Iit Jam Mathematics Answer Key 2016

Set s = { 1 , 2 , 3 , 4 , 5 } and let a permutation a = (1 2 3 ) (1 4 5 ) then a⁹⁹ is equal to __________.

A.
a
B.
I
C.
a³
D.
a⁴

Solution:

Set S = {1 , 2,3 ,4 ,5 } then by the definition of identity permutation a⁵ = I

where a = (1 2 3) (1 4 5)

Find the smallest subgroup of z containing 8 and 14 both.

A.
z
B.
2z
C.
4z
D.
6z

Solution:

Let H be a subgroup of z containing 8 and 14. since z is a cyclic group so every subgroup of cyclic group is also cyclic.

i.e. H = [ m ] , where m ∈ H

Now 8 and 14∈ H = [ m ]

=> m | 8 and m | 14

=> m divides g.c.d ( 8 , 1 4 )

=> m | 2

=> either m = 1 or m = 2

If m = 1 then H = Z if m = 2 then H = [ 2 z ]

Let G be a group of order 200, then the no. of subgroups of G of order 50 is

A.
1
B.
2
C.
4
D.
does not exists

Solution:

0(G) = 200

then integral divisor of 200 is 1 ,2 , 4, 10 ,20, 25,50,100, 200. which shows that G has all these subgroups with orders 1 , 2, 4, 10, 20, 25, 50,100 and 200 where order 1 and 200 subgroups are improper subgroups and remaining all are proper subgroups of group G, which show that G has only one subgroup of order 50.

Given axa = b in a group G , where a,b ∈ G then x is equal to

A.
a^-1b
B.
a^-1b^-1
C.
a^-1b^-1b^-1
D.
a^-1ba^-1

Solution:

If G be a group and a,b ∈ G

then

The generators of a group are

A.
a and a⁵
B.
a² and a⁴
C.
a³ and a⁶
D.
a and a³

Solution:

Here 0 ( G ) = 6 and G = [ a ]

then a cyclic group has generators of the type a^m where m < 6 [ m is co-prime to 6 ]

i.e. m = 1 , 5

Hence generators are a and a⁵.

If H be a subgroup of G, then H is normal in G, if

A.
H is a subgroup of index 2 in G.
B.
H is a subgroup of index 3 in G.
C.
H is a subgroup of index 4 in G.
D.
H is a subgroup of index 5 in G.

Solution:

By the theorem of normal subgroup. We know that if H is a subgroup of group G and H is normal in G then H is a subgroup of index 2 in G.

An element a^p of a finite cyclic group G of order n is a generator of G if 0 < p < n and also

A.
p is prime to n
B.
p is multiple of n
C.
n is the multiple of n
D.
None of these

Solution:

G be a finite cyclic group and a ∈ G be the generator of G then by theorem we know that an element a^p ∈ G 's also generator of finite cyclic group G if and only if p is co-prime with n [ order of the group G ] "If G = [ a] and G is finite with order n. then the other generators of G are of the form a^r, where r is relatively prime to n.

Let G = {e, a, a², a³, b, ab, a²b, a³b} with a⁴ = e, b² = e, ab = a^-1b and H = {e, a², ab, a³b}, K = {e, a³b}then which one of the following is false statement ?

A.
H is normal subgroup of G.
B.
K is normal subgroup of G.
C.
K is normal subgroup of H.
D.
[G : H] [H : K] = 4

Solution:

Clearly, H and K are subgroups of H.

Hence, H is normal subgroup of G and K is normal subgroup of H.
But K is not normal subgroup of G. Since a³b ∈ K and b ∈ G We have, b(a³b)b^-1 = ba³ ∉ K.

The no. of elements of order 5 in a symmetric group S₅ is

A.
5
B.
20
C.
24
D.
12

Solution:

Order of element is 5 in the symmetric group S₅ , i. e. we have to find total no. of cycles of length 5.

Given n = 5 and r = 5

Let G be a group of order 77. then, the centre of G is isomorphic to

A.
Z₍₁₎
B.
Z₍₇₎
C.
Z₍₁₁₎
D.
Z₍₇₇₎

Solution:

0 ( G ) = 77 = 7 x 11 ( p < q ) and p f q -1 then by a theorem we can say that G must be a cyclic group , i.e. G is an abelian group.

