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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?
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.
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.
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.
Number of elements of the cyclic group of order 6 can be used as generators of the group are- a)= 3
- b)= 5
- c)= 2
- d)4
Correct answer is option 'C'. Can you explain this answer?
Number of elements of the cyclic group of order 6 can be used as generators of the group are
a)
= 3
b)
= 5
c)
= 2
d)
4
|
|
Shalini Robinson answered |
Generator of a number n is less than n and relatively prime to n. so the generators are 1 and 5 answer is 2
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
|
Scholars Academy answered |
- Identify Key Terms:
The question refers to a theorem stating that every finite group G is isomorphic to a permutation group. - Analyze the Options:
- Lagrange’s Theorem (B): Concerns subgroup orders dividing the group order. Irrelevant here.
- Liouville’s Theorem (C): Pertains to complex analysis, unrelated to group theory.
- Cayley’s Theorem (A): States that every group G is isomorphic to a subgroup of the symmetric group Sn (permutation group).
- None of these (D): Incorrect because Cayley’s Theorem directly matches the statement.
-
- Conclusion:
Cayley’s Theorem confirms the given statement.
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 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.
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.
Consider the group 2 x 2 non-singular matrices under matrix multiplication over Z5. The inverse of 
is- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Consider the group 2 x 2 non-singular matrices under matrix multiplication over Z5. The inverse of is
isa)
b)
c)
d)
|
|
Veda Institute answered |
Given the group of 2 x 2 non-singular matrices under matrix multiplication over Z5.
The inverse of
i.e.
over Z5.
The inverse of
i.e.
over Z5.
If G is a Prime order group then G has - a)No Proper Subgroup
- b)No Improper Subgroup
- c)Two Improper Subgroup
- d)None Of These
Correct answer is option 'A'. Can you explain this answer?
If G is a Prime order group then G has
a)
No Proper Subgroup
b)
No Improper Subgroup
c)
Two Improper Subgroup
d)
None Of These
|
Devika Choudhary answered |
Explanation:
A Prime order group is a group in which the order of the group is a prime number. In this case, we need to determine whether a Prime order group has proper subgroups or not.
Definition:
A proper subgroup of a group G is a subgroup that is not equal to G itself.
Proof:
To show that a Prime order group has no proper subgroups, we need to consider the definition of a subgroup and the properties of a Prime order group.
1. Definition of a subgroup:
A subgroup H of a group G is a non-empty subset of G that is closed under the group operation and the inverse operation. In other words, if a and b are elements of H, then the product ab and the inverse a^(-1) are also elements of H.
2. Properties of a Prime order group:
- A Prime order group G has only two subgroups: the trivial subgroup {e} (containing only the identity element) and G itself.
- The order of the subgroup H of a group G must divide the order of G. In other words, if the order of G is p (a prime number), then the order of any subgroup H of G must be 1 (trivial subgroup) or p (G itself).
Proof by contradiction:
Assume that there exists a proper subgroup H of a Prime order group G. Since H is a subgroup of G, the order of H must divide the order of G (which is a prime number). This means that the order of H can only be 1 or p.
- If the order of H is 1, then H is the trivial subgroup {e}.
- If the order of H is p, then H is equal to G itself.
In either case, H is not a proper subgroup, which contradicts our assumption. Therefore, a Prime order group has no proper subgroups.
Conclusion:
The correct answer is option A: No Proper Subgroup. A Prime order group does not have any proper subgroups.
A Prime order group is a group in which the order of the group is a prime number. In this case, we need to determine whether a Prime order group has proper subgroups or not.
Definition:
A proper subgroup of a group G is a subgroup that is not equal to G itself.
Proof:
To show that a Prime order group has no proper subgroups, we need to consider the definition of a subgroup and the properties of a Prime order group.
1. Definition of a subgroup:
A subgroup H of a group G is a non-empty subset of G that is closed under the group operation and the inverse operation. In other words, if a and b are elements of H, then the product ab and the inverse a^(-1) are also elements of H.
2. Properties of a Prime order group:
- A Prime order group G has only two subgroups: the trivial subgroup {e} (containing only the identity element) and G itself.
- The order of the subgroup H of a group G must divide the order of G. In other words, if the order of G is p (a prime number), then the order of any subgroup H of G must be 1 (trivial subgroup) or p (G itself).
