Mathematics Exam  >  Mathematics Tests  >  Topic-wise Tests & Solved Examples for Mathematics  >  Test: Group Theory - 11 - Mathematics MCQ

Test: Group Theory - 11 - Mathematics MCQ


Test Description

20 Questions MCQ Test Topic-wise Tests & Solved Examples for Mathematics - Test: Group Theory - 11

Test: Group Theory - 11 for Mathematics 2024 is part of Topic-wise Tests & Solved Examples for Mathematics preparation. The Test: Group Theory - 11 questions and answers have been prepared according to the Mathematics exam syllabus.The Test: Group Theory - 11 MCQs are made for Mathematics 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Group Theory - 11 below.
Solutions of Test: Group Theory - 11 questions in English are available as part of our Topic-wise Tests & Solved Examples for Mathematics for Mathematics & Test: Group Theory - 11 solutions in Hindi for Topic-wise Tests & Solved Examples for Mathematics course. Download more important topics, notes, lectures and mock test series for Mathematics Exam by signing up for free. Attempt Test: Group Theory - 11 | 20 questions in 60 minutes | Mock test for Mathematics preparation | Free important questions MCQ to study Topic-wise Tests & Solved Examples for Mathematics for Mathematics Exam | Download free PDF with solutions
Test: Group Theory - 11 - Question 1

 Let σ = (125) (36) and t = (1456) (23) be two elements of the permutation group on 6 symbols. Then the product  where is

Detailed Solution for Test: Group Theory - 11 - Question 1

Given σ = (1 2 5) • (3 6) 

Test: Group Theory - 11 - Question 2

 The remainder of 831 when it is divided by 13 is

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

We need to find out the remainder when 8^31 is divided by 13

We should try and reduce the numerator in this case.

Let us have a look at the method below.

Remainder [8^31 / 13] = Remainder [8*8^30 / 13]

Remainder [8*64^15 / 13] = Remainder [8*(-1)^15 / 13] {Here We divide 64 by 13 so remanider is (-1)}

So

= Remainder [(-8) / 13]= 5

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Group Theory - 11 - Question 3

 The number of distinct group homomorphisms from (Z, +) onto (Z, +) is

Detailed Solution for Test: Group Theory - 11 - Question 3

In (Z , + ) total number of automorph is m is 2. So, Number of distinct group homomorphism form (Z, +) onto (Z, +) is 2.

Test: Group Theory - 11 - Question 4

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

Detailed Solution for Test: Group Theory - 11 - Question 4

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’.

Test: Group Theory - 11 - Question 5

Let S denote the group of all permutations on the finite set {1, 2, 3, 4, 5} under the operation of permutation multiplication. Then the order of the subgroup of S5 generated by  is 

Detailed Solution for Test: Group Theory - 11 - Question 5

Given 
The order of the subgroup of S5 generated by
= L.C.M. ( 3 ,2 ) = 6

Test: Group Theory - 11 - Question 6

Which of the following is a field?

Detailed Solution for Test: Group Theory - 11 - Question 6

Z6 is not a field.
Here, p = 3
So, Z2p;p is prime is false.
Hence, option (a) is discarded.
So, 
If a, b ε Z3 then a2 + b2 cannot be zero. So , it is invertible so set is field.
But, this is not the case if a, b ε Z4.

Test: Group Theory - 11 - Question 7

Let H be a subgroup Of ordet 60 of a group G of order 120. If a ε G/H, then which of the following is Not, a subgroup of G?

Detailed Solution for Test: Group Theory - 11 - Question 7

H is a subgroup of order 60 of group G of orders 120.
Thus, |
will contain only two distinct elements.
These are H and Ha (or aH)
If a is not identity, then aH will not have identity element.

Test: Group Theory - 11 - Question 8

G is a group of order 51. Then which one of the following statements is false?

Detailed Solution for Test: Group Theory - 11 - Question 8

Given G is a group of order 51.
i.e. o(G) = 51 = 3 x 17
All subgroup of group is cyclic.
The subgroups are of odd order so each element will have its inverse different from itself.
So, If G is abelian then there exists no proper subgroup H of G such that product of all elements of H is identity is false.

Test: Group Theory - 11 - Question 9

In the group {1, 2,........,16} under the operation of multiplication modulo 17, the order of the element 3 is

Detailed Solution for Test: Group Theory - 11 - Question 9

{1 , 2 , . . . , 16} are non-zero elements of 34 = 81 = 13(mod 17)
38 = 13 x 13(mod 17)
    = 16(mod 17)
312 = 13 x 16(mod 17)
= 208(mod 17)
= 4(mod 17)
316 = 13 x 4(mod 17)
= 52(mod 17) =l(mod 17)
The order o f the element 3 is 16 i.e. o(3) = 16

Test: Group Theory - 11 - Question 10

Let S be a non empty symmetric and transitive binary operation on a non empty set A. Consider any pair (a, b) ε S. Since S is symmetric, (b, a) ε S. Further, since S is transitive, (a, b) ε S. Which one of the following statements is true?

