Burnside Theorem Questions and Answers

This set of Discrete Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Groups – Burnside Theorem”.

1. Which of the following is not an abelian group?
a) semigroup
b) dihedral group
c) trihedral group
d) polynomial group
View Answer

Answer: b
Explanation: The dihedral group(Dih₄) of order 8 is a non-abelian p-group. But, every group of order p² must be abelian group.

2. If we take a collection of {∅, {2}, {3}, {5}} ordered by inclusion. Which of the following is true?
a) isomorphic graph
b) poset
c) lattice
d) partially ordered set
View Answer

Answer: b
Explanation: This is a poset. Since {2}, {3} and {5} have no common upper bound, it is not a lattice.

3. _______ characterizes the properties of distributive lattices.
a) Congruence Extension Property
b) Algebraic extension property
c) Poset
d) Semigroup
View Answer

Answer: b
Explanation: An algebra A describes the congruence extension property (CEP) if for every B≤A and θ∈Con(B) there exists a φ∈Con(A) such that θ = φ∩(B×B). A class M of algebras has the CEP if every algebra in the class has the CEP. The Congruence Extension Property specifically characterizes the distributive lattices among all lattices.

4. Suppose that H be an X-set and suppose that a∼b and |X_a|=|X_b|, the which of the following is true?
a) X_a is powerset of X_b
b) X_a is isomorphic to X_b
c) X_a is homomorphic to X_b
d) X_b is the subset of X_a
View Answer

Answer: b
Explanation: According to Burnside theorem, X_a is isomorphic to X_b and in particular |X_a|=|X_b|.

5. If he 4 sides of a square are to be colored by colors. How many different colourings with 50 colours are there if two arrangements that can be obtained from each other by rotation are identical?
a) 773762
b) 363563
c) 4536822
d) 1563150
View Answer

Answer: d
Explanation: There are m⁴ + m² + 2m elements after performing all rotations. Dividing this by the number of transformations 4 produces the desired number of distinct colorings \(\frac{m^4 + m^2 + 2m}{4}\). Hence, the number of distinct colorings with 50 colors is 1563150.

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6. Let H be a finite group. The order of Sylow p-subgroup of H for every prime factor p with multiplicity 9 is?
a) p+6
b) p⁹
c) p^p
d) 3!*p²
View Answer

Answer: b
Explanation: We know that, for a finite group H, there exists a Sylow p-subgroup of H having order p⁹ for every prime factor p with multiplicity 9.

7. How many indistinguishable necklaces can be made from beads of 4 colors with exactly 9 beads of each color where each necklace is of length 16?
a) 76967234
b) 5652209
c) 14414400
d) 8686214
View Answer

Answer: c
Explanation: If B is the set of all possible permutations of these 16 beads, then the required answer is |B| = 16!/(9!)4 = 14414400.

8. Invariant permutations of two functions can form __________
a) groups
b) lattices
c) graphs
d) rings
View Answer

Answer: a
Explanation: Suppose, there are two functions f₁ and f₂ which belong to the same equivalence class since there exists an invariant permutation say, π(a permutation that does not change the object itself, but only its representation), such that: f₂*π≡f₁. So, invariant permutations can form a group, as the product (composition) of invariant permutations is again an invariant permutation.

9. Suppose P(h) is a group of permutations and identity permutation(id) belongs to P(c). If ϕ(c)=c then which of the following is true?
a) ϕ^-1∈P(h)
b) ϕ^-1∈P(h)
c) ϕ^-1∈P(h)
d) ϕ^-1∈P(h)
View Answer

Answer: b
Explanation: Let, ϕ and σ both can fix h, then we can have ϕ(σ(h)) = ϕ(h) = h. Hence, ϕ∘σ fixes h and ϕ∘σ∈P(h). Now, all colorings can be fixed by the identity permutation. So id∈P(h) and if ϕ(h) = h then ϕ^-1(h) = ϕ^-1(ϕ(h)) = id(h) = h which implies that ϕ^-1∈P(h).

10. An isomorphism of Boolean algebra is defined as _______
a) order isomorphism
b) unordered isomorphism
c) order homomorphism
d) hyper-morphism
View Answer

Answer: a
Explanation: We know that very σ-complete Boolean algebra is a Boolean algebra. An isomorphism of Boolean algebras is termed as an order isomorphism. All meets and joins present in an order isomorphism domain is preserved. Hence, a Boolean algebra isomorphism preserves all meets and joins in its domain.

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