Proof by contradiction:
Assume that there exists a proper subgroup H of a Prime order group G. Since H is a subgroup of G, the order of H must divide the order of G (which is a prime number). This means that the order of H can only be 1 or p.
- If the order of H is 1, then H is the trivial subgroup {e}.
- If the order of H is p, then H is equal to G itself.
In either case, H is not a proper subgroup, which contradicts our assumption. Therefore, a Prime order group has no proper subgroups.
Conclusion:
The correct answer is option A: No Proper Subgroup. A Prime order group does not have any proper subgroups.
The multiplicative group {1, -1} is a subgroup of the multiplicative group- a){1, i, -i}
- b){1,-1, i, -i}
- c){1,0, -1, -i}
- d){-1, i, -i}
Correct answer is option 'B'. Can you explain this answer?
The multiplicative group {1, -1} is a subgroup of the multiplicative group
a)
{1, i, -i}
b)
{1,-1, i, -i}
c)
{1,0, -1, -i}
d)
{-1, i, -i}
|
Kabir Shah answered |
The given question belongs to the field of abstract algebra, specifically the study of groups. Let's break down the question and explain the answer in detail.
Understanding the problem:
We are given a set of elements and we need to determine if it forms a subgroup of another given group. In this case, we have two sets: the multiplicative group {1, -1} and the multiplicative group {1,-1, i, -i}. We need to determine if the first set is a subgroup of the second set.
Definition of a subgroup:
A subgroup is a subset of a group that is itself a group under the same operation. In other words, a subgroup is a smaller group contained within a larger group.
Checking if {1, -1} is a subgroup of {1,-1, i, -i}:
To determine if {1, -1} is a subgroup of {1,-1, i, -i}, we need to check the following conditions:
1. Closure: The subgroup must be closed under the group operation. In this case, the group operation is multiplication. We can see that multiplying any two elements from {1, -1} always gives us an element from {1, -1}, so closure is satisfied.
2. Identity: The subgroup must contain the identity element of the larger group. The identity element of the multiplicative group {1,-1, i, -i} is 1. Since 1 is present in {1, -1}, this condition is satisfied.
3. Inverses: Every element in the subgroup must have an inverse in the subgroup. Both 1 and -1 have themselves as inverses, so this condition is satisfied.
Conclusion:
All the conditions for a subgroup are satisfied by {1, -1} in the larger group {1,-1, i, -i}. Therefore, {1, -1} is a subgroup of {1,-1, i, -i}.
Answer:
The correct answer is option 'B' - {1,-1, i, -i}.
Understanding the problem:
We are given a set of elements and we need to determine if it forms a subgroup of another given group. In this case, we have two sets: the multiplicative group {1, -1} and the multiplicative group {1,-1, i, -i}. We need to determine if the first set is a subgroup of the second set.
Definition of a subgroup:
A subgroup is a subset of a group that is itself a group under the same operation. In other words, a subgroup is a smaller group contained within a larger group.
Checking if {1, -1} is a subgroup of {1,-1, i, -i}:
To determine if {1, -1} is a subgroup of {1,-1, i, -i}, we need to check the following conditions:
1. Closure: The subgroup must be closed under the group operation. In this case, the group operation is multiplication. We can see that multiplying any two elements from {1, -1} always gives us an element from {1, -1}, so closure is satisfied.
2. Identity: The subgroup must contain the identity element of the larger group. The identity element of the multiplicative group {1,-1, i, -i} is 1. Since 1 is present in {1, -1}, this condition is satisfied.
3. Inverses: Every element in the subgroup must have an inverse in the subgroup. Both 1 and -1 have themselves as inverses, so this condition is satisfied.
Conclusion:
All the conditions for a subgroup are satisfied by {1, -1} in the larger group {1,-1, i, -i}. Therefore, {1, -1} is a subgroup of {1,-1, i, -i}.
Answer:
The correct answer is option 'B' - {1,-1, i, -i}.
Consider the following statements:
1. A group of order 289 is abelian.
2. In S3 there are four elements satisfying x2= e and six elements satisfying y3 = e.
3. Every proper subgroup of S3 is cyclic.
4. (Z30, 
has 4 subgroups of order 15 .
Correct statements are:- a)l and 4
- b)1 ,3 and 4
- c)1 and 2
- d)1 and 3
Correct answer is option 'D'. Can you explain this answer?
