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Test: Group Theory - 2 - Mathematics MCQ


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Test: Group Theory - 2 - Question 1

An algebraic structure _________ is called a semigroup.

Detailed Solution for Test: Group Theory - 2 - Question 1

An algebraic structure (P,*) is called a semigroup if a*(b*c) = (a*b)*c for all a,b,c belongs to S or the elements follow associative property under “*”. (Matrix,*) and (Set of integers,+) are examples of semigroup.

Test: Group Theory - 2 - Question 2

A non empty set A is termed as an algebraic structure ________

Detailed Solution for Test: Group Theory - 2 - Question 2

A non empty set A is called an algebraic structure w.r.t binary operation “*” if (a*b) belongs to S for all (a*b) belongs to S. Therefore “*” is closure operation on ‘A’.

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Test: Group Theory - 2 - Question 3

A monoid is called a group if _______

Detailed Solution for Test: Group Theory - 2 - Question 3

A monoid(B,*) is called Group if to each element there exists an element c such that (a*c)=(c*a)=e. Here e is called an identity element and c is defined as the inverse of the corresponding element.

Test: Group Theory - 2 - Question 4

Condition for monoid is __________

Detailed Solution for Test: Group Theory - 2 - Question 4

A Semigroup (S,*) is defined as a monoid if there exists an element e in S such that (a*e) = (e*a) = a for all a in S. This element is called identity element of S w.r.t *.

Test: Group Theory - 2 - Question 5

Matrix multiplication is a/an _________ property.

Detailed Solution for Test: Group Theory - 2 - Question 5

The set of two M*M non-singular matrices form a group under matrix multiplication operation. Since matrix multiplication is itself associative, it holds associative property.

Test: Group Theory - 2 - Question 6

A group (M,*) is said to be abelian if ___________

Detailed Solution for Test: Group Theory - 2 - Question 6

A group (M,*) is said to be abelian if (x*y) = (x*y) for all x, y belongs to M. Thus Commutative property should hold in a group.

Test: Group Theory - 2 - Question 7

How many properties can be held by a group?

Detailed Solution for Test: Group Theory - 2 - Question 7

A group holds five properties simultaneously –
i) Closure
ii) associative
iii) Identity element
iv) Inverse element.

Test: Group Theory - 2 - Question 8

A cyclic group can be generated by a/an ________ element.

Detailed Solution for Test: Group Theory - 2 - Question 8

A singular element can generate a cyclic group. Every element of a cyclic group is a power of some specific element which is known as a generator ‘g’.

Test: Group Theory - 2 - Question 9

{1, i, -i, -1} is __________

Detailed Solution for Test: Group Theory - 2 - Question 9

The set of complex numbers {1, i, -i, -1} under multiplication operation is a cyclic group. Two generators i and -i will covers all the elements of this group. Hence, it is a cyclic group.

Test: Group Theory - 2 - Question 10

A cyclic group is always _________

Detailed Solution for Test: Group Theory - 2 - Question 10

A cyclic group is always an abelian group but every abelian group is not a cyclic group. For instance, the rational numbers under addition is an abelian group but is not a cyclic one.

Test: Group Theory - 2 - Question 11

Let G be a group order 6, and H be a subgroup of G such that 1 < |H| < 6. Which one of the following options is correct?

Detailed Solution for Test: Group Theory - 2 - Question 11

Concept

  • According to the Lagrange theorem order of subgroups must divide the order of the group.
  • Property of group says if a group has prime order then it is cyclic.

Explanation:

  • Since the order of G is 6. Therefore  its subgroup may have order 1,2,3,6
  • H is one of its subgroups with condition 1< |H| <6 so H may be of order 2 or 3  which is prime 
  • Hence H must be cyclic
  • The order of G is 6 which is not prime and hence it may or may not  be cyclic 

Therefore option 2 is correct

Test: Group Theory - 2 - Question 12

The number of generators of a cyclic group of order 10 is

Detailed Solution for Test: Group Theory - 2 - Question 12

Concept:
Let G be a cyclic group of order 10 generated by an element a, then o(a) = o(G) = 10
Evidently G = {a, a2, a3, a4, a5, a6, a7, a8, a9, a10 = e }
If the HCF of m and n is d, then we write (m, n) = d.
An element am ∈ G is also a generator of G if (m, 10) = 1.
Thus there are four generators of G namely, a, a3, a7, a9

Test: Group Theory - 2 - Question 13

Given:
Statement A: All cyclic groups are an abelian group.
Statement B: The order of the cyclic group is the same as the order of its generator.

Detailed Solution for Test: Group Theory - 2 - Question 13

Concept:

Abelian Group: Let {G=e, a, b} where e is identity. The operation 'o' is defined by the following composition table. Then(G, o) is called Abelian if it follows the following property-

  • Closure Property
  • Associativity
  • Existence of Identity
  • Existence of Inverse
  • Commutativity

Cyclic Group- A group a is said to be cyclic if it contains an element 'a' such that every element of G can be represented as some integral power of 'a'. The element 'a' is then called a generator of G, and G is denoted by (or [a]). 