If G is an abelian then centre of group G is equal to G i.e. Z (G ) = G => 0 [ Z ( G ) ] = 7 7

if O [ Z (G) ] is 77 then Z(G) is isomorphic to Z₍₇₇₎

S₅ be the permutation group on 5 symbols, then number of element in S₅ such that a⁵ = a²

A.
10
B.
15
C.
20
D.
30

Solution:

Given symmetric group is S₅.

Then we have to find total no. of elements which satisfied the condition a⁵ = a²

i.e. a⁵ = a² => a³ = e

so, find total no. of elements of order 3. i.e. total no. of cycles of length 3 in S₅

Here r = 3 , n = 5

Which one of the following is smallest group ?

A.
(,*)
B.
({ }, *)
C.
({e}, *)
D.
(({}, {e}, *)

Solution:

In group ({e}, *) the set G = {e} has only one element,so it is smallest group. If G = (, *) then it is called empty set and ({}, *) is not a group.

Number of elements of order 3 in Z₄₅ is

A.
3
B.
4
C.
5
D.
2

Solution:

By the theorem "Total no. of elements of order p in, where p is prime and p , q are distinct primes is p -1"

According the theorem Z₄₅ = i.e. p = 3 and total no. of elements in Z₄₅ of order 3 is 3 -1 = 2

These element are 15,30.

For any integer n > 2, how many elements in ∪(n) that satisfy X² = 1

A.
Only one element
B.
Exactly two elements
C.
atleast two elements
D.
any element in U(n) which satisfy X² = 1

Solution:

∪(n) is a group which contains only those elements which are co - prime with n.

then it is result that U(n) where n > 2 satisfy X² = 1 contains atleast two elements

The no. o f cosets of H in G ,where G = ( Z , + ) and H = ( 8Z , + ) is

A.
0
B.
2
C.
4
D.
8

Solution:

Here total no. of cosets are 8. Because by general result subgroup mZ has m different total cosets.

different cosets are

these ail 8 cosets are distinct cosets 0 + 8Z.

The no. of element of order 10 in Z₃₀ is

A.
2
B.
3
C.
4
D.
5

Solution:

Since the number of elements of order d in a cyclic group of order n is d,where d is the divisor of n.

So no. of elements of order 10 in Z₃₀ is

Which one of the following is a cyclic group ?

A.
G , where 0(G) = 71
B.
G , where 0(G) = 72
C.
G , where 0(G) =75
D.
G , where 0(G) = 77

Solution:

We know that every group of prime order is cyclic , and simple. Here 0(G) = 71 is prime no.Hence it is a cyclic group.

The order of abelian simple group should be

A.
even
B.
odd
C.
prime
D.
composite

Solution:

Let G be an abelian and simple group then each subgroup of G is normal (as G is abelian).It being simple.lt contains no non-trivial subgroups. Hence G contains only trivial subgroups i.e. G is finite of prime order.

If H be normal in G such that 0 (H) and are co-prime then

A.
H is only subgroup of G of 0(H)
B.
H is trivial subgroup of G
C.
G have exactly normal 2 non-trivial normal subgroups of given order
D.
H ≠ φ

Solution:

Let 0(H) = m

Suppose K is a subgroup of G of order m

Since H is normal

Thus 0(HK)|0(G)

Thus d = m and hence

Let a be an element of a group G and 0(a) = 75 then find the order of the element a⁴⁵.

A.
2
B.
5
C.
9
D.
15

Solution:

Let a be an element of a group G and 0(a) = 75 , then by the theorem if 0(a) = n

then

, where d =gcd (75, 45) = 15

Let G be a group with identity e such that for some a ∈ G , a²≠ e and a⁶ = e then which of the following is true ?