Consider the following statements:
1. A group of order 289 is abelian.
2. In S3 there are four elements satisfying x2= e and six elements satisfying y3 = e.
3. Every proper subgroup of S3 is cyclic.
4. (Z30,has 4 subgroups of order 15 .
Correct statements are:
1. A group of order 289 is abelian.
2. In S3 there are four elements satisfying x2= e and six elements satisfying y3 = e.
3. Every proper subgroup of S3 is cyclic.
4. (Z30,
has 4 subgroups of order 15 .Correct statements are:
a)
l and 4
b)
1 ,3 and 4
c)
1 and 2
d)
1 and 3
|
|
Chirag Verma answered |
every group of order p2 (p -> prime) is always abelian group.
(1) true
(1) true
In S3 → total no. of elements are 3 of order 2 i.e. x2 = e and identity element also satisfying this condition total elements are 4. but only 3 elements are exists which satisfies the condition y3 = e.
(3) A3 is proper subgroup of S3 which is cyclic.
(3) A3 is proper subgroup of S3 which is cyclic.
The set of all positive rational numbers forms an abelian group under the composition defined by- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The set of all positive rational numbers forms an abelian group under the composition defined by
a)
b)
c)
d)
|
Sakshi Jain answered |
Correct ans is A as the
2) don't hold associativity
3) commutative doesn't hold
4) commutative doesn't hold
If n is the order of element a of group G, then am = e, an identity element if- a)m | n
- b)n | m
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
If n is the order of element a of group G, then am = e, an identity element if
a)
m | n
b)
n | m
c)
d)
|
Scintifik World answered |
If order of a is n. and a^m= e. then n divieds m.
example.- U(10)={1,3,79}
3^4=81=1(mod10)
3^8=6561=1(mod10)
4 divided 8
i.e.8/4
example.- U(10)={1,3,79}
3^4=81=1(mod10)
3^8=6561=1(mod10)
4 divided 8
i.e.8/4
If a, b ∈ G, a group of order m, then order of ab and ba are- a)same
- b)equal to m
- c)unequal
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
If a, b ∈ G, a group of order m, then order of ab and ba are
a)
same
b)
equal to m
c)
unequal
d)
None of these
|
Akanksha Kapoor answered |
If a, b are variables, there are many possible mathematical operations that can be performed with them. Some common operations include addition (a + b), subtraction (a - b), multiplication (a * b), and division (a / b). Additionally, variables can be used in more complex operations such as exponentiation (a^b) or taking the square root (√a). The specific operation that should be performed with a and b depends on the context and what you are trying to achieve.
The number of elements of S5 (the symmetric group on 5 letters) which are their own inverses equals
- a)26
- b)10
- c)11
- d)25
Correct answer is option 'A'. Can you explain this answer?
The number of elements of S5 (the symmetric group on 5 letters) which are their own inverses equals
a)
26
b)
10
c)
11
d)
25
|
|
Chirag Verma answered |
Le S5 = {a1, a2, a3, a4, a5} be a symmetric group of order 5. the elements of S5 which are their own inverses are of the type (a1, a2) or (a1, a2) (a2, a4).
The number of elements of the type (a1, a2) are
The number of elements of the type (a3, a4) are
So the total number of elements which are their own inverses is equal to 25.
The number of elements of the type (a3, a4) are
So the total number of elements which are their own inverses is equal to 25.
Let A = {1,2,3,4}, B = { 2,4,5 } and C = A ∩ B then number of symmetric relation on C is- a)3
- b)8
- c)2
- d)4
Correct answer is option 'B'. Can you explain this answer?
Let A = {1,2,3,4}, B = { 2,4,5 } and C = A ∩ B then number of symmetric relation on C is
a)
3
b)
8
c)
2
d)
4
|
Bhavya Reddy answered |
It seems you have written the statement incomplete. Please provide the complete statement.
If H is a subgroup of finite group G and order of H and G are respectively, m and n, then- a)m | n
- b)n | m
- c)
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
If H is a subgroup of finite group G and order of H and G are respectively, m and n, then
a)
m | n
b)
n | m
c)
d)
None of these
|
Rex Awomi answered |
Lagrange theorem states that if G is finite and H is a subgroup of G then O(H) | O(G)
that means m divides n therefore option a is correct
that means m divides n therefore option a is correct
Which of the following group is cyclic ?- a)S3
- b)Z5
- c)D4
- d)K4
Correct answer is option 'B'. Can you explain this answer?
