IIT JAM Mathematics Practice Test- 7 - Mathematics MCQ

(N, *) where 

closure property a* b = a ∈ N

Associativity (a * b) * c = a * c = a

a * (b * c) = a * b = a

a * (b * c) = a * b = a

identity

a * e = a

a = a

⇒ there does not exist identity

⇒ (N, *) is not a group.

(B) (Z, *) where a * b = a - b for all a, b ∈Z (Z, *) is a group if all four properties of the group are satisfied. These properties are

(1) Closure property : If a, b ∈ Z, then a * b = a - b∈Z => closure property is satisfied

(2) Associativity => If a, b, c ∈ Z, then (a * b) * c = (a - b) * c = a - b - c ...(i)

and a*(b*c) = a * ( b - c ) = a - ( b - c ) = a - b + c ...(ii)

Hence by Eqs. (i) and (ii) (a * b) * c≠a*(b*c)

Hence the property of associativity is not satisfied.

=> (Z, *) is not a group.

(C) (Q, *) where  for all a, b ∈ Q. This is not a group since inverse of element 0 ∈ Q does not exist.

For; identity => let identity is e, then a * e = a, let a ≠ o .

⇒ if a ≠ 0 then e = 2, Thus if a = 0 then e cannot be 2

⇒ inverse of a = 0 does not exist. Hence it is not a group

(D) (R, *) w here a * b = a + b + 1 ,a , b ∈ R This is a g roup with identity given as a *

e = a + e +1 = a

⇒ a + e + 1 = a

⇒ e + 1 = a - a

⇒ e + 1 =0

⇒ e = - 1

Since - 1 ∈ Q, so e ∈ Q, So (R, *) is a group with identity - 1 .

Let (1) and (2) hold. To show G is a group, we need prove existence of identity and

Let a ∈ G be any element inverse.

By (2) the equations ax = a

ya = a

have solutions in G.

Let x = e and y = f be the solutions.

 ae = a

fa = a

Let now b e G be any element then again by (2) some x, y in G s.t.,

ax = b

ya = b.

Now ax = b ⇒ f.(a.x) = f.b

⇒ (f.a).x = f.b

⇒ a.x = f.b

⇒ b = f.b

Again y.a = h

⇒ (y.a).e = b.e

⇒y.(a.e) = b.e

⇒ y.a = be

⇒ b = be

thus we have b = fb ... (i)

b = be ... (ii)

for any b ∈ G

Putting b = e in (i) and b = f in (ii) we get

e = fe

f = fe

⇒ e = f.
Hence ae = a = fa = ea

i.e., e ∈ G, s.t., ae = ea = a

⇒ e is identity.
Again, for any a ∈ G, and (the identity) e ∈ G, the equations ax = e and ya = e have solutions.
Let the solutions be x = a₁, and y = a₂

then aa₁ = e,

a₂a = e

Now a₁ = ea₁ = (a₂a)a₁ = a₂(aa₁) = a₂e = a₂.

Hence aa₁ = e = a₁a for any a ∈ G

i.e., for any a ∈ G, some a₁ ∈ G satisfying the above relations ⇒ a has an inverse. Thus each element has inverse and, by definition, G forms a group.

Since Z₁₆ = {0, 1 , 2 ..... . 15}

The numbers 1, 3, 5, 7, 9, 11, 13,15 are the elements of Z₁₆ that are relatively prime to 16. Each of these elements generates Z₁₆.
So no. of generators = 8

Centre of G is 

Since 0(G) = p³, p is a prime number 

But Z is a subgroup of G => 0 ( Z ) / 0 ( G ) i.e. 0 ( Z ) 

divides 0(G) ⇒ 0 (Z) / p³

⇒ 0 ( Z ) = p or p² or p³.
If 0 (Z) = p³ = 0 (G) ⇒ Z = G => is abelian co ntradiction.

If 0 ( G ) = p² ⇒ 0(G /E ) 

i.e ., G/Z is a group of prime order 'p' and so is cyclic ⇒ G is 

obelian ⇒ contradiction  ⇒ O(Z) ≠ p² hence O(Z) = p]

Since ∼r is true, therefore, r is false. Also, q ⇒ r is true, therefore, q is false. (Therefore, a true statement cannot imply a false one) Also, p ⇒ q is true, therefore, p must be false.

Group of symmetries S_2n = { ....... r³, r², r, e, f, fr, fr² ......}

this group of symmetric is non-abelian for n ≥ 3 so if S_n be the symmetry group and it is abelian then n = 1 or n = 2.

Cyclic group - A group G is said to be cyclic, if, for some a ∈ G, every element x ∈ G is of the form aⁿ, where n is some integers. The element a is called a generator of G.

There may be more than one generators of a cyclic group. If G is a cyclic group generated by a, then we shall write G = {a} or G = (a). The elements of G will be of the form 

..., a^-3, a^-2, a^-1, a° = e, a, a², a³, ...

Some properties 

(i) Every cyclic group is an abelian group. 