A.
a³ = e , a⁴ ≠ e
B.
a⁴ ≠ e , a⁵ ≠ e
C.
a⁴ ≠ e , a⁵ = e
D.
a³ ≠ e , a⁴ = e

Solution:

Given that Gbe a group with identity e such that a² ≠ e and a⁶ = e , for some a ∈ G.

if a² ≠ e ⇒ a⁴ ≠ e and a⁶ = e ⇒ a⁵ = a ⇒ a⁵ ≠ e

Let Gwhere a , b , c, d are integers modulo 2 } then

A.
G is a group of order 4
B.
G is a group of order 2
C.
G is a group of order 6
D.
G is not a group

Solution:

The members of G are

Clearly, G satisfies all group axioms Hence G is a group of order 6.

If H₁ and H₂ are two right cosets of subgroup H , then

A.
H₁ ∩ H₂ = φ or H₁ = H₂
B.
H₁ ∩ H₂ = φ or H₁ ≠ H₂
C.
H₁ ≠ H₂and H₁ ∩ H₂ = φ
D.
H₁ ∪ H₂ = φ or H₁ ⊂ H₂

Solution:

Let G be a group and H be a subgroup of group G , then by the well known result "Any two left (Right) cosets of a subgroup are either identical or disjoint.

i.e. Either H₁ ∩ H₂ = φ or H₁ = H₂

Statement A: All cyclic groups are abelian.

Statement B: Every simple groups are cyclic.

A.
A and B are false
B.
A is true B is false
C.
B is true A is false
D.
A and B both are true

Solution:

By the properties of cyclic group it is clear that all cyclic group are abelian and every group of prime order is simple group i.e. Every simple group must be cyclic.

If H and K are subgroups of order 6 and 8 respectively, then the minimum no. of elements of product set HK is

A.
12
B.
16
C.
24
D.
48

Solution:

Given 0(H) = 6 and 0(K) = 8

Here H∩K is subgroup of H and K both i.e

so O(H∩K) is possible 1 and 2

minimum no. of elements of product set HK is 24.

Find the total no. of non -isomorphic abelian group of order 32.

A.
5
B.
6
C.
7
D.
8

Solution:

By the result "total no. of non - isomorphic abelian group of order pⁿ , where p is a prime no. is equal to the no. of partitions of n."

Here O(G) = 32 = 2⁵

Here n = 5 and p = 2

total no. of partitions of 5 are 7.

{1 + 1+1 + 1 + 1 , 1 + 1+1 + 2 , 2 + 3 , 2 + 2 + 1 , 3 + 1+1,4+1,5}

How many elements of order 5 in A₆ ?

A.
288
B.
72
C.
144
D.
720

Solution:

A₆ is group of all the even permutation and a cycle of odd length is called even permutation. Here 5 length cycle is even permutation, 5 is prime then total no. of elements of 5 in A₆ is

If order of any group G is 144 then find total no. of subgroup of group G.

A.
20
B.
12
C.
15
D.
10

Solution:

0(G) = 144 (given)

⇒ 0(G) = 2⁴ x 3² = 144

∴144 = 2⁴ x 3²

no. of divisor = (4 + 1) (2 + 1)

= 5 x 3 = 15

Let G be a simple group of order 168. What is the no. of subgroups of G of order 7 ?

A.
1
B.
7
C.
8
D.
28

Solution:

0(G) = 168

=> 0(G) = 2³ x3x7

By sylow's third theorem we know that total no. of subgroup of order p (1 + k p) where k= 0 , 1 , 2 ......

Here p = 7 then total no. of subgroups of order 7 are (1 + 7 k )

if k = 0 then 1 + 7k= 1 and 1|168.

if k = 0 then 1+7k = 8 and 8 | 168

so , either k = 0 or k = 1

If k = 0 then subgroup of order 7 is unique and unique subgroup is always normal .
But given that G is simple group and simple group does not contains any proper normal subgroup.

Hence k ≠ 0 , so , k = 1

i.e. G has 8 subgroups of order 7.

Find the total no. of elements of order 5 in a abelian group of order 10.

A.
1
B.
5
C.
4
D.
8

Solution:

Given 0 ( G ) = 1 0 = 2 x 5

Here 2 < 5 and 2 | 5 -1

but also given that G is an abelian group , so it is cyclic also

[ By the result if G is abelian and 0 ( G ) = pq (P < q) and p | q -1 then G is cyclic ]

Here G be a cyclic group of order 10.

So , total no. of elements are 4.

