Cyclic Groups MCQ Quiz in मराठी - Objective Question with Answer for Cyclic Groups - मोफत PDF डाउनलोड करा

Explanations:
The external direct product of the additive group of integers Z with itself, denoted Z × Z, is not a cyclic group.

In a cyclic group, every element can be expressed as a power (or multiple, in the case of additive groups) of a single element. However, in Z × Z, no single ordered pair of integers (a, b) can generate all possible ordered pairs through addition, so Z × Z is not cyclic.

However, it is abelian, because the group operation (addition, in this case) is commutative: (a, b) + (c, d) = (a+c, b+d) = (c, d) + (a, b).

Hence, Option 2 is Correct.

Concept Used:

A group G is a cyclic group if there exists an element a ∈ G such that G=[a]
i.e. every element of G can be expressed as some integral power of a.
a is called the generator of G.
G={ . . . ,a^-3, a^-2, a^-1, a⁰, a1 ,a2, a3. . .}

De Moivre's theorem :  ,where n ∈ Z

Euler's Identity: 

Calculation:

G is a Cyclic group: 
 

[By De Moivre's theorem

[By Euler's Identity]

 is the generator of the cyclic group G.

S₁: If a group (G, *) is of order n, and a ∈ G is such that a^m = e for some integer m ≤ n, then m must divide n.

Given statement is correct. As a ∈ G , is such that a^m = e for some integer m

S₂: If a group (G, *) is of even order, then there must be an element a ∈ G such that a ≠ e and a * a = e

This statement is correct.  Consider an example for this : Consider G is of order 2n. There exists a ∈ G such that a^p = e and p divides 2n. Let n = pq. So, (aⁿ)² = (a^pq)² = ((a^p)^q)² = (e^q)² = e . It means aⁿ is an element which satisfy the condition and a =! e. Here, a is non trivial subgroup of G.

Explanation:

The external direct product of additive group of integers Z with itself, commonly denoted as Z × Z, is a mathematical structure where the group operation is defined componentwise.

If we adopt addition as the operation in Z, then the operation in Z × Z will be pair addition, i.e., (a, b) + (c, d) = (a + c, b + d).

A cyclic group is a group that is generated by a single element. In Z × Z, no individual pair of integers can generate all pairs through repeated addition, because each pair only generates a "line" of pairs, so to speak, and cannot cover the entire "plane" of pairs.

So Z × Z is not cyclic. Thus, the answer is 1) not cyclic.

The group Z₄₉₅ under addition modulo 495

{0, 99, 198, 307, 406} is the unique subgroup of Z₄₉₅ of order 5.

It is true.

{0, 55, 110, 165, 220, 275, 330, 385, 440} is the unique subgroup of Z₄₉₅ of order 9.

It is true.

Explanation:

A cycle of length two is called a transposition.

Cyclic permutation:  A permutation of the form (m₁m₂ ... m_k) is called a cyclic permutation (or cycle) of length k. 

If (m₁m₂ ... m_k) denotes the permutation that carries m₁ into m₂, m₂ into m₃, ... , m_k-1 into m_k and m_k into m₁ and leaves all the remaining elements of the set in question unchanged.

Transposition:  A cyclic permutation of length 2.

Disjoint cycles. Two or more cycles which have no element in common are called (mutually) disjoint.

• It follows closure property
• It is associative
• 1 is the identify element
• Inverse exists for every element
•  i and -i are generators

Since group contains generator it is a cyclic group

Note:

Structure is semi group, group and cyclic group but best possible option is cyclic group since it semi group as well as a group.

Identity (e) = 1

From 1 we cannot generate ω and ω²

∴ it is not a generator

(ω)¹ = ω, (ω)² = ω² and (ω)³ = 1

∴ it is a generator

(ω²)¹ = ω², (ω²)² = ω and (ω²)³ = 1

∴ it is generator

The order of an element of a group is the smallest positive integer such that . Where is the identity element of the group.

The elements of G are .

Hence

OR

O(a⁸) =  =  = 5

If G = ({1, 2... 12}, ×13 )

It is a multiplicative modulo 13

∴ G is group

Identity element = e = 1

a × a^-1 = e (where e is identity element)

×₁₃  is given operator 

(2 × 2^-1 )mod13 = 1 

∴ 2^-1 = 7

(3 × 3-1 )mod13 = 1 

∴ 3^-1 = 9