Normal Subgroup and Quotient Groups MCQ Quiz - Objective Question with Answer for Normal Subgroup and Quotient Groups - Download Free PDF

Last updated on May 19, 2025

Latest Normal Subgroup and Quotient Groups MCQ Objective Questions

Normal Subgroup and Quotient Groups Question 1:

Let G be a group and N a subgroup of G . Which of the following statements is true about N being a normal subgroup of G ?

  1. If N is a normal subgroup of G , then N is necessarily cyclic.  
  2. If N is a normal subgroup of G , then G/N is always an abelian group.  
  3.  If N is a normal subgroup of G , then \( gNg^{-1} = N \)  for all  \(g \in G \) .  
  4. If N is a normal subgroup of G , then N must be the trivial group \(\{e\} \) or the whole group G .

Answer (Detailed Solution Below)

Option 3 :  If N is a normal subgroup of G , then \( gNg^{-1} = N \)  for all  \(g \in G \) .  

Normal Subgroup and Quotient Groups Question 1 Detailed Solution

Explanation:

Option (1) is incorrect because normality does not imply that the subgroup is cyclic

Option (2)  is incorrect because G/N is not necessarily abelian for all normal subgroups N

Option (3)  is correct because the definition of a normal subgroup is  \(g N g^{-1} = N \)  for all  \(g \in G \)

Option (4)  is incorrect because a normal subgroup does not have to be trivial or the whole group;

normal subgroups can have other non-trivial forms

Hence Option(3) is the Correct Answer.

 

Normal Subgroup and Quotient Groups Question 2:

If F is homomorphism of a group G into another group G' with kernel k, then which of the following is true? 

  1. k is a normal subgroup of G 
  2. k is anormal subgroup of G' 
  3. k is not a normal subgroup of G 
  4. k is complex of G 

Answer (Detailed Solution Below)

Option 1 : k is a normal subgroup of G 

Normal Subgroup and Quotient Groups Question 2 Detailed Solution

Explanation:

If F is homomorphism of a group G into another group G' with kernel k.

Since e ∈ k so k ≠ ϕ 

Let a, b ∈ k

f(ab-1) = f(a)f(b-1)

         = f(a)f(b)-1

        = e1e1-1 = e1 

So, ab-1 ∈ k

Hence k is a subgroup of G.

Let a ∈ G and h ∈ K

Then

f(aha-1) = f(a)f(h)f(a-1)

         = f(a)f(h)f(a)-1

        = f(a)e1f(a)-1 = e1 

So, aha-1 ∈ k

So k is a normal subgroup of G.

Option (1) is true.

 

Top Normal Subgroup and Quotient Groups MCQ Objective Questions

Normal Subgroup and Quotient Groups Question 3:

Let G be a group and N a subgroup of G . Which of the following statements is true about N being a normal subgroup of G ?

  1. If N is a normal subgroup of G , then N is necessarily cyclic.  
  2. If N is a normal subgroup of G , then G/N is always an abelian group.  
  3.  If N is a normal subgroup of G , then \( gNg^{-1} = N \)  for all  \(g \in G \) .  
  4. If N is a normal subgroup of G , then N must be the trivial group \(\{e\} \) or the whole group G .

Answer (Detailed Solution Below)

Option 3 :  If N is a normal subgroup of G , then \( gNg^{-1} = N \)  for all  \(g \in G \) .  

Normal Subgroup and Quotient Groups Question 3 Detailed Solution

Explanation:

Option (1) is incorrect because normality does not imply that the subgroup is cyclic

Option (2)  is incorrect because G/N is not necessarily abelian for all normal subgroups N

Option (3)  is correct because the definition of a normal subgroup is  \(g N g^{-1} = N \)  for all  \(g \in G \)

Option (4)  is incorrect because a normal subgroup does not have to be trivial or the whole group;

normal subgroups can have other non-trivial forms

Hence Option(3) is the Correct Answer.

 

Normal Subgroup and Quotient Groups Question 4:

If F is homomorphism of a group G into another group G' with kernel k, then which of the following is true? 

  1. k is a normal subgroup of G 
  2. k is anormal subgroup of G' 
  3. k is not a normal subgroup of G 
  4. k is complex of G 

Answer (Detailed Solution Below)

Option 1 : k is a normal subgroup of G 

Normal Subgroup and Quotient Groups Question 4 Detailed Solution

Explanation:

If F is homomorphism of a group G into another group G' with kernel k.

Since e ∈ k so k ≠ ϕ 

Let a, b ∈ k

f(ab-1) = f(a)f(b-1)

         = f(a)f(b)-1

        = e1e1-1 = e1 

So, ab-1 ∈ k

Hence k is a subgroup of G.

Let a ∈ G and h ∈ K

Then

f(aha-1) = f(a)f(h)f(a-1)

         = f(a)f(h)f(a)-1

        = f(a)e1f(a)-1 = e1 

So, aha-1 ∈ k

So k is a normal subgroup of G.

Option (1) is true.

 

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