*Multiple options can be correct

Let cyclic group and group G' ={x : x is a solution of xⁿ = 1} , then which of the following is/are true?

A.
Both G and G'are abelian.
B.
G ' is not a cyclic group.
C.
f : G -» G ' is one -one and onto.
D.
f is an isomorphism.

Solution:

By the definition of cyclic group we know that "A group is said to be cyclic if it can be generated by a element a ∈ G,

i. e. every elements of group G can be expressed as a some integral power of an element a ∈ G .

Herebe a cyclic group.

Similarly the multiplicative of n^throots of unity is also a cyclic group.

*Multiple options can be correct

Which statements is/are true ?

A.
Every cyclic group is abelian.
B.
Every group of prime order is cyclic
C.
Every abelian group is cyclic.
D.
None of these

Solution:

It is a very well known result that "Every cyclic group is an abelian group but converse of the theorem is not necessary true",

i.e. Every abelian group is cyclic is not necessary true it may be or not. Similarly Every group of prime order is cyclic but a group of composite order cyclic it is not necessary true.

*Multiple options can be correct

Let S_n be symmetric group on n symbols and A_n be the alternating subgroup of S_n , then

A.
A_n is normal subgroup of S_n
B.
[ S_n : A_n ] =2
C.
Order of A_n is n!
D.
A_n is the set of all odd permutations of S_n.

Solution:

The alternating group A_n of all even permutations of degree n is a normal subgroup of the symmetric group

Let

Here two cases are arise,

case I : If α is an even permutation then α^-1 is also even,

Hence

case II : if α is an odd permutation then α^-1 is also, then

So , Here we can see that

we know that

then

*Multiple options can be correct

Consider the symmetric group S₅ then which one of the following is/are true ?

A.
The no. of elements of order 3 is 20.
B.
The no. of elements of order 2 is 10.
C.
The no. of elements of order 6 is 20
D.
The no. of elements of order 6 is 10.

Solution:

An element of S₅ is of order 6 if and only if it is product of two disjoint cycles of one of length 2 and another of length 3.

lf S = { a , b , c , d , e } then an element of order 6 is of the type (a b c) (d e)

Similarly , an element of S₅ is of order 2 is a cycle of length 2.

and 3 length cycles are

*Multiple options can be correct

If G be a group of order p q r , p < q < r all three no. are primes then which of the following is/are true?

A.
G has a normal subgroup of order r.
B.
G has a normal subgroup of order qr.
C.
G has a normal subgroup of order q , where qX( r - 1)
D.
G has a normal subgroup of order p

Solution:

It is a very well known theorem that " If G be a group o f order p q r ,

p < q < r being primes , then all the subgroup of order r is normal in G.

Hence option (A) is correct.

Again group G has a normal subgroup of order pq .

and , if q + (r -1) then G has a normal subgroup of order q.

But it is not necessary that the subgroup of order p is normal, it may be or not.

*Multiple options can be correct

If in a group G , a⁵ = e , aba^-1 = b² for a,b∈ G then ,

A.
b⁴ = a²ba^-2
B.
b¹⁶ = a⁸ba^-6
C.
0 (b ) = 31 , b ≠ e
D.
0 (b ) = 32 , b ≠ e

*Multiple options can be correct

If G is an abelian group , then which of the following is/are true?

A.
B.
for any three consecutive integers n.
C.
D.
No. of elements in a group G is less than or equal to four.

Solution:

If G is an abelian group , then

*Multiple options can be correct

Let a mapping f : G→ G defined by f (x) = x^-1 , then which of the following is / are true ?

A.
f is one - one
B.
f is on to
C.
f is homomorphism as well as automorphism
D.
G is a non - abelian group.

Solution:

F : G→ G defined by

Let G be a commutative group and

Hence F : G→G is a isomorphism.

So f is an automorphism.