Which of the following group is cyclic ?
a)
S3
b)
Z5
c)
D4
d)
K4
|
|
Veda Institute answered |
Concept used:
A group is cyclic iff there exists an element of order whose order is equal to the order of the group
Calculations:
o(S3) = 3!
contains 6 elements of the Symmetric group S(3), of degree 3, written in cycles are : {(1), (12), (13), (23), (123), (321)}.
o(1) = 1
o(12), o(23) and o(13) = 2
o(123), and o(321) = 3
no element having order equal to order of group
∴ option 1 is incorrect.
o(D4) = 8
D4 = {1, r, r2, r3 , s, sr, sr2, sr3}
o(s), o(sr), o(sr2), o(sr3) = 2
o(1) = 1, o(r) = 2 , o(r2) = 3, o(r3) = 4
no element having order equal to order of the group
∴ option 3 is incorrect.
o(K4) = 4
K4 = {e, a, b, ab}
o(e) = 1 , o(a) = 2, o(b) = 2, o(ab) = 2
No element having order equal to the order of group
∴ option 4 is incorrect.
o(Z5) = 5
Z5 = {0, 1, 2, 3, 4}
o(0) = 1 o(1) = 5, o(2) = 5 , o(3) = 5 , o(4) = 5
all elements having order = order of the group
all are the generator of the group.
∴ option 2 is correct.
If a, b ∈ G, a group, then b is conjugate to a, if there exist c ∈ G, such that- a)b = c-1ac
- b)a = cb
- c) b = ac-1
- d)b = cc-1a
Correct answer is option 'A'. Can you explain this answer?
If a, b ∈ G, a group, then b is conjugate to a, if there exist c ∈ G, such that
a)
b = c-1ac
b)
a = cb
c)
b = ac-1
d)
b = cc-1a
|
Omkar Rana answered |
And c are real numbers, then the equation ax^2 + bx + c = 0 represents a quadratic equation.
If A = {1,2,3} then the number of equivalence relation on A is- a)3
- b)7
- c)5
- d)8
Correct answer is option 'A'. Can you explain this answer?
If A = {1,2,3} then the number of equivalence relation on A is
a)
3
b)
7
c)
5
d)
8
|
|
Chirag Verma answered |
ANSWER :- a
Solution :- R1 = {(1, 1), (2, 2), (3, 3)}
R2= {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
R3 = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)}
R4= ((1, 1), (2, 2), (3, 3), (2, 3), (3, 2)}
R5 = {(1,2,3)4=.AxA =A2} Maximum number of equivalence relation is '5'
A set contains 2n + 1 elements. The number of subsets of this set containing more than n elements is equal to- a)2n-1
- b)2n
- c)2n+1
- d)22n
Correct answer is option 'D'. Can you explain this answer?
A set contains 2n + 1 elements. The number of subsets of this set containing more than n elements is equal to
a)
2n-1
b)
2n
c)
2n+1
d)
22n
|
Aditi Singh answered |
Explanation:
To solve this problem, we need to understand the concept of subsets and combinatorics.
Understanding Subsets:
A subset is a set that contains elements from another set. For example, if we have a set A = {1, 2, 3}, then the subsets of A are {}, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, and {1, 2, 3}. The number of subsets of a set with n elements is 2^n.
Understanding the Problem:
We are given a set with 2n+1 elements. We need to find the number of subsets of this set that contain more than n elements.
Solution:
Let's consider the set S with 2n+1 elements. We can divide this set into two parts: the first part contains n elements and the second part contains n+1 elements.
Case 1: Choosing n+1 elements from the second part
In this case, we need to select n+1 elements from the second part of the set. The number of ways to do this is given by the binomial coefficient (n+1)C(n+1) = 1.
Case 2: Choosing n elements from the first part
In this case, we need to select n elements from the first part of the set. The number of ways to do this is given by the binomial coefficient nCn = 1.
Case 3: Choosing k elements from the first part, where k > n
In this case, we need to select k elements from the first part of the set, where k > n. The number of ways to do this is given by the sum of all binomial coefficients from n+1 to 2n+1, inclusive.