(ii) Every subgroup of a cyclic group is cyclic. 

(iii) Every proper subgroup of an infinite cyclic group is infinite.

Hence by property III, if H is a subgroup of an infinite cyclic group, then H is a also infinite.

Let D = H ∩ n K then D is a subgroup of K and as in the proof of Lagrange’s theorem.  a decomposition of K into disjoint right cosets of D in K and

and also

Again  and since  

Thus  

now no two of  Hk₁, Hk₂,....., Hk_t can be equal as if Hk_i = Hk_j for i, j

then 

which is not true

Hence 

which proves the result

By Lagranges theorem the order of every subgroup of a finite group is a divisor of the group.So n|m.

Since,in a group there must be an identity element.Also it is its own inverse. Therefore,minimum number of elements of G,which are their own inverse=1

A non empty subset H of a group G is said to be a subgroup of G. If under the operation defined on G. H itself forms a group. According to result on sub-group

(i) Every subgroup of a cyclic group is also cyclic.

(ii) Every proper subgroup of an infinite cyclic group is infinite.
Both are true.

(i) true only since every quotient group of a cyclic group is cyclic.

If T = {z ∈ s c : | z | = 1} then (T, x) is said to be a circle group All though the circle group has infinite order it has many finite subgroups as the complex numbers satisfying the equation zⁿ = 1 are called the nth root of unity. They are

Since prime order group have no proper subgroup as well as simple group have no proper subgroup so (1) is false.

Let G be the group of nonzero real numbers under multiplication,

H = {x ∈ G | x = 1 or x is irrational} and

K ={x ∈ G | x ≥ 1}then H is not a subgroup of G since 

Also, K is not a subgroup since  

We know that every cycle is permutation but every permutation is not a cycle and it is also true that S_n is not cyclic for all n and every permutation s ∈ S_n can be written as product of (n -1) transposition.

Every subgroup of an abelian group is normal as for every x ∈ G and h ∈ H

Every subgroup of cyclic group is normal and intersection of two normal subgroups is normal subgroup again.
If N is a normal subgroup of G and H is any subgroup of G then NH is a subgroup of G but NH is necessarily not normal.

If H, K be two subgroups of a group G then union of subgroups is not necessarily a group of G.
But intersection of subgroups is a subgroup of G but H ∩ K may not be a subgroup of H. HK is subgroup of G when HK = KH. By Lagrange's theorem we can say that order of H and K are divisors of order of G.

(A), (B), (C) are correct statements. If two cosets aH and bH are equal then aH and bH are identical.

In the given set G of all 2 x 2 matrices 

where ad ≠ 0 under matrix multiplication and   Where a, b, d are real numbers, the N will be a subgroup of G and according to the definition of normal subgroup. A subgroup H of a group G is said to be a normal subgroup of G if for every x ∈ G and for every h ∈ H, xHx^-1 ∈ H, so here, N is a subgroup of G and N will be normal in G.

Hence smallest positive integer is n = 7

Group G has 35 elements, i.e. its order is 35

So, possible subgroup sizes can be 1, 5, 7, 35.

Thus the largest possible size of subgroup other than G itself (proper subgroup) is 7.

Given system of linear equations:
x + y + z = 1 ….(1)
2x + 3y + 2z = 1 ….(2)
2x + 3y + (a² - 1)z = a + 1 …..(3)
Consider a² - 1  = 2
then LHS of (2) and (3) are same but RHS are not.
Hence a² = 3 => |a| = √3
For |a| = √3, system is inconsistence.
So option (b) is correct.

The identity of the multiplicative group of integers modulo 5 is {1}. 

Here

{3}.₅{3} = 4 * {1} [•, 3.3 0 4 (mod 5)]

∴ {3}.₅{3}.₅{3} = {4}.₅{3} = {2} * {1} [∵, 4.3⁰ 2 (mod 5)]

∴{3}.₅{3}.₅{3}.₅{3} = {2}.₅{3} = {1} [∵, 2.3⁰ (mod 5)]

Hence, the order of {3} is 4.

(1)⁶ = 1 + 1 + 1 + 1 + 1 +1 = 6 (mod 6) = 0
∴ Older of 1 is 6.

The number of proper subgroups of the group (Z, +) of integers is ∞

Partitions of 3 are three namely 

3 = 3,2 + 1, 1 +1 +1.

Partition of 4 are five namely 

4 = 4, 3 + 1, 2 + 2, 2 + 1 +1, 1 +1 +1 +1. 

Partition of 1 is only one 1 = 1 

Thus there are (I) non isomorphic abelian group of order 2. 

(II) five non isomorphic abelian group of order 3⁴. and 

(III) One abelian group of order 5.

Hence total number of non abelian group of order

2³. 3⁴. 5 = 3 x 5 x 1 = 1 

=15

Because 0(1245) = 4 and o(36) = 2

We know that all the subgroup of quotient group

where H₁,H₂..........H_m are subgroup of 

Here ’ then its subgroups are 

two proper subgroups, two improper.