ({1,2,3,4,5,6} x 7). This is determined by calculating the Euler's totient function φ(7), which is equal to 6. The number of generators in the group is equal to φ(7), so there are 6 generators.
f = (1 2 3) (1 2) is- a)odd permutation
- b)even permutation
- c)Both (a) and (b)
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
f = (1 2 3) (1 2) is
a)
odd permutation
b)
even permutation
c)
Both (a) and (b)
d)
None of these
|
|
Chirag Verma answered |
we have (1 2 3) (1 2) = (1 2) (1 3) (1 2) which is a product of odd number of transpositions.
Let G be a group of order 7 and φ(x) = x4, x ∈ G. Then f is - a)not one-one
- b)not onto
- c)not a homomorphism
- d)one-one, onto and a homomorphism
Correct answer is option 'D'. Can you explain this answer?
Let G be a group of order 7 and φ(x) = x4, x ∈ G. Then f is
a)
not one-one
b)
not onto
c)
not a homomorphism
d)
one-one, onto and a homomorphism
|
|
Chirag Verma answered |
A group of prime order must be cyclic and every cyclic group is abelian. Then we can show that φ: G → G s.t. φ(x) = xn is an isomorphism if 0(G) and n and are co-prime.
If G is a group and H is a subgroup of index 2 in G then choose the correct statement.- a)H is a normal subgroup of G
- b)H is not a normal subgroup of G
- c)H is a subgroup of G
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
If G is a group and H is a subgroup of index 2 in G then choose the correct statement.
a)
H is a normal subgroup of G
b)
H is not a normal subgroup of G
c)
H is a subgroup of G
d)
None of the above
|
|
Aakriti Yadav answered |
Explanation:
To understand why option A is the correct answer, we need to first understand the concept of a normal subgroup.
A subgroup H of a group G is said to be a normal subgroup if and only if for every element g in G, the conjugate of H by g is also a subset of H. In other words, for any h in H and g in G, the element ghg^(-1) is also in H.
Now, let's consider the given information: G is a group and H is a subgroup of index 2 in G.
What does it mean for H to have index 2 in G?
The index of a subgroup H in G is the number of distinct left cosets of H in G. In this case, since H has index 2 in G, there are exactly two distinct left cosets of H in G.
Proof that H is a normal subgroup of G:
Since H has index 2 in G, there are two distinct left cosets of H in G, denoted by H and gH, where g is an element of G but not in H.
Consider an element h in H and an arbitrary element g in G. We want to show that the element ghg^(-1) is also in H.
Case 1: gh is in H
If gh is in H, then ghg^(-1) is also in H since H is a subgroup. In this case, H is a normal subgroup of G.
Case 2: gh is not in H
If gh is not in H, then gh is in the coset gH. Since there are only two distinct cosets, the other coset must be H itself. Therefore, gH = H, which implies that g is in H.
Now, consider ghg^(-1). Since g is in H, we can write g = h' for some h' in H. Therefore, ghg^(-1) = h'hg^(-1). Since H is a subgroup, h'h is also in H. Thus, ghg^(-1) is in H.
In both cases, we have shown that for any h in H and any g in G, ghg^(-1) is in H. Therefore, H is a normal subgroup of G.
Hence, the correct statement is option A: H is a normal subgroup of G.
To understand why option A is the correct answer, we need to first understand the concept of a normal subgroup.
A subgroup H of a group G is said to be a normal subgroup if and only if for every element g in G, the conjugate of H by g is also a subset of H. In other words, for any h in H and g in G, the element ghg^(-1) is also in H.
Now, let's consider the given information: G is a group and H is a subgroup of index 2 in G.
What does it mean for H to have index 2 in G?
The index of a subgroup H in G is the number of distinct left cosets of H in G. In this case, since H has index 2 in G, there are exactly two distinct left cosets of H in G.
Proof that H is a normal subgroup of G:
Since H has index 2 in G, there are two distinct left cosets of H in G, denoted by H and gH, where g is an element of G but not in H.
Consider an element h in H and an arbitrary element g in G. We want to show that the element ghg^(-1) is also in H.