Sum = (n+1)C(n+1) + (n+1)C(n+2) + (n+1)C(n+3) + ... + (n+1)C(2n+1)
Using the identity (r-1)Cr + rCr = (r+1)Cr+1, we can simplify the above sum as follows:
Sum = (n+2)C(n+1) + (n+3)C(n+1) + ... + (2n+1)C(n+1)
= (2n+1+1)C(n+1) - (n+1)C(n+1)
= (2n+2)C(n+1) - 1C1
= (2n+2)C(n+1) - 1
Therefore, the total number of subsets that contain more than n elements is given by the sum of the cases:
Total = Case 1 + Case 2 + Case 3
= 1 + 1 + ((2n+2)C(n+1) - 1)
= (2n+2)C(n+1)
Now, we know that the number of subsets of a set with 2n+1 elements is 2^(2n+1). Therefore, the number of subsets that contain more than n elements is given by:
Total = (2
To solve this problem, we need to understand the concept of subsets and combinatorics.
Understanding Subsets:
A subset is a set that contains elements from another set. For example, if we have a set A = {1, 2, 3}, then the subsets of A are {}, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, and {1, 2, 3}. The number of subsets of a set with n elements is 2^n.
Understanding the Problem:
We are given a set with 2n+1 elements. We need to find the number of subsets of this set that contain more than n elements.
Solution:
Let's consider the set S with 2n+1 elements. We can divide this set into two parts: the first part contains n elements and the second part contains n+1 elements.
Case 1: Choosing n+1 elements from the second part
In this case, we need to select n+1 elements from the second part of the set. The number of ways to do this is given by the binomial coefficient (n+1)C(n+1) = 1.
Case 2: Choosing n elements from the first part
In this case, we need to select n elements from the first part of the set. The number of ways to do this is given by the binomial coefficient nCn = 1.
Case 3: Choosing k elements from the first part, where k > n
In this case, we need to select k elements from the first part of the set, where k > n. The number of ways to do this is given by the sum of all binomial coefficients from n+1 to 2n+1, inclusive.
Sum = (n+1)C(n+1) + (n+1)C(n+2) + (n+1)C(n+3) + ... + (n+1)C(2n+1)
Using the identity (r-1)Cr + rCr = (r+1)Cr+1, we can simplify the above sum as follows:
Sum = (n+2)C(n+1) + (n+3)C(n+1) + ... + (2n+1)C(n+1)
= (2n+1+1)C(n+1) - (n+1)C(n+1)
= (2n+2)C(n+1) - 1C1
= (2n+2)C(n+1) - 1
Therefore, the total number of subsets that contain more than n elements is given by the sum of the cases:
Total = Case 1 + Case 2 + Case 3
= 1 + 1 + ((2n+2)C(n+1) - 1)
= (2n+2)C(n+1)
Now, we know that the number of subsets of a set with 2n+1 elements is 2^(2n+1). Therefore, the number of subsets that contain more than n elements is given by:
Total = (2
is
If G is a finite group of order n, a ∈ G and order of a is m 7M, if G is cylic, then- a)m = n
- b)m > n
- c)m < n
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
If G is a finite group of order n, a ∈ G and order of a is m 7M, if G is cylic, then
a)
m = n
b)
m > n
c)
m < n
d)
None of these
|
|
Arjun Mehta answered |
If G is a finite group of order n, a subgroup H of G is a subset of G that is also a group under the same operation. In other words, H is a subgroup of G if it satisfies the following conditions:
1. H is non-empty: There exists at least one element in H.
2. H is closed under the group operation: For any two elements a, b in H, their product ab is also in H.
3. H is closed under inverses: For any element a in H, its inverse a^(-1) is also in H.
4. H is closed under taking powers: For any element a in H and any positive integer k, the kth power of a, denoted by a^k, is also in H.
Furthermore, Lagrange's theorem states that the order of a subgroup H must divide the order of the parent group G. In other words, if the order of G is n and the order of H is m, then m must divide n, denoted by m | n.
For example, if G is a finite group of order 12, the possible orders of subgroups H can be 1, 2, 3, 4, 6, or 12.
