Test: Group Theory - 11 - Mathematics MCQ

Given σ = (1 2 5) • (3 6) 

We need to find out the remainder when 8^31 is divided by 13

We should try and reduce the numerator in this case.

Let us have a look at the method below.

Remainder [8^31 / 13] = Remainder [8*8^30 / 13]

Remainder [8*64^15 / 13] = Remainder [8*(-1)^15 / 13] {Here We divide 64 by 13 so remanider is (-1)}

So

= Remainder [(-8) / 13]= 5

In (Z , + ) total number of automorph is m is 2. So, Number of distinct group homomorphism form (Z, +) onto (Z, +) is 2.

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’.

Given 
The order of the subgroup of S₅ generated by
= L.C.M. ( 3 ,2 ) = 6

Z₆ is not a field.
Here, p = 3
So, Z_2p;p is prime is false.
Hence, option (a) is discarded.
So, 
If a, b ε Z₃ then a² + b² cannot be zero. So , it is invertible so set is field.
But, this is not the case if a, b ε Z₄.

H is a subgroup of order 60 of group G of orders 120.
^Thus, |
will contain only two distinct elements.
These are H and Ha (or aH)
If a is not identity, then aH will not have identity element.

Given G is a group of order 51.
i.e. o(G) = 51 = 3 x 17
All subgroup of group is cyclic.
The subgroups 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.

{1 , 2 , . . . , 16} are non-zero elements of 3⁴ = 81 = 13(mod 17)
3⁸ = 13 x 13(mod 17)
    = 16(mod 17)
3¹² = 13 x 16(mod 17)
= 208(mod 17)
= 4(mod 17)
3¹⁶ = 13 x 4(mod 17)
= 52(mod 17) =l(mod 17)
The order o f the element 3 is 16 i.e. o(3) = 16

Here, (a, b) ε S so (b, a) ε S.
Since S is transitive.
So, (a, a) ε S so it is reflexive.

The number o f elements in   is atmost .

If we take matrix A = then A^-1 does not exist.
Hence, option (a) discarded.
If we take a finite group S₃ which is not cyclic the option (e) discarded.
Also , the set o f all 2 x 2 real non singular matrices may or may not form an abelian group under matrix multiplication.
Hence, option (d) discarded.
Hence, a finite abelian group of order 6 has exactly two non-trivial subgroups, i.e. a finite abelian group of order 6 is isomorphic to Z₆ and Z₆ is cyclic and it has exactly two non-trivial subgroups.

Given that X denote the set of all real valued functions defined on Z. Define a relation in X by f ~ g if f(0) ≠ g(0).
So, f(0) ≠ f(0) which fails reflexive relation.
If wetake f(x ) = 1- x
g(x) = x - l
Here , f(0) ≠ g(x) and we take h(x) = 1 - 2x
Here g(0) ≠ h(0)
But f(0)= h(0).
This fails transitive relation.
Hence, option (d) is correct.

Z₂ = { 0 , 1 }
Ideal of R generated by x³ + x + 1
I(0) = 0 + 0 + 1 = 1 
I(1) = 1 + 1 + 1 = 1 
Thus (x³ + x + 1) is irreducible
Now R/I will contain element of the form I + g(x) By division algorithms
g(x) = f (x) x (x² + x + 1) + r(x)
deg[r(x)] < deg(x³ + x + 1)
deg[r(x)] < 3
So r(x) will be o f the form ax² + bx + c, a, b,c e Z₂
So , elements of R/I will be I + f ( x ) x (x³ + x + 1) +r(x)
or, 1+ ax² + bx + c
a, b,c each can be chosen in 2 ways
So , total number of elements in R/I will be 8.

If X Δ Z = Y Δ Z for the non empty sets X , Y and Z, where Δ represents the symmetric difference.
Put X = Y then above condition also hold.
Hence, Option (a) is correct.

If order of G is 24
i.e. o(G) =24 = 2³ x 3 and abelian.
The number of subgroups of order 3
= 1 + 3k/2³
= 1 where k = 0
Hence, exactly one subgroup of order 3.

The number of group homomorphisms form the group (Z18, ) to group (Z30, ) is equal to g.c.d.(18,30) = 6.

H is normal subgroup of G of order 2.
H will contain an identity and one more element.
Since,
Thus, 
Thus, a will belong to centre.
So, centre of G will contain atleast two element, identity and a.
Also, centre is subgroup G.
Also, H will be a subgroup of the centre, so its order will be an even number.

Let
= (1 2 10 12 7 6 3 8 11) (4 5 9)
The cardinality of the orbit 2 under σ is 9.