Group & Subgroups MCQ Quiz in मराठी - Objective Question with Answer for Group & Subgroups - मोफत PDF डाउनलोड करा

Concept:

G is finitely generated group if it has finite generating set S. So that every element of G can be written as combination of finitely many elements of finite set S.

Explanation:

G =  

Option (1)

As,  ,  ∈ G

But  +  =  ∉ G

⇒ Option (1) is false

Option (2)

As,  x =  ∈ G , y =  ∈ G

and xy =  , yx = 

Clearly, xy ≠ yx

⇒ Option (2) is false

Option (3)

A =  ∈ G 

Eigen value of A are 1,1  ⇒ Algebraic Multiplicity of 1 = 2

Now, (A-I) = 

Here, η(A-I) = 1 = Geometric Multiplicity ⇒ Geometric Multiplicity Algebraic Multiplicity ⇒ Not diagonalizable 

⇒ Option (3) is true

Option(4)

S =  ⊆ G

Clearly,  ⊆ G

Now, To show G ⊆  

Also  is either of the form  or 

Now,   & 

Clearly, G ⊆  

⇒ G =  

Hence, (4) option is true

Concept: 

One-step test for subgroup: Let H be a subset of G. Then (H, +) be a subgroup of (G, +) if and only if H is non-empty  a, b belongs to H  and a - b belongs to H 

Explanation: 

If we assume A be 2Z which is a subset of Z 

so 2A or -2A be subset of A and its addition or substraction 

is equal to A 

−2A ⊆ A, A + A = A  and 

2A ⊆ A, A + A = A 

Therefore, the correct options are (1) and (4).  

Explanation:

We need to find the first odd integer for which a non-abelian group exists- 

we can write 21 as 7 × 3 which are multiples of primes 

and 3|6 so possibility be it is isomorphic to cyclic group 

and one non abelian group 

but in case of 23 it is prime that is only isomorphic  

and In 27 it is  which is abelian 

and for 33 we can write 3.11 where (3-1) does not divide 11 

Therefore, Correct Option is Option 3).

Concept - 

Class equation of  

Explanation -

o(D5) = 1 + 2 + 2 + 5 

o(D9) = 1 + 2 + 2 + 2 + 2 + 9 

o(D7) = 1 + 2 + 2 + 2 + 7 

o(D4) = 1 + 1 + 2 + 2 + 2 

Therefore, Correct Option(s) are Option 1) , option 2), option 3) and Option 4).

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).

Solution-  

Given, U(8) = { 1, 3, 5, 7} and U(10) = {1, 3, 7, 9} 

Here, order of U(8) is 4 and order of U(10) is 4 

So, number of element in U(8) × U(10) = 4 × 4 = 16 

and elements of U(8) × U(10) = {(1,1),(1,3),(1,7),(1,9),(3,1)........(7,9)}

Therefore, Correct Option is Option 4).

Explanation:

(1) σ = (123) (45) is an element of order 6, thus ⟨σ⟩ ≤ S₅ is a cyclic subgroup of order 6.

(2) ⟨(1234),(12) (34)⟩ ≅ D₄ is non-abelian group of order 8, in S₄.

(3) K₄ = {I, (1 2) (3 4), (1 3) (2 4), (1 4) (3 2)} ≅ ℤ₂ × ℤ2 is a subgroup of S₅. So (3) is false.

(4) For a cyclic subgroup of order 7, we need an element of order 7, which is not in S₅.

Explanation:

(1) Any group with order pⁿ, n > 1 has non- trivial center.

And hence option (1) is correct.

(2) G = Q₈ is non-abelian group of order 2³.

So option (2) is false.

(3) Q₈ has five normal subgroups, so option (3) is false.

(4) ℤ₈ and ℚ₈ are two non-isomorphic subgroup of order 8.

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.