1. H is non-empty: There exists at least one element in H.
2. H is closed under the group operation: For any two elements a, b in H, their product ab is also in H.
3. H is closed under inverses: For any element a in H, its inverse a^(-1) is also in H.
4. H is closed under taking powers: For any element a in H and any positive integer k, the kth power of a, denoted by a^k, is also in H.
Furthermore, Lagrange's theorem states that the order of a subgroup H must divide the order of the parent group G. In other words, if the order of G is n and the order of H is m, then m must divide n, denoted by m | n.
For example, if G is a finite group of order 12, the possible orders of subgroups H can be 1, 2, 3, 4, 6, or 12.
If in a group G, a ∈ G, the order of that is n and order of aP is m, then- a)m < n
- b)m > n
- c)m = n
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
If in a group G, a ∈ G, the order of that is n and order of aP is m, then
a)
m < n
b)
m > n
c)
m = n
d)
None of these
|
Mohit Chauhan answered |
If in a group G, a
If in a group G, a is an element of G and a^2 = e, where e is the identity element of G, then a is an involution in G.
If in a group G, a is an element of G and a^2 = e, where e is the identity element of G, then a is an involution in G.
A necessary and sufficient condition for a non-empty subset H o f a finite group G to be a subgroup is that
- a)a ∈ H, b ∉ H which implies a, b ∈ H
- b)a ∈ H, b ∈ H ⇒ (a + b) ∈ H
- c)a, b ∈ H ⇒ ab-1 ∈ H
- d)a ∈ H, h ∈ H ⇒ (a - h) ∈ H
Correct answer is option 'C'. Can you explain this answer?
A necessary and sufficient condition for a non-empty subset H o f a finite group G to be a subgroup is that
a)
a ∈ H, b ∉ H which implies a, b ∈ H
b)
a ∈ H, b ∈ H ⇒ (a + b) ∈ H
c)
a, b ∈ H ⇒ ab-1 ∈ H
d)
a ∈ H, h ∈ H ⇒ (a - h) ∈ H
|
|
Ananya Patel answered |
A) a) H is closed under the group operation of G.
b) H contains the identity element of G.
c) H is closed under taking inverses.
d) H is closed under the group operation of G and contains the identity element of G.
e) H is closed under the group operation of G, contains the identity element of G, and is closed under taking inverses.
The correct answer is e) H is closed under the group operation of G, contains the identity element of G, and is closed under taking inverses. This is because a subgroup must satisfy all three conditions in order to be considered a subgroup.
b) H contains the identity element of G.
c) H is closed under taking inverses.
d) H is closed under the group operation of G and contains the identity element of G.
e) H is closed under the group operation of G, contains the identity element of G, and is closed under taking inverses.
The correct answer is e) H is closed under the group operation of G, contains the identity element of G, and is closed under taking inverses. This is because a subgroup must satisfy all three conditions in order to be considered a subgroup.
If G = {1, -1, i, -i} is a multiplicative group, then order of - i is- a)one
- b)two
- c)three
- d)four
Correct answer is option 'D'. Can you explain this answer?
If G = {1, -1, i, -i} is a multiplicative group, then order of - i is
a)
one
b)
two
c)
three
d)
four
|
|
Vanya Singh answered |
Explanation:
Multiplicative Group:
A multiplicative group is a set of elements together with an operation (usually denoted as *) that satisfies the following properties:
1. Closure: For any two elements a and b in the set, their product a * b is also in the set.
2. Associativity: For any three elements a, b, and c in the set, the product (a * b) * c is equal to a * (b * c).
3. Identity Element: There exists an element e in the set such that for any element a in the set, the product a * e is equal to a.
4. Inverse Element: For any element a in the set, there exists an element a^-1 in the set such that the product a * a^-1 is equal to the identity element e.
Givens:
The given multiplicative group IfG = {1, -1, i, -i} consists of four elements: 1, -1, i, and -i.
Order of an Element:
The order of an element in a group is the smallest positive integer n such that raising the element to the power n gives the identity element.
Determining the Order of -i:
To determine the order of -i in the multiplicative group IfG, we need to find the smallest positive integer n such that (-i)^n = 1.
Calculating the powers of -i:
(-i)^1 = -i
(-i)^2 = -1
(-i)^3 = i
(-i)^4 = 1
From the calculations, we can see that (-i)^4 = 1, which means that the order of -i is 4.
Therefore, the correct answer is option D, which states that the order of -i is four.
