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Test: Group Theory - 3 - Mathematics MCQ


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Test: Group Theory - 3 - Question 1

 If N is a set of natural numbers, then under binary operation  a · b = a + b, (N, ·) is

Detailed Solution for Test: Group Theory - 3 - Question 1

Test: Group Theory - 3 - Question 2

The number of generators in cyclic group of order 10 are

Detailed Solution for Test: Group Theory - 3 - Question 2

The number of generators in a cyclic group of order 10 can be determined by identifying the elements that can generate the entire group through their powers.

  • A cyclic group of order 10 has elements represented as integers modulo 10: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
  • To be a generator, an element must be relatively prime to 10. This means the greatest common divisor (GCD) of the element and 10 must be 1.
  • The integers that meet this criterion are:
    • 1 (GCD(1, 10) = 1)
    • 3 (GCD(3, 10) = 1)
    • 7 (GCD(7, 10) = 1)
    • 9 (GCD(9, 10) = 1)
  • Thus, the generators of the cyclic group of order 10 are 1, 3, 7, and 9.

In conclusion, there are four generators for a cyclic group of order 10.

Test: Group Theory - 3 - Question 3

The set of all positive rational numbers forms an abelian group under the composition defined by

Test: Group Theory - 3 - Question 4

Set (1,2,3,4} is a finite abelian group of order... under multiplication modulo ... as composition.

Detailed Solution for Test: Group Theory - 3 - Question 4
The set {1, 2, 3, 4} forms a finite abelian group under multiplication modulo 5. Each element has an inverse in the set: - 1-1 ≡ 1 mod 5 - 2-1 ≡ 3 mod 5 (since 2×3=6≡1) - 3-1 ≡ 2 mod 5 - 4-1 ≡ 4 mod 5 (since 4×4=16≡1). Thus, the order is 4 and modulus 5.
Test: Group Theory - 3 - Question 5

Let G be a group of order 7 and φ(x) = x4, x ∈ G. Then f is 

Detailed Solution for Test: Group Theory - 3 - Question 5

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.

Test: Group Theory - 3 - Question 6

HK is a sub-group of G iff

Detailed Solution for Test: Group Theory - 3 - Question 6
For HK to be a subgroup, every element must belong to both HK and KH, implying equality.
Test: Group Theory - 3 - Question 7

Check the correct statement.

Detailed Solution for Test: Group Theory - 3 - Question 7
Statement A is true because every subgroup of a cyclic group is cyclic. Statement B is correct as an infinite cyclic group has exactly two generators and is isomorphic to the integers under addition. Statement C holds because any finite group with composite order has proper subgroups. Therefore, all statements are correct, making D the right answer.
Test: Group Theory - 3 - Question 8

If a, b ∈ G, a group, then b is conjugate to a, if there exist c ∈ G, such that

Detailed Solution for Test: Group Theory - 3 - Question 8
- Conjugation Definition: In a group \( G \), element \( b \) is conjugate to \( a \) if there exists an element \( c \) in \( G \) such that \( b = c^{-1}ac \).
- Option 1 Analysis:
- \( b = c^{-1}ac \) directly matches the conjugation definition.
- It correctly shows how \( b \) is obtained by conjugating \( a \) with \( c \).
- Conclusion: Therefore, Option 1 is the correct expression for \( b \) being conjugate to \( a \).
Test: Group Theory - 3 - Question 9

If H1 and H2 are two subgroups of G, then following is also a subgroups of G

Detailed Solution for Test: Group Theory - 3 - Question 9
  • Intersection H₁ ∩ H₂: Always a subgroup. It contains elements common to both H₁ and H₂, satisfying closure and inverses.
  • Union H₁ ∪ H₂: Not necessarily a subgroup unless one is contained in the other.
  • Product H₁H₂: Not always a subgroup unless specific conditions are met.
  • None of these: Incorrect since the intersection is always a subgroup.
Test: Group Theory - 3 - Question 10

If (G, *) is a group and for all a, b ∈ G, b-1 * a-1* b * a = e, then G is

Detailed Solution for Test: Group Theory - 3 - Question 10
Given the condition b-1a-1ba = e for all a, b in G, we recognize this as the commutator of a and b being the identity. This implies that ab = ba for all a, b ∈ G, meaning G is an abelian group.
Test: Group Theory - 3 - Question 11

Number of elements of the cyclic group of order 6 can be used as generators of the group are

