Test: Group Theory - 10 - Mathematics MCQ

Order of (1 2) (5 4 6) (3 9 7 8), order of (1 2) = 2, order of (5 4 6) = 3, order of (3 9 7 8) = 4
Thus,order of composite = 1.c.m. of 2,3 and 4=12

G is a group of order 7 which is prime. So group will be cyclic and each element can generate it. Now, let x generate the group, then x1 = e.

Thus, φ is one-one and onto.
Also, 

Given S₁ = {2} ,
S₂ = {4 , 6} ,
S₃ = {8, 10, 12},
S₄ = {14,16,18,20}
S₁₀ = {9 2 , 94, 96, 98, 100, 102, 104, 106, 108, 110}
So, The sum of elements of S₁₀ is equal to 10/2 x 202 = 1010

x*y = 3xy
Let e is identity then
x*e =x implies 3xe = x implies e = 1/3
Let inverse of 2 is a then
  implies 3 x 2 x a = 1/3, So, a = 1/18. 

Given G be a group with identity e such that for some a ε G, a² ≠ e and a⁶ = e.
Hence a⁴ ≠ e , a⁵ ≠ e .

If we take G = { 1 , - 1 , i, - i } then G has a non-trivial proper subgroup H = {1, - 1}
So, option (a) is discarded.
option (b) is discarded because identity is self inverse element in any group, option (c) is discarded because if we take Z₆ = {0,1,2, 3,4, 5} then Z₆ is a group but it is not power of 2.
For even order group apart from identity there will odd number of elements. Since inverse element exist in pairs so at least one element will be it own inverse including identity such elements will be two.

Number of disjoint set will be chosen from remaining 7 elements.
So, number of disjoint set
= ⁷C₀ + ⁷C₁ + ⁷C₂ + . . . + ⁷C₇ = 2⁷ = 128.

Let P be a set having n > 10 elements. The number of subsets of P having odd number of elements is ⁿC1 + ⁿC₃ + ⁿC₅
+ⁿC₇ + ⁿC₉ + ⁿC₁₁ = 2^n-1.

Given the set {1, 3, 7, 9} under the operation of multiplication modulo 10. i.e. U(10)= {1,3, 7, 9}
This is cyclic group because 3 and 7 are generators.
Hence, It has a unique generator is false.

If P ∪ Q = P ∪ R then Q = R is incorrect.
For example, Let P = { 1,2 ,3 } ,Q = { 1,2 } ,R = {1} P ∪ Q = { 1 ,2 , 3} , P ∪ R = {1, 2, 3}. But Q ≠ R. 

Number of subset having odd number of elements
=¹⁰C₁ + ¹⁰C₂ + ... + ¹⁰C₉ = = 512
Since Z_I2 under addition modulo 12 is a cyclic group
So, Total number of non-trivial proper subgroups
= Z(12) - 2 
= 6 - 2 = 4

F is field with 5 elements K = {{a, b )| a, b ε F}
Since, operation is component-wise thus all properties of group will be satisfied component-wise.
All elements of K will possess inverse and there is identity thus K is a field with 5 x 5 = 25 elements.

We know that

where p(r) =Number of partitions of the set.

Option (a) is incorrect 
Option (b) is incorrect because every group of finite order may or may not be cyclic.
For example K₄ is finite grup but it is not cyclic.
Option (c) is correct because every cyclic group is abelian.
Option (d) is incorrect because A4 is a group and 6/o(A₄)
But A₄ has not a subgroup of order 6.
 

α = (1 3) (2 5 4)
|order of (1 3) = 2
order of (2 5 4) = 3
order of α = 1.c.m. of 2 and 3 = 6

So, α = 1
Thus α⁶⁰ = 1
Thus α⁶⁵ = α⁶⁵ . α⁶⁵ = α⁵ = α^-1
So, inverse of a will be (4 5 2) (3 1) or (2 4 5) (3 1)
Thus (d) is the correct option.

For a, b ε z, define a relation aRb if ab ≥ 0 then the relation R is an equivalence relation.
Hence, the relation R is symmetric, reflexive and transitive.

Given R be a commutative ring with unity of characteristics 3.
So a + a + a = 3a = 0
Hence, (a + b)⁶
=⁶C₀a⁶ + ⁶C₁a⁵b + ⁶C₂a⁴b²+ ⁶C₃a³b³+ ⁶C₄a²b⁴+ ⁶C₅ab⁵ + ⁶C₆b⁶
=a⁶ + 0 + 0 + 0 + 0 + 0 + b⁶
= a⁶ + b⁶

 A non empty set A is called an algebraic structure w.r.t binary operation “*” if (a*b) belongs to S for all (a*b) belongs to S. Therefore “*” is closure operation on ‘A’.