Multiplicative Group:
A multiplicative group is a set of elements together with an operation (usually denoted as *) that satisfies the following properties:
1. Closure: For any two elements a and b in the set, their product a * b is also in the set.
2. Associativity: For any three elements a, b, and c in the set, the product (a * b) * c is equal to a * (b * c).
3. Identity Element: There exists an element e in the set such that for any element a in the set, the product a * e is equal to a.
4. Inverse Element: For any element a in the set, there exists an element a^-1 in the set such that the product a * a^-1 is equal to the identity element e.
Givens:
The given multiplicative group IfG = {1, -1, i, -i} consists of four elements: 1, -1, i, and -i.
Order of an Element:
The order of an element in a group is the smallest positive integer n such that raising the element to the power n gives the identity element.
Determining the Order of -i:
To determine the order of -i in the multiplicative group IfG, we need to find the smallest positive integer n such that (-i)^n = 1.
Calculating the powers of -i:
(-i)^1 = -i
(-i)^2 = -1
(-i)^3 = i
(-i)^4 = 1
From the calculations, we can see that (-i)^4 = 1, which means that the order of -i is 4.
Therefore, the correct answer is option D, which states that the order of -i is four.
If set A is empty then cardinality of the set P ( p ( P ( A ) ) ) is- a)6
- b)16
- c)2
- d)4
Correct answer is option 'D'. Can you explain this answer?
If set A is empty then cardinality of the set P ( p ( P ( A ) ) ) is
a)
6
b)
16
c)
2
d)
4
|
|
Ganu Dhaliwal answered |
Power set of a set contains 2 raise to power n where n is the no. of elements in the set.
so here set A is emply so n= 0
P(A)= 2⁰= 1
P(P(A))=2¹=2
P(P(P(A)))=2²=4
so here set A is emply so n= 0
P(A)= 2⁰= 1
P(P(A))=2¹=2
P(P(P(A)))=2²=4
A survey shows that 63 % of Americans like cheese whereas 76% like apples. If x % of the Americans like both cheese and apples, then- a)x = 39
- b)x = 63
- c)39 ≤ x ≤ 63
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
A survey shows that 63 % of Americans like cheese whereas 76% like apples. If x % of the Americans like both cheese and apples, then
a)
x = 39
b)
x = 63
c)
39 ≤ x ≤ 63
d)
none of these
|
|
Ayush Gupta answered |
To find the percentage of Americans who like both cheese and apples, we can add the percentages of Americans who like cheese and apples and then subtract the total percentage of Americans who like either cheese or apples.
Let's assume there are 100 Americans in total.
63% of Americans like cheese, so that means 63 out of 100 Americans like cheese.
76% of Americans like apples, so that means 76 out of 100 Americans like apples.
To find the percentage of Americans who like both cheese and apples, we add 63 and 76:
63 + 76 = 139.
However, we have counted some Americans twice, so we need to subtract the percentage of Americans who like either cheese or apples.
The total percentage of Americans who like either cheese or apples is given by the equation:
63 + 76 - x = 100.
Rearranging the equation, we have:
139 - x = 100.
Simplifying the equation, we have:
-x = -39.
Multiplying both sides by -1, we have:
x = 39.
Therefore, x = 39.
So the answer is option c) 39.
Let's assume there are 100 Americans in total.
63% of Americans like cheese, so that means 63 out of 100 Americans like cheese.
76% of Americans like apples, so that means 76 out of 100 Americans like apples.
To find the percentage of Americans who like both cheese and apples, we add 63 and 76:
63 + 76 = 139.
However, we have counted some Americans twice, so we need to subtract the percentage of Americans who like either cheese or apples.
The total percentage of Americans who like either cheese or apples is given by the equation:
63 + 76 - x = 100.
Rearranging the equation, we have:
139 - x = 100.
Simplifying the equation, we have:
-x = -39.
Multiplying both sides by -1, we have:
x = 39.
Therefore, x = 39.
So the answer is option c) 39.
Let G be a group order 6, and H be a subgroup of G such that 1 < |H| < 6. Which one of the following options is correct?
- a)G is always cyclic, but H may not be cyclic.
- b)
Both G and H are always cyclic.
- c)G may not be cyclic, but H is always cyclic.
- d)Both G and H may not be cyclic.
