Test: Group Theory - 5 - Mathematics MCQ

The inverse of an even permutation is always an even permutation.

Here's why:

• A permutation is called even if it can be expressed as the product of an even number of transpositions (two-element swaps).
• A permutation is called odd if it can be expressed as the product of an odd number of transpositions.

Now, when you take the inverse of a permutation:

• If a permutation is even, the number of transpositions it takes to express it is even.
• Reversing the sequence of transpositions (which makes up the inverse) does not change the parity (even or odd) of the number of transpositions. Hence, the inverse will also consist of an even number of transpositions, meaning it remains even.

Thus, the inverse of an even permutation is also an even permutation.

By the theorem, the order of any permutation ζ in Sn is equal to the l.c.m. of the orders of the disjoint cycles in ζ. Here, = (1 2 4 5)(3 6).
The order of the permutation is l.c.m. of (1 2 4 5) and (3 6).
l.c.m. (4, 2) = 4
as o(1 2 4 5) = 4 and o(3, 6) 

• A cyclic group G of order n has elements generated by powers of a single element a, called the generator. For example, G = {a, a²,a³,…,aⁿ = e}, where e is the identity.

• For a^p to be a generator of G, the order of a^p must be equal to n, the order of the group. This happens when p is coprime (relatively prime) to n.

  • If p shares any common divisor with n other than 1, the order of a^p would be a divisor of n, and a^p would fail to generate the entire group.
  • Conversely, if p is relatively prime to nnn, a^p will traverse all elements of the group and thus be a generator.

• Therefore, the condition p is prime to n ensures that a^p is a generator.

Key Concepts:

1. Symmetric Group S_m​: The group of all permutations of mmm elements.
2. Order of a Permutation (∣P∣): The least common multiple of the lengths of disjoint cycles in the permutation.
3. Odd Prime Order: A permutation P is in K_m​ if the order of P is an odd prime number.

Subgroup Criteria:

A subset K_m​ is a subgroup of S_m​ if it satisfies the subgroup criteria:

1. Closure under composition: If P, Q ∈ K_m then P∘ Q ∈ K_m​.
2. Existence of an identity element.
3. Existence of inverses for all elements in K_m​.

Analysis:

For K_m​ to be a subgroup, the following must hold:

• The orders of permutations in K_m​ must remain odd primes after composition. However, the order of a composition of two permutations is generally not guaranteed to be a prime (or odd).

From group theory, K_m​ forms a subgroup only under specific conditions. For m ≥ 3m, K_m​ cannot satisfy the subgroup criteria unless m is prime and sufficiently structured.

Let G be a group and a ∈ G such that the order of a is n. By definition, this means:

aⁿ = e(the identity element in G).

Now, consider a^P, where P is a number prime to n (i.e., gcd⁡(P, n) = 1

Key Points:

1. The order of a^P is the smallest positive integer m such that:

   (a^P)^m = e

2. Expanding this:

   a^P⋅m = e

3. Since the order of a is n, we know a^k = e if and only if k is a multiple of n. Therefore:

   P⋅m must be a multiple of n.

4. Because P and nnn are coprime (gcd⁡(P, n) = 1 there exists a unique integer m = n that satisfies this condition.

Conclusion:

The order of a^P is exactly n.

- To find the generators of a group G = {a, a², a³, a⁴ = e} , we check which elements generate the entire group.
- The order of the group is 4. Generators are elements whose order equals the group's order.
- (a) has order 4, as (a⁴ = e ).
- (a²) has order 2, as (a²)² = e).
- ( a³) also has order 4, since ((a³)⁴ = e).
- Hence, both (a) and (a³) are generators.

The question states that G is a group of order p², where p is a prime number. The goal is to determine a property of G.

Key Points:

1. Group Order and Structure:

   • By the Sylow Theorems, if the order of a finite group G is p², then G is either cyclic or isomorphic to the direct product of two cyclic groups of order p.
   • Both of these cases result in G being Abelian.

2. Cyclic and Abelian Groups:

   • A cyclic group of any order is Abelian.
   • A direct product of cyclic groups of prime order is also Abelian because their elements commute.

3. Conclusion:

   • A group of order p² is always Abelian, regardless of its specific structure.

Correct Answer:

Option A: G is Abelian.

- The given relation is  (34 × 78) × 57 = 34 × (78 × 57).
- This equation demonstrates the associative property.
- The associative property states that the way numbers are grouped in multiplication (or addition) does not affect the result.
- In this case, changing the grouping of the numbers being multiplied does not change the product.
- Therefore, the correct answer is A: associative.

The question asks about the inverse of an odd permutation.

Key Points:

1. Parity of a Permutation:

   • A permutation is called odd if it can be expressed as the product of an odd number of transpositions (two-element swaps).
   • A permutation is even if it is expressed as the product of an even number of transpositions.

2. Properties of the Inverse:

   • The inverse of a permutation has the same parity as the original permutation.
   • This is because reversing the sequence of transpositions (to form the inverse) does not change the total count of transpositions, which determines the parity.

3. Application:

   • If a permutation is odd, its inverse must also be odd because the parity remains unchanged.

Conclusion:

The inverse of an odd permutation is always an odd permutation.

Let

Let

⇒ AB = BA ∀ A, B ∈ R

⇒ R₁ is commutative if AB = 0
⇒ A = 0 or B = 0
⇒ R₁ has no zero divisors

We are analyzing two statements about cyclic groups:

Statement A: All cyclic groups are abelian.

