Algebra MCQ Quiz in தமிழ் - Objective Question with Answer for Algebra - இலவச PDF ஐப் பதிவிறக்கவும்

Explanation:

The symmetric group defined over any set is the group whose elements are all the bijections from the set to itself, and whose group operation is the composition of functions. In particular, the finite symmetric group  defined over a finite set of  symbols consists of the permutations that can be performed on the 
 symbols. Since there are 
 such permutation operations, the order (number of elements) of the symmetric group 
 is 
.

Multiplication of Symmetric Group :

 (1 2 3) 

Images of the first Permutation element

1 --> 2, 2 --> 3, 3 --> 1, 4 --> 4, 5 --> 5, 6 --> 6

(5 6 4 1)

Images of the Second Permutation element

1 --> 5, 2 --> 2, 3 --> 3, 4 --> 1, 5 --> 6, 6 --> 4

For Multiplying the Element of the Group 

we have to process from backward to forward

1 --> 5, 5 --> 5 so, The Image of 1 under Permutation is 5

2 --> 2, 2 --> 3 so, The image of 2 under Permutation is 3

3 --> 3, 3 --> 1 so, The image of 3 under Permutation is 1

4 --> 1, 1 --> 2 so, The image of 4 under Permutation is 2

5 --> 6, 6 --> 6 so, The image of 5 under Permutation is 6

6 --> 4, 4 --> 4 so, The image of 6 under Permutation is 4

So, The multiplication of (1 2 3)(5 6 4 1) = (1 5 6 4 2 3)

Concept -

Algebraic Closure Theorem - every field has an algebraic closure, and when considering the complex numbers as the field K, this theorem implies that C is an algebraic extension of K.

Explanation:

The Algebraic Closure Theorem ensures that every field has an algebraic closure, and when considering the complex numbers as the field K, this theorem implies that C is an algebraic extension of K.

Since K is a proper subfield of ℂ and is not contained in ℝ, K must contain some complex numbers.

So, By the Algebraic Closure Theorem, ℂ is an algebraic extension of K.

Hence, Option 3 is Correct.

Concept:

Explanation:

(4): Let R is commutative

N = {x ∈ R|xn = 0 for some n ∈ ℕ}.

Now, Let x, y ∈ N then xn = 0 and yn = 0 for some n ∈ ℕ

Now, (x - y)ⁿ = xn + (-1)ⁿyn +  = 0

So, x - y ∈ N

Also, (xy)ⁿ = xⁿyⁿ = 0, (yx)n = ynxn = 0

Hence xy ∈ N

Therefore N is an ideal  

(4) is correct

(1): Let A =  and B =  then A, B ∈ N

But A + B =  ∉ N as (A + B)ⁿ ≠ 0 for any integer n

Closer property not hold

(1) is false

Concept :

The order of an element in a direct product of a finite number of finite groups is the lcm of the orders of the components of the element.

|(,,....,)| = lcm(||,||,....,||) 

Suppose  is an isomorphism from a group G onto a group G'.Then G is Abelian iff G' is Abelian.

Calculation :

G = z3⊕z3⊕z3 

|G|=27

The possible order of elements in z3 is 1 and 3, so the possible order of elements in G is 1 and 3.

All the non-identity elements are of order 3.

Thus number of elements of order 3 in G is 26.

|H| = 27

Every non-identity element in H is also of order 3 since,

Thus, number of elements of order 3 in H are 26.

We know that direct sum ⊕....⊕ is abelian iff  are abelian.

Thus, G is abelian.

But H is not abelian. (Take a=1, c=1 in one element and a=1,c=0 in another element and check they do not commute.)

So G and H are not isomorphic.

Thus options (1), (2) and (4) are correct.

Explanation:
The external direct product of two cyclic groups is not always cyclic.

For instance, the direct product of Zn and Zm (representing the cyclic groups of orders n and m respectively) is cyclic if and only if n and m are coprime.

If n m are not coprime (i.e., they share a common divisor greater than 1), the resulting group is not cyclic, although it is still abelian, because all cyclic groups are abelian. For example, Z₂ × Z₂ is isomorphic to the Klein four-group, which is not cyclic, but is abelian

Hence, option 3 is correct.

Concept:

(i) General Linear Group :

The group of n ×  n invertible matrices with entries from  has order given by:

(ii) Euler’s Totient Function  ϕ : For p  a prime and k  a positive integer,

Explanation:

Calculate the order of  G :

Given G is the group of 3 × 3 invertible matrices with entries from , we need to calculate

For n = 3 and  p = 3,  k = 3:

Now, we multiply these terms along with (3³)³:

So, the order of G is:

Option 1)

a is divisible by 3⁷ :

Since  3¹² is a factor of |G| , a  is divisible by 3⁷ .

This statement is true.

Option 2)

a  is divisible by  2⁴:

Since  2⁵ is a factor of  |G| , a  is divisible by 2⁴.

This statement is true.

Option 3)

 a  is divisible by 6⁴

Since  6⁴ = 2⁴ × 3⁴  and 2⁵ × 3¹² are factors of |G| ,  a is divisible by 6⁴. This statement is true.

option 4)  a is not divisible by 54:

Since 54 = 2 × 3³

So a is divisible by 54.

Hence this statement is not true.

Concept -

Cauchy Theorem - Let G be a finite group and let p be a prime number that divide the order of G then G contains a element of order p.

Explanation -

By Cauchy Theorem, any group of order 6 has an element of order 2 as well as order 3.

So, subgroup generated by these elements are cyclic of order 2 and 3, say H and K ,respectively.

Subgroup of order 3, i.e. K has index 2 and hence normal.

So, product of H and K is defined and is equal to G as it contains both H and K. So, either H is normal or not.

If H is normal, we get cyclic group of order 6, and H is not normal we get semi direct product of H and K which is isomorphic to S₃.

So, only two groups are there upto isomorphism.

Concept:

Explanation:

Given 𝜎 = 𝜎1 𝜎2 where 𝜎1 is a 4-cycle and 𝜎2 is a 3-cycle in 𝑆8.

𝜎1 and 𝜎2 are disjoint cycles then there exists a 𝜎3 of 1- cycle

So 𝜎 = 𝜎1 𝜎2 𝜎3  

conjugate to 𝜎 means their cycle decomposition will be the same with 𝜎.

Hence the number of elements in 𝑆8 that are conjugate to 𝜎 

=  = 3360

Answer is 3360

Explanation -

We know that some results about the subgroup 

(i) H is proper normal subgroup of 

(ii) H is infinitely generated subgroup of 

(iii) H is non cyclic subgroup of 

Hence option (3) is true.

Explanation:

If G is abelian then G is cyclic
 A group G of order 21 has order as product of two primes (in this case, 7 and 3). By the Fundamental Theorem of Finite Abelian Groups, a finite group of order n = p × q, where p and q are different primes, is always abelian and, moreover, it is always cyclic. So, if G is an abelian group of order 21, it must also be cyclic.

Hence. option 2 is correct.