Case 1: gh is in H
If gh is in H, then ghg^(-1) is also in H since H is a subgroup. In this case, H is a normal subgroup of G.
Case 2: gh is not in H
If gh is not in H, then gh is in the coset gH. Since there are only two distinct cosets, the other coset must be H itself. Therefore, gH = H, which implies that g is in H.
Now, consider ghg^(-1). Since g is in H, we can write g = h' for some h' in H. Therefore, ghg^(-1) = h'hg^(-1). Since H is a subgroup, h'h is also in H. Thus, ghg^(-1) is in H.
In both cases, we have shown that for any h in H and any g in G, ghg^(-1) is in H. Therefore, H is a normal subgroup of G.
Hence, the correct statement is option A: H is a normal subgroup of G.
Consider the following relations, which of the following is an equivalence relation?
- a)m~n if and only if mn ≥ 0
- b)m~n if and only if mn > 0
- c)m~n if and only if |m| = |n|
- d)m~n if and only if mn ≤ 0
Correct answer is option 'A,C'. Can you explain this answer?
Consider the following relations, which of the following is an equivalence relation?
a)
m~n if and only if mn ≥ 0
b)
m~n if and only if mn > 0
c)
m~n if and only if |m| = |n|
d)
m~n if and only if mn ≤ 0
|
|
Chirag Verma answered |
The equivalence class of α under ~, denoted [α] [, is defined as [α] = { x ∈ X | x ~ α }
Set of rational number of the form m/2n (.m, n integers) is a group under- a)addition
- b)subtraction
- c)multiplication
- d)division
Correct answer is option 'A'. Can you explain this answer?
Set of rational number of the form m/2n (.m, n integers) is a group under
a)
addition
b)
subtraction
c)
multiplication
d)
division
|
|
Aryan Verma answered |
Group under Addition:
To show that the set of rational numbers of the form m/2n is a group under addition, we need to prove the following properties:
1. Closure:
For any two rational numbers m/2n and p/2q, their sum is (mq + np)/(2nq), which is also in the form m/2n. Therefore, the set is closed under addition.
2. Associativity:
For any three rational numbers m/2n, p/2q, and r/2s, the sum (m/2n + p/2q) + r/2s is equal to (mq + np)/(2nq) + r/2s, which can be simplified to (2nsq(mq + np) + 2nqrs)/(4nsq), and further simplified to (2n^2sqmq + 2n^2snp + 2nqrs)/(4nsq). Similarly, m/2n + (p/2q + r/2s) simplifies to (2nqrs(mq + np) + 2n^2snp + 2nqrs)/(4nsq). Since addition is associative for integers, the two expressions are equal. Therefore, the set is associative under addition.
3. Identity Element:
The identity element in this set is 0/1, since for any rational number m/2n, m/2n + 0/1 = (mq + 0)/(2n) = m/2n. Therefore, the set has an identity element.
4. Inverse Element:
For any rational number m/2n, its inverse is -m/2n, since m/2n + (-m/2n) = (mq + (-mq))/(2n) = 0/2n = 0/1, which is the identity element. Therefore, every element in the set has an inverse.
Since the set satisfies all the properties of a group under addition, we can conclude that the set of rational numbers of the form m/2n is a group under addition.
Non-Applicability to Other Operations:
The set of rational numbers of the form m/2n does not form a group under subtraction, multiplication, or division. This is because not all elements have inverses under these operations, which violates the property of having an inverse. For example, if we consider division, the element 1/2 does not have an inverse in the set, as there is no rational number m/2n such that (m/2n) * (1/2) = 1/1. Similarly, the set does not satisfy the closure property for subtraction and multiplication, as the result of these operations may not be in the form m/2n.
Therefore, the correct answer is option 'A' - the set of rational numbers of the form m/2n is a group under addition.
To show that the set of rational numbers of the form m/2n is a group under addition, we need to prove the following properties:
1. Closure:
For any two rational numbers m/2n and p/2q, their sum is (mq + np)/(2nq), which is also in the form m/2n. Therefore, the set is closed under addition.
