Cyclic Groups MCQ Quiz in தமிழ் - Objective Question with Answer for Cyclic Groups - இலவச PDF ஐப் பதிவிறக்கவும்

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 : \((\cos x+i \sin x)^n=\cos n x+i \sin n x\) ,where n ∈ Z

Euler's Identity: \(e^{i\theta}=\cos \theta + i\sin \theta\)

Calculation:

G is a Cyclic group: \( \therefore 1=\cos 2 m \pi+\sin 2 m \pi, \quad m ∈ Z\)
\( \therefore 1^{1 / n}=\cos \frac{2 m \pi}{n}+i \sin \frac{2 m \pi}{n}, \) \(\quad m=0,1,2,3, \ldots .(n-1)\)

[By De Moivre's theorem
\(\left.=e^{2 m \pi i /n}, \quad m=0,1,2,3, \ldots,(n-1)\right]\)
[By Euler's Identity]

\(\therefore G=\left\{e^{2 \pi / n}, e^{4 \pi i/ n}, e^{6 \pi i/ n}, \ldots \ldots, e^{2 n \pi i / n}\right\}\)
\(=\left[e^{2 \pi i /n}\right] \) 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<=n, then it means it is a subgroup of G and m is the order of a. where order of given group is n. By the property of subgroup or Lagrange’s theorem, a order of subgroup divides the order of a group. So, here m must divide n is true.

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.

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 \(m\) such that \(a^m = e\). Where \(e\) is the identity element of the group.

The elements of G are \({e, a^1, a^2, a^3, a^4, a^5, a^6, a^7, a^8, a^9}\).

\(O(e) = 1, O(a)=10, O(a^2)=5\ and \ so \ on\)

Hence \((a^8)^5=a^{10}a^{10}a^{10}a^{10}=e\)

OR

O(a⁸) = \(\frac{10}{gcd(8,10)}\) = \(\frac{10}{2}\) = 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