Correct answer is option 'C'. Can you explain this answer?
Let G be a group order 6, and H be a subgroup of G such that 1 < |H| < 6. Which one of the following options is correct?
a)
G is always cyclic, but H may not be cyclic.
b)
Both G and H are always cyclic.
c)
G may not be cyclic, but H is always cyclic.
d)
Both G and H may not be cyclic.
|
Dhruv Kapoor answered |
Based on the given information, we know that the order of group G is 6. This means that G has 6 elements.
We also know that H is a subgroup of G. Since H is a subgroup, it must also be a group.
Now, let's consider the possible orders of H. Since H is a subgroup of G, the order of H must divide the order of G. In this case, the order of G is 6.
The possible orders of H are therefore 1, 2, 3, or 6.
If the order of H is 1, then H contains only the identity element of G. However, H cannot be the trivial subgroup since the given condition states that H is nontrivial. Therefore, the order of H cannot be 1.
If the order of H is 2, then H contains 2 elements. Since H is a subgroup of G, it must also contain the identity element of G. Therefore, H must contain one more element. However, this is not possible since the order of G is 6, and H can have at most 2 elements. Therefore, the order of H cannot be 2.
If the order of H is 3, then H contains 3 elements. Since H is a subgroup of G, it must also contain the identity element of G. Therefore, H must contain two more elements. However, this is not possible since the order of G is 6, and H can have at most 3 elements. Therefore, the order of H cannot be 3.
If the order of H is 6, then H contains 6 elements. Since H is a subgroup of G, it must also contain the identity element of G. Therefore, H must contain five more elements. However, this is not possible since the order of G is 6, and H can have at most 6 elements. Therefore, the order of H cannot be 6.
Since none of the possible orders of H satisfy the given condition, it is not possible to have a subgroup H of G such that the order of H is 1.
We also know that H is a subgroup of G. Since H is a subgroup, it must also be a group.
Now, let's consider the possible orders of H. Since H is a subgroup of G, the order of H must divide the order of G. In this case, the order of G is 6.
The possible orders of H are therefore 1, 2, 3, or 6.
If the order of H is 1, then H contains only the identity element of G. However, H cannot be the trivial subgroup since the given condition states that H is nontrivial. Therefore, the order of H cannot be 1.
If the order of H is 2, then H contains 2 elements. Since H is a subgroup of G, it must also contain the identity element of G. Therefore, H must contain one more element. However, this is not possible since the order of G is 6, and H can have at most 2 elements. Therefore, the order of H cannot be 2.
If the order of H is 3, then H contains 3 elements. Since H is a subgroup of G, it must also contain the identity element of G. Therefore, H must contain two more elements. However, this is not possible since the order of G is 6, and H can have at most 3 elements. Therefore, the order of H cannot be 3.
If the order of H is 6, then H contains 6 elements. Since H is a subgroup of G, it must also contain the identity element of G. Therefore, H must contain five more elements. However, this is not possible since the order of G is 6, and H can have at most 6 elements. Therefore, the order of H cannot be 6.
Since none of the possible orders of H satisfy the given condition, it is not possible to have a subgroup H of G such that the order of H is 1.
If (G, *) is a group and for all a, b ∈ G, b-1 * a-1* b * a = e, then G is- a)abelian group
- b)non group
- c)ring
- d)field
Correct answer is option 'A'. Can you explain this answer?
If (G, *) is a group and for all a, b ∈ G, b-1 * a-1* b * a = e, then G is
a)
abelian group
b)
non group
c)
ring
d)
field
|
|
Sahil Singhania answered |
If (G, *) is a group and for all a, b in G, the equation a * x = b has a unique solution for x in G, then the group (G, *) is called a group with unique left inverses.
The number of elements in the set {m : 1 ≤ m ≤ 1000, m and 1000 are relatively prime} is- a)1000
- b)250
- c)300
- d)400
Correct answer is option 'D'. Can you explain this answer?
The number of elements in the set {m : 1 ≤ m ≤ 1000, m and 1000 are relatively prime} is
a)
1000
b)
250
c)
300
d)
400
|
|
Veda Institute answered |
A non empty set A is called an algebraic structure w.r.t binary operation “*” if (a*b) belongs to S for all (a*b) belongs to S. Therefore “*” is closure operation on ‘A’.
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