Group & Subgroups MCQ Quiz - Objective Question with Answer for Group & Subgroups - Download Free PDF

Concept:

• Divisible Group: A group G is said to be divisible if for every element and every positive integer , there exists an element such that .
• Group   under Addition: For any rational and any , satisfies ⇒ divisible.
• Group   under Multiplication: Every non-zero complex number has an root in ⇒ divisible.
• Finite Groups: A finite group cannot be divisible, because if has finite order, then not all equations like are solvable for arbitrary .
• Cyclic Group of Order 5: Has only 5 elements. For , not every element has an such that ⇒ not divisible.
• Symmetric Group  : Non-abelian finite group of permutations of 5 elements. Not divisible for same reason: finite, and not all roots exist for all elements.

Calculation:

Given:

Group must satisfy: for every and every , there exists such that

Step 1: Check    under addition

⇒ Let

⇒ divisible

Step 2: Check   under multiplication

⇒ Let

⇒ Take any root of

⇒ Roots exist in

⇒ divisible

Step 3: Check cyclic group of order 5

⇒ Order is 5

⇒ For roots might not exist

⇒ Not divisible

Step 4: Check symmetric group  

⇒ Finite group of order 120

⇒ Not every element has an root

⇒ Not divisible

∴ Final Answer: Only options 1 and 2 are correct. 

Explanation:

H be an abelian non-trivial proper subgroup of G with the property that H ∩ gHg−1 = {e} for all g / ∉ H

K = ,

H is normal in K , as K centralizes H

Option (1): K is a proper subgroup of H

Since    by definition, K cannot be a proper subgroup of H

Instead, K either equals H or is strictly larger.

Option(1) is not correct 

Option (2): H is a proper subgroup of K

K may equal H because the centralizer of H may consist solely of H itself, depending on the group structure

Option (3): K = H

Since   

no element outside H can commute with all elements of H

Therefore, the centralizer of H in G is exactly H , implying K = H

Option (4): There exists no abelian subgroup    such that K is a proper subgroup of L

Since K = H , and H is abelian and isolated (no elements outside H commute with all elements of H ),

there cannot exist a larger abelian subgroup L containing K as a proper subgroup.

Hence Option(3) and Option(4) are correct 

Explanation:  

Given |G| =  57 = 3 × 19

All Subgroup of Group is Cyclic

The Subgroup 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 

Hence Option(4) is the correct answer.

Solution - Sn denote the group of all permutations on n symbols.  

In  possible Order be lcm (3,1) so maximum possibility be 3 

Therefore, Option 1) is wrong 

In,   maximum possibility be 4 

Therefore, Option 2) and Option 3) is also wrong 

In,  has maximum possibility be lcm (3,2) =6 

Therefore, Correct Option is Option 4).

Explanation:

Let the operation, Δ i.e., A, B ∈ P(X) ⇒ A Δ B = (A ∪ B) \ (A ∩ B) for this,

(i) Closer: Let A, B ∈ P(x) then A Δ B = (A ∪ B) \ (A ∩ B) ∈ P(X)
So, P(x) is closed under Δ. 

(ii) Associativity: let A, B, C ∈ P(x), then (A Δ B) ΔC = ([(A ∪ B) \ (A ∩ B))] ∪ C) \[([A ∪ B) \ ((A ∩ B))] ∩ C)

A Δ (B Δ C) = (A ∪[(B ∪ C) | (B∩C)]) \ (A∩[(B∪C) | (B∩C)])

form, figures you can see,

(A Δ B) ΔC = A Δ (B Δ C)

(iii) Identity:

AΔϕ = (A ∪ ϕ) \ (A ∩ ϕ) = A \ ϕ = A

So, ϕ ∈ P(x) such that A Δ ϕ = A

(iv) Inverse:

A Δ A = (A ∪ A) \ (A ∩ A) = A \ A = ϕ

So, for A ∈ P(x),  A-1 = A.

∴ P(x) is group under Δ.

