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

Explanation: The dihedral group(Dih

_{4}) of order 8 is a non-abelian p-group. But, every group of order p

^{2}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

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

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

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

Explanation: There are m

^{4}+ m

^{2}+ 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.

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) p^{9}

c) p^{p}

d) 3!*p^{2}

View Answer

Explanation: We know that, for a finite group H, there exists a Sylow p-subgroup of H having order p

^{9}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

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

Explanation: Suppose, there are two functions f

_{1}and f

_{2}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: f

_{2}*π≡f

_{1}. 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

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

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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