Detailed Solution for Test: Group Theory - 11 - Question 10

Here, (a, b) ε S so (b, a) ε S.
Since S is transitive.
So, (a, a) ε S so it is reflexive.

Test: Group Theory - 11 - Question 11

Let L be an equivalence relation on a set S of n elements, Consider the set Sa = {x : x ≠ a and aLx). Further, let D be such that no two elements in D are related under L. The number of elements in is at most

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

The number o f elements in   is atmost .

Test: Group Theory - 11 - Question 12

Which of the following is true?

Detailed Solution for Test: Group Theory - 11 - Question 12

If we take matrix A = then A-1 does not exist.
Hence, option (a) discarded.
If we take a finite group S3 which is not cyclic the option (e) discarded.
Also , the set o f all 2 x 2 real non singular matrices may or may not form an abelian group under matrix multiplication.
Hence, option (d) discarded.
Hence, a finite abelian group of order 6 has exactly two non-trivial subgroups, i.e. a finite abelian group of order 6 is isomorphic to Z6 and Z6 is cyclic and it has exactly two non-trivial subgroups.

Test: Group Theory - 11 - Question 13

Let X denote the set of all real-valued functions defined on Z. Define a relation -in X by f ~ g if f(0)1 g(0). Then the relation ~ is

Detailed Solution for Test: Group Theory - 11 - Question 13

Given that X denote the set of all real valued functions defined on Z. Define a relation in X by f ~ g if f(0) ≠ g(0).
So, f(0) ≠ f(0) which fails reflexive relation.
If wetake f(x ) = 1- x
g(x) = x - l
Here , f(0) ≠ g(x) and we take h(x) = 1 - 2x
Here g(0) ≠ h(0)
But f(0)= h(0).
This fails transitive relation.
Hence, option (d) is correct.

Test: Group Theory - 11 - Question 14

Let R be the ring o f p olynom ials over Z2 and let I be the ideal of R generated by the polynomial x2 + x + 1. Then the number of elements in the quotient ring R/ I is

Detailed Solution for Test: Group Theory - 11 - Question 14

Z2 = { 0 , 1 }
Ideal of R generated by x3 + x + 1
I(0) = 0 + 0 + 1 = 1 
I(1) = 1 + 1 + 1 = 1 
Thus (x3 + x + 1) is irreducible
Now R/I will contain element of the form I + g(x) By division algorithms
g(x) = f (x) x (x2 + x + 1) + r(x)
deg[r(x)] < deg(x3 + x + 1)
deg[r(x)] < 3
So r(x) will be o f the form ax2 + bx + c, a, b,c e Z2
So , elements of R/I will be I + f ( x ) x (x3 + x + 1) +r(x)
or, 1+ ax2 + bx + c
a, b,c each can be chosen in 2 ways
So , total number of elements in R/I will be 8.

Test: Group Theory - 11 - Question 15

If X Δ Z = Y Δ Z for the non-empty sets X, Y and Z, where Δ represents the symmetric difference then.

Detailed Solution for Test: Group Theory - 11 - Question 15

If X Δ Z = Y Δ Z for the non empty sets X , Y and Z, where Δ represents the symmetric difference.
Put X = Y then above condition also hold.
Hence, Option (a) is correct.

Test: Group Theory - 11 - Question 16

The number of elements of S5 (the symmetric group on 5 letters) which are their own inverses equals.

Detailed Solution for Test: Group Theory - 11 - Question 16


Test: Group Theory - 11 - Question 17

An abelian group of order 24 has

Detailed Solution for Test: Group Theory - 11 - Question 17

If order of G is 24
i.e. o(G) =24 = 23 x 3 and abelian.
The number of subgroups of order 3
= 1 + 3k/23
= 1 where k = 0
Hence, exactly one subgroup of order 3.

Test: Group Theory - 11 - Question 18

The number of group homomorphisms from the group ( Z18 , ) to the group ( Z30,  )

Detailed Solution for Test: Group Theory - 11 - Question 18

The number of group homomorphisms form the group (Z18, ) to group (Z30, ) is equal to g.c.d.(18,30) = 6.

Test: Group Theory - 11 - Question 19

Let G be a finite group and H be a normal subgroup G of order 2. Then the order of the center of G is

Detailed Solution for Test: Group Theory - 11 - Question 19

H is normal subgroup of G of order 2.
H will contain an identity and one more element.
Since,
Thus, 
Thus, a will belong to centre.
So, centre of G will contain atleast two element, identity and a.
Also, centre is subgroup G.
Also, H will be a subgroup of the centre, so its order will be an even number.

Test: Group Theory - 11 - Question 20

Let

The cardinality of the orbit of 2 under σ is

Detailed Solution for Test: Group Theory - 11 - Question 20

Let
= (1 2 10 12 7 6 3 8 11) (4 5 9)
The cardinality of the orbit 2 under σ is 9.

27 docs|150 tests
Information about Test: Group Theory - 11 Page
In this test you can find the Exam questions for Test: Group Theory - 11 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Group Theory - 11, EduRev gives you an ample number of Online tests for practice
Download as PDF