• A cyclic group is generated by a single element g, and every element in the group can be written as a power of g, i.e., gⁿ.
• In a cyclic group, the group operation involves powers of g, and the order of multiplication does not matter because g^a ⋅ g^b = g^a+b = g^b ⋅ g^a 
• Thus, all cyclic groups are commutative (abelian).
• Statement A is true.

Statement B: The order of a cyclic group is the same as the order of its generator.

• The order of a group is the number of elements in it.
• The order of the generator g is the smallest positive integer n such that gⁿ = e (identity element).
• In a cyclic group, the order of the generator g is exactly equal to the order of the group because g generates all elements of the group.
• Statement B is true.

Key Points:

1. Definition of Composition:

   • The composition g∘fmeans applying f first and then g. In other words, for an element x, find f(x) first, and then apply g to the result.
2. Step-by-Step Computation:
   • Start with each symbol 1,2,3,4,5 and determine where it maps under f and then g.
   • For 1:
     • Under f: 1 remains 1 (since 1 is not involved in (2 3)
     • Under g: 1 remains 1 (since 1 is not involved in (4 5).
     • Final result: 1→1.
   • For 2:
     • Under f: 2→3.
     • Under g: 3 remains 3 (since 3 is not involved in (4, 5).
     • Final result: 2→3.
   • For 3:
     • Under f: 3→2.
     • Under g: 2 remains 2 (since 2 is not involved in (4. 5).
     • Final result: 3→2.
   • For 4:
     • Under f: 4 remains 444 (since 4 is not involved in (2, 3).
     • Under g: 4→5.
     • Final result: 4→5.
   • For 5:
     • Under f: 5 remains 5 (since 5 is not involved in (2, 3).
     • Under ggg: 5→4.
     • Final result: 5→4.
3. Resulting Permutation: Combining all results, we get:
   

we have = (135) (24) = (13) (15) (24)

- A cyclic group is generated by a single element.
- An isomorphic image of a group maintains structural similarity, including cyclic nature.
- A homomorphic image is formed by applying a homomorphism; for cyclic groups, the image of the generator will generate the image group.
- Thus, both isomorphic and homomorphic images of a cyclic group are cyclic.
- Therefore, both statements A and B are true.
- Answer: D (Both A and B are true).

In the context of groups, an idempotent element is defined as an element e such that:

e ⋅ e = e

Key Points:

1. Identity Element in a Group:

   • The identity element e of a group satisfies e ⋅ e = e by definition, as e ⋅ e = e for all operations with e.

2. Other Elements in Groups:

   • In general, no other elements in a group satisfy the property a ⋅ a = a unless a = e, because groups require invertibility and associativity.

Conclusion:

The identity element is the only idempotent element in a group.

The correct answer is A: m = n.
- In group theory, a left coset of a subgroup (H) in a group (G) is a set formed by multiplying each element of (H) by a fixed element from the left. Similarly, a right coset is formed by multiplying each element of (H) by a fixed element from the right.
- The number of left cosets is equal to the number of right cosets, both equal to the index of (H) in G), denoted as ([G : H]).
- This is a fundamental result of the Lagrange's theorem in group theory.

The given permutation is written in two-line notation:

This describes how each element in the first row (top) maps to the corresponding element in the second row (bottom). We need to convert this to cycle notation, which groups elements into cycles that describe the mapping.

Step-by-Step Conversion:

1. Start with the smallest unused number, 1:

   • 1→6,
   • 6→3,
   • 3→2,
   • 2→1 (cycle closes).
   • First cycle: (1 6 3 2).

2. Move to the next smallest unused number, 4:

   • 4→5,
   • 5→4 (cycle closes).
   • Second cycle: (4 5)

3. Combine the cycles: The permutation in cycle notation is:
   (1 6 3 2)(4 5)

Key Concept:

A group of prime order has a very specific structure due to the properties of groups and Lagrange's theorem.

Details:

1. Order of a Group:

   • The order of a group is the number of elements in the group.
   • If a group has prime order p, the only divisors of p are 1 and p itself.

2. Lagrange's Theorem:

   • The order of any subgroup of a group must divide the order of the group.
   • For a group of prime order, the only possible subgroups are the trivial subgroup ({e}) and the group itself.

3. Cyclic Nature:

   • In a group G of prime order, every non-identity element generates the entire group. Thus, G is cyclic.

4. Abelian Nature:

   • A cyclic group is always abelian, as the group operation can be expressed in terms of powers of the generator.

Conclusion:

Every group of prime order is cyclic.

 we have C = (1234567) = 
which implies C² = 
which implies C³ = 
= 1 4 7 3 6 2 5 
Hence C³ = (1 4 7 3 6 2 5)

The question is about the properties of cosets of a subgroup H in a group G. Specifically, it addresses the intersection of two right cosets.

Key Points:

1. Definition of Right Cosets:

   • A right coset of a subgroup H in a group G is of the form Hx = {hx∣h ∈ H} in H}, where x ∈ G
2. Properties of Right Cosets:
   • Right cosets partition the group G. This means:
     1. Any two right cosets are either disjoint or identical.
     2. Formally, for two right cosets H₁ = Hx₁ and H₂ = Hx₂: H₁ ∩ H₂ = ∅ or H₁ = H₂.​
3. Reasoning:
   • If H₁ ∩ H2 ≠ ∅, then there exists some element g ∈ H₁ ∩ H₂​. This implies:
     g ∈ Hx₁ and g ∈ Hx₂.
   • Therefore, the two cosets must be identical: H₁ = H₂.
   • Otherwise, the two cosets have no common elements: H₁ ∩ H₂ = ∅.
4. Conclusion:
   • For any two right cosets H_1​ and H₂​, the intersection H₁ ∩ H₂​ is either empty (∅) or they are identical (H₁ = H₂​).