Detailed Solution for Test: Group Theory - 3 - Question 11

Here, 6 = 2 x 3

Test: Group Theory - 3 - Question 12

The multiplicative group {1, -1} is a subgroup of the multiplicative group

Detailed Solution for Test: Group Theory - 3 - Question 12
To determine which group contains {1, -1} as a subgroup, we analyze each option: - Option A: {1, i, -i}. This set is not closed under multiplication (e.g., i2 = -1, which isn't in the set), so it's not a group. - Option B: {1, -1, i, -i}. This set satisfies all group properties: - Identity: 1 is present. - Inverses: Each element has its inverse within the set. - Closure: All products remain within the set. Thus, it's a valid multiplicative group. - Option C: {1, 0, -1, -i}. The presence of 0 disqualifies it as a multiplicative group because 0 lacks an inverse. - Option D: {-1, i, -i}. It lacks the identity element (1), so it's not a group. Only Option B forms a valid multiplicative group and contains {1, -1} as a subgroup.
Test: Group Theory - 3 - Question 13

A set G with a binary composition denoted multiplicative is a group, if

Detailed Solution for Test: Group Theory - 3 - Question 13
To determine if a set G with a binary operation forms a group, four properties must be satisfied: closure, associativity, identity, and inverses. The question does not mention closure, which is essential for a group. Without closure, even if conditions A (associativity) and B (implying identity and inverses) are met, G cannot be a group. Thus, the correct answer is D: None of the above.
Test: Group Theory - 3 - Question 14

In the additive group of integers, the order of every elements a ≠  0 is

Detailed Solution for Test: Group Theory - 3 - Question 14

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.

Test: Group Theory - 3 - Question 15

Let Z be a set of integers, then under ordinary multiplication (Z, ·) is

Detailed Solution for Test: Group Theory - 3 - Question 15
The set of integers Z under ordinary multiplication satisfies the properties of a monoid. Here's why: 1. Closure: For any two integers a and b, their product a·b is also an integer. 2. Associativity: For all a, b, c in Z, (a·b)·c = a·(b·c). 3. Identity Element: The integer 1 serves as the multiplicative identity since a·1 = a for any integer a. While every element does not have a multiplicative inverse in Z, the presence of an identity element and associativity make (Z, ·) a monoid rather than just a semi-group.
Test: Group Theory - 3 - Question 16

Set of all n, nth roots of unity from a finite abelian group of order n with respect to

Detailed Solution for Test: Group Theory - 3 - Question 16

The set of all nth roots of unity from a finite abelian group of order n can be understood in the context of different operations.

  • Addition: In this operation, the roots combine to form new elements, but they do not maintain the properties of the roots of unity.
  • Subtraction: Similar to addition, subtracting roots does not preserve the unity characteristic.
  • Multiplication: This is the key operation. When you multiply the nth roots of unity, you obtain another nth root of unity, preserving the structure of the group.
  • Division: While division can be performed, it does not yield a consistent result that maintains the property of being a root of unity.

The multiplication of these roots is significant because it allows us to remain within the set of nth roots of unity, whereas addition and subtraction do not. Therefore, the most appropriate operation to consider with respect to the nth roots of unity is multiplication.

Test: Group Theory - 3 - Question 17

The generators of a group G = {a, a2, a3, a4, a5, a6 = e) are

Detailed Solution for Test: Group Theory - 3 - Question 17

we have o(G) = 6 and prime to 6 are 1 and 5

Test: Group Theory - 3 - Question 18

If G is a group such that a2 = e, for all a ∈ G, then G is

Detailed Solution for Test: Group Theory - 3 - Question 18
Given that G is a group where every element satisfies a2 = e, we know each element is its own inverse, i.e., a = a-1. To check if G is abelian, consider any two elements a and b in G: (ab)2 = abab = e. Multiplying both sides on the left by a and then by b, we get: a-1b-1 = ba. Since each element is its own inverse (a-1 = a), this simplifies to: ab = ba. Thus, G is abelian.
Test: Group Theory - 3 - Question 19

The value of k for which kx + 3y - k + 3 = 0 and 12x + ky = k have infinite solution is:

Detailed Solution for Test: Group Theory - 3 - Question 19



Test: Group Theory - 3 - Question 20

If G is a group, then for all a, b ∈ G

Detailed Solution for Test: Group Theory - 3 - Question 20
In group theory, the inverse of the product of two elements a and b is given by reversing the order of multiplication and taking the inverses individually. Therefore, (ab)-1 = b-1a-1.
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