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

1. An infinite cyclic group does not have a ______ series.

a) AP

b) GP

c) Composite

d) Finite

View Answer

Explanation: Suppose that any finite group of order less than n has a composition series. Let G be a finite group of order n. If G is simple, then G⊳{e}, where e is the identity element of G and hence, it is a composition series. However, any infinite cyclic group does not have a composite series.

2. Every cyclic group is a/an ______

a) infinite subgroup

b) abelian group

c) monoid

d) commutative semigroup

View Answer

Explanation: Let C be a cyclic group with a generator g∈C. Namely, we have G={g.Let x and y be arbitrary elements in C. Then, there exists n, m∈Z such that x=gn and y=gm. It follows that x*y = gn*gm = gn+m = gm*gn = yx. Hence, we find that xy=yx for any x,y belongs to G.Thus, G is an abelian group.

3. What is an irreducible module?

a) A cyclic module in a ring with any non-zero element as its generator

b) A cyclic module in a ring with any positive integer as its generator

c) An acyclic module in a ring with rational elements as its generator

d) A linearly independent module in a semigroup with a set of real numbers

View Answer

Explanation: A nonzero R-module M is irreducible if and only if M is a cyclic module with any nonzero element as its generator. Suppose that M is an irreducible module. Let a∈M be any nonzero element and consider the submodule (a) generated by the element a. Since a is a nonzero element, the submodule (a) is non-zero. Since M is irreducible, this implies that M=(a). Hence M is a cyclic module generated by a. Since a is any nonzero element, the module M is a cyclic module with any nonzero element as its generator.

4. A finite group G of order 219 is __________

a) a semigroup

b) a subgroup

c) a commutative inverse

d) a cyclic group

View Answer

Explanation: The prime factorization 219=3⋅73. By the definition of Sylow’s theorem, determine the number np of Sylow p-group for p=3,73. np≡1(mod p) and np divides n/p. Thus, n3 could be 1, 4, 7, 10, 13,… and n3 needs to divide 219/3=73. Hence the only possible value for n3 is n3=1. So there is a unique Sylow 3-subgroup P3 of G. By Sylow’s theorem, the unique Sylow 3-subgroup must be a normal subgroup of G. Similarly, n73=1, 74,… and n73 must divide 219/73=3 and hence we must have n73=1. Thus, G has a unique normal Sylow 73-subgroup P73.

5. The number of generators of cyclic group of order 219 is __________

a) 144

b) 124

c) 56

d) 218

View Answer

Explanation: The number of generators of a cyclic group of order n is equal to the number of integers between 1 and n that are relatively prime to n.Namely, the number of generators is equal to ϕ(n), where ϕ is the Euler totient function. We know that G is a cyclic group of order 219. Hence, the number of generators of G is ϕ(219) = ϕ(3)ϕ(73) = 3⋅73 = 144.

6. The order of a simple abelian group is __________

a) infinite

b) real number

c) finite

d) prime

View Answer

Explanation: Let p be the order of g (hence the order of G). As a contradiction, assume that p=ab is a composite number with integers a > 1, b > 1. Then (ga) is a proper normal subgroup of G. This is a contradiction since G is simple. Thus, p must be a prime number.

Therefore, the order of G is a prime number.

7. The Number of Elements Satisfying g7=e in a finite Group F is ______

a) even

b) not a number

c) odd

d) rational

View Answer

Explanation: Let g≠e be an element in the group F such that g7=e. As 7 is a prime number, this yields that the order of g is 7. Consider, the subgroup (g) is generated by g. As the order of g is 7, the order of the subgroup (g) is 7. Hence, the order must be odd.

8. All the rings of order p2 is ____________

a) associative

b) cyclic

c) inverse

d) commutative

View Answer

Explanation: Let R be a ring with unit 1. Suppose that the order of R is |R|=p2 for some prime number p. Then it has been proven that R is a commutative ring.

9. An element of a commutative ring R(1≠0) is nilpotent if __________

a) a+1=0

b) a^{n} = 0, for some positive integer n

c) a^{n} = 1, for some integer n

d) a^{2} = 0

View Answer

Explanation: Since a is nilpotent in a commutative ring R, we have an=0 for some positive integer n. since R is commutative, for any m∈R, we have (am)n=anmn=0. Then we have the following equality: (1−am)(1+(am)+(am)2+⋯+(am)n−1)=1. Hence, 1−am is a unit in R.

10. A group G of order 20 is __________

a) solvable

b) unsolvable

c) 1

d) not determined

View Answer

Explanation: The prime factorization of 20 is 20=2⋅5. Let n5 be the number of 5-Sylow subgroups of G. By Sylow’s theorem, we have, n5≡1(mod 5)and n5|4. Thus, we have n5=1. Let P be the unique 5-Sylow subgroup of G. The subgroup P is normal in G as it is the unique 5-Sylow subgroup. Then consider the subnormal series G▹P▹{e}, where e is the identity element of G. Then the factor groups G/P, P/{e} have order 4 and 5 respectively. Hence these are cyclic groups(in particular abelian). Hence, the group G of order 20 has a subnormal series whose factor groups are abelian groups, and thus G is a solvable group.

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