Test: Group Theory - 1

The equivalence class of  under ~, denoted , is defined as 

The order of subgroups divides the Order of Group but if some m divide the order of the group that doesn't mean that group has the subgroup of order m 

Proper subgroup :  H ≠ e and H ≠ G is the subgroup of G

Calculations:

G is a Prime order Group of p 

so, it has only 2 divisors 1 and itself that divides the order of Group 

⇒ Only 2 subgroups are generated by 1 and p

⇒ <1> = G and

= {e} where e is the identity 

∴ option 1 is correct, No proper Subgroup 

Correct Answer :- d

Explanation : Reflexive: For any a∈N, we have a−a=0=0×n⇒−a is divisible by n

⇒(a,a)∈R

Thus (a,a)∈R for all a∈Z. So, R is reflexive.

Symmetry: Let (a,b)∈R. Then, 

⇒(a,b)∈R

⇒(a−b) is divisible by n.

⇒(a−b)=np for some p∈Z

⇒b−a=n(−p)

⇒b−a is divisible by n

Thus, (a,b)∈R

⇒(b,a)∈R for all a,b∈Z

So, R is symmetric on Z.

Transitive: Let a,b,c∈Z such that (a,b)∈R and (b,c)∈R. Then (a,b)∈R

⇒(a−b) is divisible by n.

⇒a−b=np for some p∈Z

(b,c)∈R

⇒(b−c) is divisible by n.

⇒b−c=nq for some q∈Z

therefore,(a,b)∈R and b−c∈R

⇒a−b=npb−c=nq

⇒(a−b)+(b−c)=np+nq

a−c=n(p+q)

⇒−c is divisible by n.

⇒(a−c)∈R

Thus, (a,b)∈R and (b,c)∈R

⇒(a,c)∈R for all a,b,c∈Z

So, R is a transitive relation on Z.

Thus, R being reflexive, symmetric and transitive, is an equivalence relation.

(a, b) R (c, d) ⇒ a + d = b + c

(a, a) R (a, a) ⇒ 2a = 2a

It is reflexive.

(a, b) R (c, d) ⇒ (c, d) R (a, b)

a + d = b + c

b + c = a + d

It is symmetric.

Transitive:

(a, b) R (c, d) ⇒ (c, d) R (e, f) ⇒ (a, b) R (e, f)

a + d = b + c, c + f = d + c

a – e = b – f

a + f = b + c

(a, b) R (e, f)

It is transitive.

Concept used:

Improper subgroup means The subgroup is Group itself 

Proper Subgroup of Z_n = The subgroups which are not improper

Number of subgroups of Z_n  = Number  of divisors of n 

Calculations: 

The number of divisors of 6 = 1, 2, 3, 6 

Total subgroups is 4

but One subgroup is equal to Group that is improper subgroup generated by <1> = {0, 1, 2, 3, 4, 5}

∴ Total number of proper subgroup of Z₆  = 3

The empty relation is the subset  ∅. It is clearly irreflexive, hence not reflexive. To check symmetry, we want to know whether  aRb⇒bRa  for all  a,b∈A.

Concept used:

Order of generator = Order of group

Every element in the group must divide the order of group and satisfies the property d⁷ = e

Explanations:

g7 = e

o(g)\7 

Two number divides 7 that is 1 or 7 

Now, 

g ≠ e

⇒ o(g) = 7 

Let us consider that the subgroup generated by g. 

As the order of g is 7

The order of the subgroup generated by g is 7 (because order of generator = order of group)

⇒ no of elements satisfies g⁷ = 7 is odd (every element in the group must divide the order of group and satisfies the property d⁷ = e}

∴ option 2 is correct 

R= {(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4),(4,5),(5,4),(5,5)}R={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4),(4,5),(5,4),(5,5)} 

As, a<b doesn't imply b<a ,which make the relation asymmetric and thus it is not an equivalence relation .

Let
So, Z₁₀ is isomorphic to G/Ker f (by fundamental theorem of homomorphism)
implies 

implies 

Group (Z₆₀, +) of integer modulo 60.
Order of Subgroup will divide order of the group.
60 = 2² • 3 • 5
So, total number of divisor = 3 x 2 x 2 = 12
So, 12 subgroups are possible.
Since operation is addition modulo 60 thus each divisor will form a subgroup.
Thus, there are 12 subgroups.

Remember: There is essentially only one cyclic group of each order (upto isomorphism).

Let G = ( 1 , 2 , . . . , p - 1 } be the group w.r.t. multiplication modulo p.
Given, the inverse o f 110 is 4
i.e. 110 x 4 = 1 (mod p)
implies 440 = (1 mod p)
implies p = 440 - 1
p = 439 —> which is of the form of 5n+4

Given the group of 2 x 2 non-singular matrices under matrix multiplication over Z₅.
The inverse of 
i.e. 
over Z₅.

[Remember o(H ∩ K) divides o(H) and o(K) where H and K are groups]

every group of order p² (p -> prime) is always abelian group.
(1) true

In S₃ → total no. of elements are 3 of order 2 i.e. x² = e and identity element also satisfying this condition total elements are 4. but only 3 elements are exists which satisfies the condition y₃ = e.
(3) A₃ is proper subgroup of S₃ which is cyclic.

Concept used: 

A group is cyclic iff there exists an element of order whose order is equal to the order of the group 

Calculations:

o(S₃) = 3!

contains 6 elements of the Symmetric group S(3), of degree 3, written in cycles are : {(1), (12), (13), (23), (123), (321)}.

o(1) = 1

o(12), o(23) and o(13) = 2

o(123),  and  o(321) = 3 

no element having order equal to order of group 

∴ option 1 is incorrect.

o(D4) = 8 

 D₄ = {1, r, r², r³ , s, sr, sr², sr³}

o(s), o(sr), o(sr2), o(sr3) = 2

o(1) = 1, o(r) = 2 , o(r2) = 3, o(r3) = 4

no element having order equal to order of the group 

∴ option 3 is incorrect.

o(K₄) = 4 

K₄  = {e, a, b, ab}

o(e) = 1 , o(a) = 2, o(b) = 2, o(ab) = 2 

No element having order equal to the order of group 

∴ option 4 is incorrect.

o(Z₅) = 5 

Z₅ = {0, 1, 2, 3, 4}

o(0) = 1 o(1) = 5, o(2) = 5 , o(3) = 5 , o(4) = 5 

all elements having order = order of the group 

all are the generator of the group.

∴ option 2 is correct.

Concept

According to the Lagrange theorem order of subgroups must divide the order of the group.

Property of group says if a group has prime order then it is cyclic.

Explanation:

Since the order of G is 6. Therefore  its subgroup may have order 1,2,3,6

H is one of its subgroups with condition 1< |H| <6 so H may be of order 2 or 3  which is prime 

Hence H must be cyclic

The order of G is 6 which is not prime and hence it may or may not  be cyclic 

Therefore option 2 is correct

A group is said to be abelian if (a*b) = (b*a) ∀a,b ∈G

Since (G, ⋅) is a group

∴ (ab)^-1 = (b^-1a^-1)             (1)

Given: (ab)^–1 = a^–1b^–1       (2)

From (1) and (2)

 (a*b) = (b*a)

Therefore it is abelian