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 \( y \in G \) and every positive integer \( n \), there exists an element \( x \in G \) such that \( x^n = y \).
• Group \(\mathbb{Q} \)  under Addition: For any rational \( y \in \mathbb{Q} \) and any \( n \in \mathbb{Z}^+ \), \( x = \frac{y}{n} \in \mathbb{Q} \) satisfies \( nx = y \) ⇒ divisible.
• Group \(\mathbb{C} \setminus \{0\} \)  under Multiplication: Every non-zero complex number has an \( n^{th} \) root in \( \mathbb{C} \setminus \{0\} \) ⇒ divisible.
• Finite Groups: A finite group cannot be divisible, because if \( G \) has finite order, then not all equations like \( x^n = y \) are solvable for arbitrary \( y \in G \).
• Cyclic Group of Order 5: Has only 5 elements. For \( n = 10 \), not every element has an \( x \in G \) such that \( x^{10} = y \) ⇒ not divisible.
• Symmetric Group \(S_5 \) : 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 \( y \in G \) and every \( n \in \mathbb{Z}^+ \), there exists \( x \in G \) such that \( x^n = y \)

Step 1: Check  \(\mathbb{Q} \)  under addition

⇒ Let \( y \in \mathbb{Q}, n \in \mathbb{Z}^+ \Rightarrow x = \frac{y}{n} \in \mathbb{Q} \)

⇒ \( nx = y \Rightarrow \) divisible

Step 2: Check \(\mathbb{C} \setminus \{0\} \)  under multiplication

⇒ Let \( y \in \mathbb{C} \setminus \{0\}, n \in \mathbb{Z}^+ \)

⇒ Take any \( n^{th} \) root of \( y \)

⇒ Roots exist in \( \mathbb{C} \)

⇒ divisible

Step 3: Check cyclic group of order 5

⇒ Order is 5

⇒ For \( n = 2,3,4,\dots \) roots might not exist

⇒ Not divisible

Step 4: Check symmetric group \(S_5 \) 

⇒ Finite group of order 120

⇒ Not every element has an \( n^{th} \) 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 = \(\{g \in G: g h=h g \text { for all } h \in H\}\),

H is normal in K , as K centralizes H

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

Since  \(H \subseteq K \)  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  \(H \cap gHg^{-1} = \{e\} for all g \notin H , \) 

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  \(L \subseteq G \)  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 \(S_3\) possible Order be lcm (3,1) so maximum possibility be 3 

Therefore, Option 1) is wrong 

In, \(S_4 \)  maximum possibility be 4 

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

In, \(S_5\) 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 = \(\rm \begin{bmatrix}0&-1\\ 1&0\end{bmatrix}\) So S² = \(\rm \begin{bmatrix}0&-1\\ 1&0\end{bmatrix}\)\(\rm \begin{bmatrix}0&-1\\ 1&0\end{bmatrix}\) = \(\rm \begin{bmatrix}-1&0\\ 0&-1\end{bmatrix}\) = - I, S3 = -S, S⁴ = I

Also, 𝑇1 = \(\rm \begin{bmatrix}i&0\\ 0&i\end{bmatrix}\)

So, 𝑇1 = i𝑇1, \(𝑇_1^2 =-I\), \(𝑇_1^3 =-iI\), \(𝑇_1^4 =I\)

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 = < 𝑆, 𝑇1 >, where 𝑇1 = \(\rm \begin{bmatrix}i&0\\ 0&i\end{bmatrix}\),

𝐺2 = < 𝑆, 𝑇2 >, where 𝑇2 = \(\rm \begin{bmatrix}i&0\\ 0&-i\end{bmatrix}\),

𝐺3 = < 𝑆, 𝑇3 >, where 𝑇3 = \(\rm \begin{bmatrix}0&1\\ 1&0\end{bmatrix}\).

Then 𝐺1 = \(\{S^iT_1^j|i, j\in\mathbb Z\}\) is non-abelian group of order 8

Similarly, 𝐺2 = \(\{S^iT_2^j|i, j\in\mathbb Z\}\) is non-abelian group of order 8

𝐺3 = \(\{S_1, T_3|S^4=I, T_3^2=I, T_3S=T_3S^{-1}\}\) ≈  D₄

So, |𝑍(𝐺3)| = |𝑍(D4)| = 2

Hence 𝑍(𝐺3) = {\(\rm \begin{bmatrix}1&0\\ 0&1\end{bmatrix}\)} 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 \(\tau(n)\)

Explanation- 

Number of subgroups of a cyclic group of order 50 is \(\tau (50) \) 

\(\tau (50) = \tau(5^2.2)\) = (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 \(\mathbb{Z}_{45} \ \ and \ \ \mathbb{Z}_{5} \times \mathbb{Z}_{3} \times \mathbb{Z}_{3}\)

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 \(p| q-1\) 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 ∈ \(\mathbb Z\) 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