Group & Subgroups MCQ Quiz in मल्याळम - Objective Question with Answer for Group & Subgroups - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 17, 2025
Latest Group & Subgroups MCQ Objective Questions
Top Group & Subgroups MCQ Objective Questions
Group & Subgroups Question 1:
Which of the following is not true?
Answer (Detailed Solution Below)
Group & Subgroups Question 1 Detailed Solution
Concept:
(i) Let G be a group. A group H is said to be a normal subgroup of G if ghg-1 ∈ H for all g ∈ G and h ∈ H.
(ii) A group is abelian if (ab)3 = a3b3 and no elements of order 3.
Explanation:
Option (4): let G be a group and H be a commutative subgroup.
We have to prove that H is normal subgroup of G,
ghg-1 = hgg-1 = he = h ∈ H
So, H is normal subgroup of G
Option (4) is true.
Option (2): G is a group in which (ab)3 = a3 b3 for all a, b ∈ G,
Then G is abelian.
H = {x3 : x ∈ G}
Let h3 ∈ H and g ∈ G
then gh3g-1 = h3gg-1 = he = h ∈ H (as G is abelian)
So, H is normal subgroup of G.
Option (2) is true
Similarly, we can prove that option (1) is true.
Hence option (3) is not true
Group & Subgroups Question 2:
Consider the set of matrices
G = \(\left\{\left(\begin{array}{ll} s & b \\ 0 & 1 \end{array}\right): b \in \mathbb{Z}, s \in\{-1,+1\}\right\}\)
Then which of the following is true?
Answer (Detailed Solution Below)
Group & Subgroups Question 2 Detailed Solution
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 = \(\left\{\left(\begin{array}{ll} s & b \\ 0 & 1 \end{array}\right): b ∈ \mathbb{Z}, s ∈\{-1,+1\}\right\}\)
Option (1)
As, \(\begin{bmatrix} -1 & 1 \\ 0 & 1 \end{bmatrix}\) , \(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) ∈ G
But \(\begin{bmatrix} -1 & 1 \\ 0 & 1 \end{bmatrix}\) + \(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) = \(\begin{bmatrix} 0 & 2 \\ 0 & 2 \end{bmatrix}\) ∉ G
⇒ Option (1) is false
Option (2)
As, x = \(\begin{bmatrix} -1 & 1 \\ 0 & 1 \end{bmatrix}\) ∈ G , y = \(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) ∈ G
and xy = \(\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\) , yx = \(\begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix}\)
Clearly, xy ≠ yx
⇒ Option (2) is false
Option (3)
A = \(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) ∈ G
Eigen value of A are 1,1 ⇒ Algebraic Multiplicity of 1 = 2
Now, (A-I) = \(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\)
Here, η(A-I) = 1 = Geometric Multiplicity ⇒ Geometric Multiplicity < Algebraic Multiplicity ⇒ Not diagonalizable
⇒ Option (3) is true
Option(4)
S = \(\left\{\begin{bmatrix} -1 & 1 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \right\}\) ⊆ G
Clearly, ⊆ G
Now, To show G ⊆
Also \(\begin{bmatrix} s & b \\ 0 & 1 \end{bmatrix}\) is either of the form \(\begin{bmatrix} -1 & b \\ 0 & 1 \end{bmatrix}\) or \(\begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix}\)
Now, \(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^b = \begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix}\) & \(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^{1-b}. \begin{bmatrix} -1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -1 & b \\ 0 & 1 \end{bmatrix}\)
Clearly, G ⊆
⇒ G =
Hence, (4) option is true
Group & Subgroups Question 3:
Let A ⊆ ℤ with 0 ∈ A. For r, s ∈ ℤ, define
rA = {ra : a ∈ A}, rA + sA = {ra + sb : a, b ∈ A}.
Which of the following conditions imply that A is a subgroup of the additive group ℤ?
Answer (Detailed Solution Below)
Group & Subgroups Question 3 Detailed Solution
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 \(\forall\) 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).
Group & Subgroups Question 4:
Which is the first odd integer for which a non-abelian group exists?
Answer (Detailed Solution Below)
Group & Subgroups Question 4 Detailed Solution
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 \(Z_{23}\)
and In 27 it is \(3^3\) which is abelian
and for 33 we can write 3.11 where (3-1) does not divide 11
Therefore, Correct Option is Option 3).
Group & Subgroups Question 5:
Which of the following is/are true?
Answer (Detailed Solution Below)
Group & Subgroups Question 5 Detailed Solution
Concept -
Class equation of \(D_n\)
\(O(D_n) = 1+1+2+..2(\frac{n-2}{2}) \ times +\frac{n}{2} +\frac{n}{2}; n \ is\ even\)
\(1+2+..2(\frac{n-1}{2})\ times +n ; (\ n \ odd)\)
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).
Group & Subgroups Question 6:
For n ≥ 1, let Sn denote the group of all permutations on n symbols. Which of the following statement is true?
Answer (Detailed Solution Below)
Group & Subgroups Question 6 Detailed Solution
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).
Group & Subgroups Question 7:
If U(8) = {1, 3, 5, 7} and U(10) = {1, 3, 7, 9}, then find how many element are there in U(8) × U(10)?
Answer (Detailed Solution Below)
Group & Subgroups Question 7 Detailed Solution
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).
Group & Subgroups Question 8:
Let S5 be the symmetric group on five symbols. Then which of the following statements is false?
Answer (Detailed Solution Below)
Group & Subgroups Question 8 Detailed Solution
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.
Group & Subgroups Question 9:
Let G be a group of order pn, p a prime number and n > 1. Then which of the following is true?
Answer (Detailed Solution Below)
Group & Subgroups Question 9 Detailed Solution
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.
Group & Subgroups Question 10:
Let X be a non-empty set and P(X) be the set of all subsets of X. On P(X), define two operations ⋆ and Δ as follows: for A, B ∈ P(X), A ⋆ B = A ∩ B; AΔB = (A ∪ B)\(A ∩ B).
Which of the following statements is true?
Answer (Detailed Solution Below)
Group & Subgroups Question 10 Detailed Solution
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.