Test: Group Theory - 6 - Mathematics MCQ

• understanding Element Order:
  The order of an element a in G is the smallest positive integer m such that a^m = e.
  Similarly, the order of a^-1 is the smallest positive integer n such that (a^-1)ⁿ = e.

• Key Relationship:

  • If a^m = e, then taking the inverse of both sides gives (a^-1)^m = e. Thus, n divides m.
  • Similarly, since a = (a^-1)^-1, the order of a divides n.
  • From these two, m = n.

• Examples to Verify:

  • In additive groups (e.g., Z₅) or multiplicative groups (e.g., symmetric group S₃), the order of an element and its inverse are always equal.
  • For instance, in S₃, a 3-cycle and its inverse both have order 3; a transposition (self-inverse) has order 2.

• Conclusion:
  The order of a and a^-1 must be equal in any group G. Option C is invalid because orders are positive integers (never 0).

In a finite group G, the order (number of elements) of G is finite. By Lagrange's Theorem, the order of any element a in G (the smallest positive integer n such that a^n = e, where e is the identity) must divide the order of G. Since G is finite, every element a in G must have a finite order.

Conclusion:
All elements in a finite group have finite orders. The correct answer is A.

A one-to-one (injective) mapping of a finite group onto itself is called an automorphism.

• Automorphism: An automorphism is a bijective (both one-to-one and onto) homomorphism from a group to itself. Since the mapping is one-to-one and onto, it satisfies the definition of an automorphism.

• Isomorphism: An isomorphism is a bijective homomorphism between two groups, but in this case, we are dealing with a group mapping onto itself, so "automorphism" is the more specific term.

• Homomorphism: A homomorphism is a function between two groups that preserves the group operation, but it doesn't have to be bijective. Since we are specifically talking about a bijective mapping, the term "automorphism" is more appropriate.

Thus, the correct answer is:

C: automorphism.

In the additive group of integers G = (Z, +), the inverse element of an integer a is -a, since adding a and -a gives the identity element 0, i.e., a + (-a) = 0.

Order of an element in a group

• The order of an element a in a group is the smallest positive integer n such that n * a = 0.
• In the additive group of integers Z, no integer other than 0 has a finite order because for any nonzero integer a, n * a = 0 does not happen for any finite n.

Since the inverse of any element a in Z is -a, and the operation is addition, the inverse element -a also has infinite order (because no nonzero integer can add up to zero except for zero itself).

Thus, the correct answer is:

C: infinity.

⇒ AB = BA ∀ A, B ∈ R
⇒ R1 is commutative if AB = 0
⇒ A = 0 or B = 0
⇒ R1 has no zero divisors
 

The normalizer of an element a in a group G, denoted as N(a), is the set of elements x in G such that xax⁻¹ = a. This means that x conjugates a to itself.

In other words, if x belongs to the normalizer N(a), the relationship that holds is:

x * a = a * x

This is the condition for x being in the normalizer of a, because conjugation by x does not change a.

Thus, the correct answer is:

A: xa = ax.

(1 2 3)(1 2)= (1 3) odd permutation

(1 2 3 4 5)(1 2 3)(4 5)=(4 1 3 2)= (4 2)(4 3)(4 1) odd permutation

(1 2)(1 3)(1 4)(2 5)= even permutation.

correct option should (1 2)(1 3)(1 4)(2 5)

The product of even permutation is even permutation 

eg : 4 * 6 = 24

Correct answer is C. H is not isomorphic to K, since there is no homomorphism from H to K. 

Since wk that every group of order p² is abelian,
where p is a prime integer.
∴ Group of order p² is abelian also.
∴ Group of order 7²

 i.e. 49 is abelian and cyclic.

The correct statement is:

D: Every quotient group of an abelian group is abelian but the converse is not true.

Explanation

• Every quotient group of an abelian group is abelian: This is true. If the original group G is abelian, then any normal subgroup N of G will also make the quotient group G/N abelian. This is because the operation in G/N is induced by the operation in G, and if G is abelian, the commutative property carries over to the quotient group.

• The converse is not true: This means that even though a quotient group is abelian, the original group may not necessarily be abelian. For example, the group S3 (the symmetric group on 3 elements) is non-abelian, but its quotient by a normal subgroup (like the alternating group A3) is abelian.