2. Associativity:
For any three rational numbers m/2n, p/2q, and r/2s, the sum (m/2n + p/2q) + r/2s is equal to (mq + np)/(2nq) + r/2s, which can be simplified to (2nsq(mq + np) + 2nqrs)/(4nsq), and further simplified to (2n^2sqmq + 2n^2snp + 2nqrs)/(4nsq). Similarly, m/2n + (p/2q + r/2s) simplifies to (2nqrs(mq + np) + 2n^2snp + 2nqrs)/(4nsq). Since addition is associative for integers, the two expressions are equal. Therefore, the set is associative under addition.
3. Identity Element:
The identity element in this set is 0/1, since for any rational number m/2n, m/2n + 0/1 = (mq + 0)/(2n) = m/2n. Therefore, the set has an identity element.
4. Inverse Element:
For any rational number m/2n, its inverse is -m/2n, since m/2n + (-m/2n) = (mq + (-mq))/(2n) = 0/2n = 0/1, which is the identity element. Therefore, every element in the set has an inverse.
Since the set satisfies all the properties of a group under addition, we can conclude that the set of rational numbers of the form m/2n is a group under addition.
Non-Applicability to Other Operations:
The set of rational numbers of the form m/2n does not form a group under subtraction, multiplication, or division. This is because not all elements have inverses under these operations, which violates the property of having an inverse. For example, if we consider division, the element 1/2 does not have an inverse in the set, as there is no rational number m/2n such that (m/2n) * (1/2) = 1/1. Similarly, the set does not satisfy the closure property for subtraction and multiplication, as the result of these operations may not be in the form m/2n.
Therefore, the correct answer is option 'A' - the set of rational numbers of the form m/2n is a group under addition.
If in a group a5 = e, aba-1 = b2 for a,b ε G, then o(b) is - a)30
- b)31
- c)32
- d)33
Correct answer is option 'B'. Can you explain this answer?
If in a group a5 = e, aba-1 = b2 for a,b ε G, then o(b) is
a)
30
b)
31
c)
32
d)
33
|
Harsh Joshi answered |
In the given group, we have a^5 = e and aba^(-1) = b^2 for elements a and b.
To find the value of a^2, we can multiply both sides of aba^(-1) = b^2 by a^(-1) on the right:
aba^(-1)a^(-1) = b^2a^(-1)
ab = b^2a^(-1)
Now, we can substitute a^5 = e into this equation:
ab = b^2a^(-1)
ab = b^2a^(-1)a^5
ab = b^2a^4
Now, we can multiply both sides of ab = b^2a^4 by a:
aba = b^2a^5
aba = b^2e
aba = b^2
Since aba = b^2, we can substitute this into aba = b^2a^(-1):
aba = b^2a^(-1)
b^2 = b^2a^(-1)
Since b^2 = b^2a^(-1), we can cancel out b^2 from both sides:
e = a^(-1)
Therefore, a^(-1) is the identity element e.
To find the value of a^2, we can multiply both sides of aba^(-1) = b^2 by a^(-1) on the right:
aba^(-1)a^(-1) = b^2a^(-1)
ab = b^2a^(-1)
Now, we can substitute a^5 = e into this equation:
ab = b^2a^(-1)
ab = b^2a^(-1)a^5
ab = b^2a^4
Now, we can multiply both sides of ab = b^2a^4 by a:
aba = b^2a^5
aba = b^2e
aba = b^2
Since aba = b^2, we can substitute this into aba = b^2a^(-1):
aba = b^2a^(-1)
b^2 = b^2a^(-1)
Since b^2 = b^2a^(-1), we can cancel out b^2 from both sides:
e = a^(-1)
Therefore, a^(-1) is the identity element e.
The product of permutation (1,2,3) (2, 4, 3) (1, 3,4) is equal to- a)l
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The product of permutation (1,2,3) (2, 4, 3) (1, 3,4) is equal to
a)
l
b)
c)
d)
|
Curiosity With Wickedmin answered |
See the question first. Do correct Edurev! there are lots and lots of mistakes.
then find BA.
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