Discrete Mathematics Questions and Answers – Group Axioms


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

1. __________ are called group postulates.
a) Group lemmas
b) Group theories
c) Group axioms
d) Group
View Answer

Answer: c
Explanation: The group axioms are also called the group postulates. A group with an identity (that is, a monoid) in which every element has an inverse is termed as semi group.

2. A subgroup has the properties of ________
a) Closure, associative
b) Commutative, associative, closure
c) Inverse, identity, associative
d) Closure, associative, Identity, Inverse
View Answer

Answer: d
Explanation: A subgroup S is a subset of a group G (denoted by S <= G) if it holds the four properties simultaneously – Closure, Associative, Identity and Inverse element.

3. If a * b = a such that a ∗ (b ∗ c) = a ∗ b = a and (a * b) * c = a * b = a then ________
a) * is associative
b) * is commutative
c) * is closure
d) * is abelian
View Answer

Answer: a
Explanation: ‘∗’ can be defined by the formula a∗b = a for any a and b in S. Hence, (a ∗ b)∗c = a∗c = a and a ∗(b ∗ c)= a ∗ b = a. Therefore, ”∗” is associative. Hence (S, ∗) is a semigroup. On the contrary, * is not associative since, for example, (b•c)•c = a•c = c but b•(c•c) = b•a = b Thus (S,•) is not a semigroup.
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4. The set of odd and even positive integers closed under multiplication is ________
a) a free semigroup of (M, ×)
b) a subsemigroup of (M, ×)
c) a semigroup of (M, ×)
d) a subgroup of (M, ×)
View Answer

Answer: b
Explanation: Let C and D be the set of even and odd positive integers. Then, (C, ×) and (D, ×) are subsemigroups of (M, ×) since A and B are closed under multiplication. On the other hand, (A, +) is a subsemigroup of (N, +) since A is closed under addition, but (B, +) is not a subsemigroup of (N, +) since B is not closed under addition.

5. If F is a free semigroup on a set S, then the concatenation of two even words is ________
a) a semigroup of F
b) a subgroup of F
c) monoid of F
d) cyclic group of F
View Answer

Answer: b
Explanation: Let F be the free semigroup on the set S = {m,n}. Let, E consist of all even words, i.e, words with even length and the concatenation of two such words is also even. Thus E is a subsemigroup of F.

6. The set of rational numbers form an abelian group under _________
a) Association
b) Closure
c) Multiplication
d) Addition
View Answer

Answer: c
Explanation: The set of nonzero rational numbers form an abelian group under multiplication. The number 1 is the identity element and q/p is the multiplicative inverse of the rational number p/q.

7. Condition of semigroup homomorphism should be ____________
a) f(x * x) = f(x * y)
b) f(x) = f(y)
c) f(x) * f(y) = f(y)
d) f(x * y) = f(x) * f(y)
View Answer

Answer: d
Explanation: Consider two semigroups (S,∗) and (S’,∗’). A function f: S -> S’ is called a semigroup homomorphism if f(a∗b) = f(a)∗f(b). Suppose f is also one-to-one and onto. Then f is called an isomorphism between S and S’ and S and S’ are said to be isomorphic semigroups.

8. A function f:(M,∗)→(N,×) is a homomorphism if ______
a) f(a, b) = a*b
b) f(a, b) = a/b
c) f(a, b) = f(a)+f(b)
d) f(a, b) = f(a)*f(a)
View Answer

Answer: b
Explanation: The function f is a homomorphism since f(x∗y)= f(ac, bd)= (ac)/(bd) = (a/b)(c/d) = f(x)f(y).

9. A function defined by f(x)=2*x such that f(x+y)=2x+y under the group of real numbers, then ________
a) Isomorphism exists
b) Homomorphism exists
c) Heteromorphic exists
d) Association exists
View Answer

Answer: b
Explanation: Let T be the group of real numbers under addition, and let T’ be the group of positive real numbers under multiplication. The mapping f: T -> T’ defined by f(a)=2*a is a homomorphism because f(a+b)=2a+b = 2a*2b = f(a)*f(b). Again f is also one-to-one and onto T and T’ are isomorphic.

10. If x * y = x + y + xy then (G, *) is _____________
a) Monoid
b) Abelian group
c) Commutative semigroup
d) Cyclic group
View Answer

Answer: c
Explanation: Let x and y belongs to a group G.Here closure and associativity axiom holds simultaneously. Let e be an element in G such that x * e = x then x+e+xe=a => e(1+x)=0 => e = 0/(1+x) = 0. So, identity axiom does not exist but commutative property holds. Thus, (G,*) is a commutative semigroup.

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