Thus, the statement D correctly captures both parts of the truth: every quotient group of an abelian group is abelian, but not every abelian quotient group implies the original group is abelian.

• If H is a subgroup of G and the index of H in G is 2 (i.e., [G : H] = 2), then there are exactly two cosets of H in G: the identity coset H itself, and one other coset, which we can denote as gH for some g ∈ G.

• A subgroup of index 2 is always normal in the group. This is because the left cosets and right cosets of H in G must coincide. In other words, for any g ∈ G, the cosets gH and Hg must be the same. This property is what defines a normal subgroup.

Therefore, if the index of H in G is 2, H is always a normal subgroup of G.

Thus, the correct answer is:

A: H is a normal subgroup of G.

A singular element can generate a cyclic group. Every element of a cyclic group is a power of some specific element which is known as a generator ‘g’.

The set of complex numbers {1, i, -i, -1} under multiplication operation is a cyclic group. Two generators i and -i will covers all the elements of this group. Hence, it is a cyclic group.

• H ∩ K is a subgroup: The intersection of two subgroups H and K is always a subgroup. This is because the intersection of two subgroups is closed under the group operation (if x, y ∈ H ∩ K, then x * y is also in both H and K), contains the identity element (since both subgroups contain the identity), and includes the inverse of each element (if x ∈ H ∩ K, then x⁻¹ ∈ H ∩ K).

• HK may not be a subgroup: The product of two subgroups H and K, denoted HK, is not necessarily a subgroup. For HK to be a subgroup, it needs to be closed under the group operation. In general, HK is only a subgroup if H and K are normal subgroups of G. If H and K are not normal subgroups, HK may fail to be closed under the group operation, meaning HK may not form a subgroup.

Thus, the correct answer is:

B: H ∩ K is a subgroup but HK may not be a subgroup.

1. Structure of (Z, +):
   The group Z under addition is an infinite cyclic group generated by 1 (or -1).

2. Automorphisms of Cyclic Groups:

   • For a cyclic group G, an automorphism is uniquely determined by its action on a generator.
   • In Z, the generators are 1 and -1.

3. Possible Automorphisms:

   • Identity Automorphism: Maps 1 → 1. This induces φ(n) = n for all n in Z.
   • Negation Automorphism: Maps 1 → -1. This induces φ(n) = -n for all n in Z.

4. No Other Automorphisms Exist:

   • Suppose φ(1) = m. For φ to be surjective, m must generate Z. The only generators of Z are ±1.
   • If φ(1) ≠ ±1, the image φ(Z) would be a proper subgroup (e.g., 2Z), violating surjectivity.

5. Conclusion:
   There are exactly 2 automorphisms of Z: the identity and negation maps.

Final Answer:
B: 2

If N1 and N2 are normal subgroups of G, then the quotient groups G/N1 and G/N2 are well-defined. The condition G/N1 = G/N2 means that the coset structure of G modulo N1 is the same as the coset structure of G modulo N2.

For the quotient groups to be equal, the normal subgroups N1 and N2 must be the same. Here's why:

• If G/N1 = G/N2, then the set of cosets of N1 in G is identical to the set of cosets of N2 in G. This implies that N1 and N2 must be the same normal subgroup because cosets are determined by the normal subgroup.

• If N1 and N2 are distinct normal subgroups, their quotient groups G/N1 and G/N2 will not be identical, because their coset structures would differ.

Thus, the condition G/N1 = G/N2 holds if and only if N1 = N2.

Therefore, the correct answer is:

C: N1 = N2.

1. Group Order and Sylow Subgroups:
   Given |G| = 200 = 2³ x  5², we seek the number of subgroups of order 25 = 5². These are Sylow 5-subgroups.

2. Sylow's Third Theorem:
   The number of Sylow 5-subgroups, n_5, must satisfy:

   • n₅ divides ( |G| / 5² ) = 200 / 25 = 8,
   • n₅ ≡ 1 mod 5.

3. Possible Values for n_5:

   • Divisors of 8: 1, 2, 4, 8.
   • Values congruent to 1 mod 5: Only 1.

4. Conclusion:

   • n₅ = 1, meaning there is exactly one Sylow 5-subgroup of order 25.
   • This subgroup is unique and normal in G.

Final Answer:
A