Now for * operation, A * B = A ∩ B, A, B ∈ P(x)

let x = {1, 2, 3} then P(x) = {ϕ, x, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}}

Here, if we take, e = x

(∵ x ∩ A = A, A ∈ P(x))

But for e = x, inverse  of any A, A ∈ P(x)

∵ A ∩ B ≠ x (for any A, B ∈ P(x)A, B ≠ x)

So, P(x) is not a group under (*).

option (3) is true.

Explanation:

S =  So S² =  =  = - I, S3 = -S, S⁴ = I

Also, 𝑇1 = 

So, 𝑇1 = i𝑇1, , , 

Let Q₄ = {S, S², S³, I, i, -i, iS} then Q₄ is a quaternion group

Given 

𝐺1, 𝐺2, 𝐺3 be three subgroups of 𝐺𝐿2 (ℂ) given by

𝐺1 = 1 >, where 𝑇1 = ,

𝐺2 = 2 >, where 𝑇2 = ,

𝐺3 = 3 >, where 𝑇3 = .

Then 𝐺1 =  is non-abelian group of order 8

Similarly, 𝐺2 =  is non-abelian group of order 8

𝐺3 =  ≈  D₄

So, |𝑍(𝐺3)| = |𝑍(D4)| = 2

Hence 𝑍(𝐺3) = {} false because then |𝑍(𝐺3)| = 1

𝑍(𝐺1) is isomorphic to 𝑍(𝐺2) also 𝐺1 is not somorphic to 𝐺3

Hence (1), (2), (3) are false

(4) is correct

Concept -  

Number of subgroups of cyclic group order n is 

Explanation- 

Number of subgroups of a cyclic group of order 50 is  

 = (2+1)(1+1) = 6

Therefore, Correct Option is Option 1 ).

Explanation:

(I) Simple group: A group of order greater than 1 which have only normal subgroups are the identity and the group itself.

(II) If O (G) = Pk.m, (p, m) = 1, then the number of p-SSG is equal to np, where np = 1 + kp s.t. 1 + pk|m,

k = 0, 1, 2, ....

Here, |G| = 2020 = 22 × 5 × 101

Then Number of 101 - SSG = n₁₀₁ = 1 + 101k

such that 1 + 101k |20

then n101 = 1

So, 101 - SSG is unique ⇒ 101-SSG is normal

∵ p-SSG is normal ⇒ gt is unique

(∵ G has a proper normal subgroup of order 101.

⇒ G is not a simple group.

opt (1) is correct.

Let G = D1010 then |G| = 2020. D1010 is non-abelian.

⇒ D1010 is non-cyclic. Also D1010 has more than four proper subgroups. opt (2), (3) and (4) - False.

Explanation:

G is a simple group.

A simple group is a nontrivial group whose only normal subgroups are the trivial group and the group itself.

By this definition, if G has only two normal subgroups {e} (the trivial group containing only the identity element) and G itself, then G is a simple group

Explanation -

O(G) = 45 = 9.5 = 3².5

Clearly  3 - SSG and  5 - SSG is unique hence normal. 

So option (1) and (2) are true.

Every group of order 45 is isomorphic to one of the group 

Hence always abelian but not cyclic.

So option (3) is true and option (4) is not true.

Concept: 

Every group of prime order is cyclic 

i.e. If G be a group of order p, where p is prime then G is cyclic

Explanation: 

In given Options, 

7 is the prime number 

So, if O(G) = 7 then G is cyclic. 

Therefore, the Correct Option is Option (2)

Concept -

if  then there exist two group of order p.q up to isomorphism, one is cyclic and another one is non cyclic.

Explanation -

Option (1) -

O(G) = 21 = 3.7 = 3 | 7-1

Hence this is true.

Option (2) -

O(G) = 22 = 2.11 = 2 | 11-1

Hence this is true.

Option (3) -

O(G) = 55 = 5.11 = 5 | 11-1

Hence this is true.

So the correct option is (4).

Concept:

A group (G, ∘) is said to be an abelian group if a ∘ b = b ∘ a for all a, b ∈ G

Explanation:

Let G be a cyclic group with generator g ∈ G i.e., G = ⟨g⟩ 

Let a, b ∈ G then there exist m, n ∈  such that

a = g^m, b = gⁿ

Now, 

ab = gmgn = gm+n = gn+m = gngm = ba

So, G is an abelian group.

Option (2) is true