Theorem: 
(i) All cyclic groups are Abelian, but an Abelian group is not necessarily cyclic. 
(ii) The order of a cyclic group is the same as the order of its generator. 
Thus it is clear that A and B both are true.

Test: Group Theory - 2 - Question 14

If (G, ⋅) is a group such that (ab)-1 = a-1 b-1, ∀ a, b ∈ G, then G is a/an

Detailed Solution for Test: Group Theory - 2 - Question 14

A group is said to be abelian if (a*b) = (b*a) ∀a,b ∈G
Since (G, ⋅) is a group
∴ (ab)-1 = (b-1a-1)             (1)
Given: (ab)–1 = a–1b–1       (2)
From (1) and (2)
 (a*b) = (b*a)
Therefore it is abelian

Test: Group Theory - 2 - Question 15

A subset H of a group (G, ∗) is a group if

Detailed Solution for Test: Group Theory - 2 - Question 15

Concept:
A non-empty subset H of a group (G, ∗) is a group of G iff,
⇒ a, b ∈ H ⇒ a ∗ b-1 ∈ H

Test: Group Theory - 2 - Question 16

Which of the following is true?

Detailed Solution for Test: Group Theory - 2 - Question 16

A group of prime order p (in this case, 31) is always cyclic. And every cyclic group is abelian is true

 While there does exist a cyclic group of order 4, namely Z4, there also exists a non-cyclic group of order 4, the Klein-4 group. Both are abelian hence, This statement is not always true.

 The Fundamental Theorem of Finite Abelian Groups tells us that for any integer n, there exists an abelian group of order n. In this particular case, 2021 = 43 * 47, which are both primes, so you could have a direct product of two cyclic groups of orders 43 and 47 as an abelian group of order 2021 hence, This statement is not true.

 While the cyclic group of order 6, Z6, is abelian, there is also a non-abelian group of order 6, the symmetric group S3 hence, This statement is not true.

Test: Group Theory - 2 - Question 17

Number of proper Subgroups of Zis 

Detailed Solution for Test: Group Theory - 2 - Question 17

The group Z6\mathbb{Z}_6, the integers modulo 6 under addition, has the following elements: {0,1,2,3,4,5}\{0, 1, 2, 3, 4, 5\}.

To find the proper subgroups of Z6\mathbb{Z}_6, we need to determine which subsets of Z6\mathbb{Z}_6 form subgroups and exclude the trivial subgroups (i.e., Z6\mathbb{Z}_6 itself and the identity element only).

The possible subgroups of Z6\mathbb{Z}_6 are:

  1. {0}\{0\} (trivial subgroup).
  2. {0,3}\{0, 3\}: This is a subgroup because it is closed under addition mod 6 and contains the identity element (0).
  3. {0,2,4}\{0, 2, 4\}: This is also a subgroup because it is closed under addition mod 6.
  4. {0,1,2,3,4,5}\{0, 1, 2, 3, 4, 5\} (the whole group, which we exclude as we are looking for proper subgroups).

Thus, the proper subgroups of Z6\mathbb{Z}_6 are:

  • {0,3}\{0, 3\}
  • {0,2,4}\{0, 2, 4\}

So, there are 2 proper subgroups of Z6\mathbb{Z}_6.

Test: Group Theory - 2 - Question 18

The number of generators of the cyclic group G of order 8 is

Detailed Solution for Test: Group Theory - 2 - Question 18

Calculation:
Let a cyclic group G of order 8 generated by an element a, then 
⇒ o(a) = o(G) = 8
To determine the number of generators of G,
Evidently, G = {a, a2, a3, a4, a5, a6, a7, a8 = e}
An element am ∈ G is also a generator of G is HCF of m and 8 is 1.
HCF of 1 and 8 is 1, HCF of 3 and 8 is 1, HCF of 5 and 8 is 1, HCF of 7 and 8 is 1.
Hence, a, a3, a5, a7 are generators of G.
Therefore, there are four generators of G.

Test: Group Theory - 2 - Question 19

In any group, the number of improper subgroups is 

Detailed Solution for Test: Group Theory - 2 - Question 19

Concept:

In group theory, improper subgroups of any given group include the group itself and the trivial group, which contains only the identity element. These are considered improper because they are trivially subgroups of every group. All other subgroups are proper subgroups.

Therefore, every group has exactly two improper subgroups:

  1. The group itself.
  2. The trivial group (the subgroup containing only the identity element).

So, the correct answer is:

Option A: 2

Test: Group Theory - 2 - Question 20

If G is a Prime order group then G has 

Detailed Solution for Test: Group Theory - 2 - Question 20

Given:
If G is a Prime order group of p
Concept used:
Lagranges theorem:

The order of subgroups divides the Order of Group but if some m divide the order of the group that doesn't mean that group has the subgroup of order m 
Proper subgroup :  H ≠ e and H ≠ G is the subgroup of G
Calculations:
G is a Prime order Group of p 
so, it has only 2 divisors 1 and itself that divides the order of Group 
⇒ Only 2 subgroups are generated by 1 and p
⇒ <1> = G and
= {e} where e is the identity 
∴ option 1 is correct, No proper Subgroup 

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