Discrete Mathematics Questions and Answers – Cyclic Groups

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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

Answer: c
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.
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2. Every cyclic group is a/an ______
a) infinite subgroup
b) abelian group
c) monoid
d) commutative semigroup
View Answer

Answer: b
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

Answer: a
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

Answer: d
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

Answer: a
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.
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6. The order of a simple abelian group is __________
a) infinite
b) real number
c) finite
d) prime
View Answer

Answer: a
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

Answer: c
Explanation: Let g≠e be an element in 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

Answer: d
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) an = 0, for some positive integer n
c) an = 1, for some integer n
d) a2 = 0
View Answer

Answer: b
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.
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10. A group G of order 20 is __________
a) solvable
b) unsolvable
c) 1
d) not determined
View Answer

Answer: a
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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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn