IIT JAM Mathematics Practice Test- 16 - Mathematics MCQ

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

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

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 ]

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.

If G be a group and a,b ∈ G 

then

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.

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.

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.

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.

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

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₎

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

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.

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.

∪(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 

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.

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

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.

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.

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 by the theorem if 0(a) = n

then 

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

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

The members of G are

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

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

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.

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.

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}

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

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

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.

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.