*Multiple options can be correct

Let S₃ be symmetric group on three number 1,2 and 3 and N(a) , Z(G) are normalizer of a and centre of G respectively, then which of the following is/are true? [ I denotes identity element of S₃ ]

A.
N(12) = {1,(1 2)}
B.
N(13) = {1,(1 2),(1 3)}
C.
Z(G) = {1}
D.
N(l) = S₃

Solution:

Definition of normalizer of a and centre of G is

N(a) ={ X∈ G|ax = xa }

Hence N(l) = S₃ always.

option B : N(13 contains (12) if (12)(13) = (13)(12)

but (1 3 )(1 2 ) ≠ (1 3 )(1 2) so (1 2 )∉ N (1 3)

i.e N(13) = {l,(13)}

option C : if S_n symmetric group then centre of symmetric group contains only identity element Z(S₃) = I

option D : N(12) = {1,(12)} is true by the definition of normalizer.

*Multiple options can be correct

Let G be a non abelian group then it's order can be

A.
25
B.
125
C.
55
D.
35

Solution:

option A : 0(G ) = 25 = 5²

if 0(G) = p² then G must be an abelian group.

option B : 0(G) =125 = 5³

if 0(G) = p³ then it is not necessary that G be a abelian group

option C : 0(G) = 55 = 5 x 11

By a result we know that if 0(G) = pq where p < q and then G be a cyclic group but, if p < q and p | q -1 then it is not necessary G be a cyclic group.

Here 5 < 11 but 5 |11-1

Hence group of order 55 is a non abelian group,

option D O(G) = 35 = 5 x 7

Here 5 < 7 and Hence group G with order 35 is cyclic and every cyclic group is abelian group also.

*Answer can only contain numeric values

Let f be a homomorphism from ( z , + ) to ( G , X ) such that,then find the value of f(6).

Solution:

Given that f ; (z, + ) ( G, X) be a homomorph is m such that then

f(6) = f(3+3) = f(3) f(3)

[ ∵ it is a homomorphism ]

*Answer can only contain numeric values

If G is a group of order 23 then find total no. of subgroups of group G.

Solution:

Since 23 is prime, it's divisors are 1 and 23. Hence if H ⊆ G then by lagrange's theorem O(H) divides O(G).

So that 0(H) = 1 or 23.

if O(H) = 1 => H ={e}

if O(H) = 23 =>H = G

Which shows that G has only trivial subgroups.

Total no. of subgroups are 2.

*Answer can only contain numeric values

If K is proper subgroup of H and H is a proper subgroup of G , where O(k) = 35 and 0(G) = 700, then how many subgroups H are possible ?

Solution:

Let 0(H) = m then by lagrange's theorem 0(H) | 0(G) and 0(K) | 0(H).

i.e. 35|m and m | 700

Thus m is multiple of 35 and divisor of 700.

Now 700 = 35 x 20

So, the possible values of m are 35, 2 x 3 5 , 4 x 3 5 , 5 x 3 5 , 10 x 3 5 , 20x35, i.e. 3 5 ,7 0 ,140 ,175 ,350 ,700 .

But K is proper subgroup of H, m ≠ 35 and H is proper subgroup of G, m ≠ 700.

Hence the possible orders of H are 70 ,140,175 and 350 total possible H are 4

*Answer can only contain numeric values

If 0(G ) = 60 and A , B are normal sub - group o f order 2 , 3 respectively then find the order of group G/AB

Solution:

and we know that

If 0(A) = m and 0(B) = n then

*Answer can only contain numeric values

Let σ be the permutation

If I be the identity permutation and m be the order of σ i.e.

m = min { positive integers then find the value of m.

Solution:

Here m is order of the permutation order of permutation = L.C.M (disjoint cycles)

*Answer can only contain numeric values

If order of a group is 125 then what is the order of center of G ?

Solution:

By a well known theorem we know that ifand O[N(a) ] = p² , where a ∈ G and p be any prime no. 5
Here given that O(G) = 125 = 5³

then O[Z(G)] = 5

*Answer can only contain numeric values

For the quotient group Z/50Z , find the total no. of subgroups of quotient group.

Solution:

Let K be subgroup of, then for some subgroup H of Z , such that

Now we have to find all subgroups of Z that contains 50Z, i.e we have to find all the +ve divisiors of 50, which are 1,2,5,10,25,50

Hence Z/50Z has 6 subgroup.

*Answer can only contain numeric values

Find total no. of all the generators of the cyclic group

Solution:

By a Result we know that "a cyclic group (Z_p ,t_p ) have (p - 1 ) generators,

have total no. of generators (7 -1) = 6 .

*Answer can only contain numeric values

Find the total no. of proper subgroups of a cyclic group G = [ a ] of order 40.

Solution:

By lagrange theorem we know that, If G be a group and H be a subgroup of group G , then order of H divides order of G.
Hence all the divisors of 40 are order of subgroup of group G.
All the divisors are → 1 , 2 , 4 , 5 , 8 , 1 0 , 2 0 , 4 0 but order 1 and order 40 are improper subgroup of group G , Hence total proper subgroup are 6.

*Answer can only contain numeric values

Find the total no. of all non - abelian groups of order 6.

Solution:

Let G be a non - abelian group of order 6 , then by Cauchy's theorem

Such that O(a) = 3 , O(b) = 2 .

Let H = < a > then 0(H) = 0(a) = 3.

Since index of H in G is 2 , Hence H is normal in G.

if b ∈ H then 2 |3 which is contradiction, so b ∈ H

H and H_b are distinct right cosets of H in G.

Hence

also H is normal in

If b^-1ab = a > then as (0(a) , 0(b)) = 1 , we get O(a b) = 0(a) 0(b) = 6 i.e. G is cyclic and so , abelian, which is not so ,

So , there is only one non - abelian group G of order 6 , namely

*Answer can only contain numeric values

Let the order of group G is 47, then find out the total no. of proper subgroup.

Solution:

Since 47 is a prime no. i. e. group G has no proper subgroup.
Note: It is a well known theorem that Every group of prime order is a simple group and simple group has no proper normal subgroup.

*Answer can only contain numeric values

How many generators are exists for the cyclic group of order 26.

Solution:

Let G = [ a ] be a cyclic group . then generators of this group is of the from a^m

where m < 26 . [ Co-prime of 26 ]

Hence m = 1, 3 ,5 , 7 , 9 , 1 1 , 15 ,17 ,1 9,2 1 ,23 ,25 total generators are 12.

*Answer can only contain numeric values

If G = [ a] be acyclic group of order 625 and H = [a⁵] be cyclic subgroup of G, then what is the order of subgroup ?

Solution:

By the theorem "If G = [ a ] is a cyclic group of order n and H = [a^p] , then H is a cyclic subgroup of G is of order n/d where d = HCF (n ,p)."

Here O(G) = 625 i.e. n = 625

*Answer can only contain numeric values

Let G be any group of order 224 and a∈ G be an element of order 56. then calculate the order of a⁴.

Solution:

[∵

G is a cyclic so order of group and it's generators both are same]

which shows that O(a⁴) = 14 .

*Answer can only contain numeric values

S₃ be a group of all permutations on three symbols, with the identity elements, then the no. o f elements in S₃ is ____ , that satisfy the equation x² = e.

Solution:

We know that O(S₃) = 6

S₃ = {(1) (12) (23) (31) (123) (132)}

Here it is clear that total no. of elements of order 2 are 3.

But we have to find all those elements in S₃ which satisfied the equation x²= e

Here identity elements is also satisfied this condition.

So total no. of elements

= 3 + 1

= 4

*Answer can only contain numeric values

If O(G) = 21 and G is non abelian then the total no. of elements in centre of G is ________.

Solution:

if O(G) = 21 and G is non abelian then by a well known result we know that

which implies G is abelian but given that G is non abelian.

Hence only possibility is 0[z(G)] = 1 i.e z(G) = {0}

*Answer can only contain numeric values

Let S₁ = {2 }, S₂ = { 4 , 6 } , S₃ = { 8 , 1 0 . 1 2 } ,S₄ = { 14 , 16 , 18 , 20 } and so on then sum of elements of S₁₀ is __________ .

Solution:

Hence sum of elements of S₁₀is equal to

*Answer can only contain numeric values

The number of distinct group homomorphisms from (z, +) onto (z, +) is _________.

Solution:

( z ,+) is infinite cyclic group.
In ( z , + ) total no. of auto morphism is 2.
So , No. of distinct group homomorphism from ( z , + ) onto ( z , + ) is 2.

*Answer can only contain numeric values

The remainder, when 72¹⁰⁰¹ is divided by 31 is ___________.

Solution: