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Consider the following relations, which of the following is an equivalence relation?
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
Let n be a fixed positive integer. Define a relation R on the set Z of integers by then R is
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.
Let R be a relation over the set and it is defined by (a, b) R (c, d) ⇔ a + d = b + c then R is
(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 Z_{6 } = 3
An integer m is said to be related to another integer n if m is a multiple of n then relation is
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.
The number of elements Satisfies g7 = e, g ≠ e in any finite group
Concept used:
Order of generator = Order of group
Every element in the group must divide the order of group and satisfies the property d^{7} = 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 g^{7 }= 7 is odd (every element in the group must divide the order of group and satisfies the property d^{7} = e}
∴ option 2 is correct
The relation R defined on the set A ={1,2,3,4,5} by P = {(x, y) : x^{2} —y^{2} < 16} is given by
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 G be a finite group and Z_{10} be a homomorphic image of G. Then
Let
So, Z_{10} is isomorphic to G/Ker f (by fundamental theorem of homomorphism)
implies
implies
The number of all subgroups of the group (Z_{60}, +) of integers modulo 60 is
Group (Z_{60}, +) of integer modulo 60.
Order of Subgroup will divide order of the group.
60 = 2^{2} • 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,................, p1} be the group with respect to multiplication modulo p. If the inverse of 110 in G is 4, then p is of the form
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
Consider the group 2 x 2 nonsingular matrices under matrix multiplication over Z_{5}. The inverse of is
Given the group of 2 x 2 nonsingular matrices under matrix multiplication over Z_{5}.
The inverse of
i.e.
over Z_{5}.
a, b ε G, a group. If o(a) = 10 and o(b) = 21, then (Where <a> means cyclic group generated by a.)
[Remember o(H ∩ K) divides o(H) and o(K) where H and K are groups]
Consider the following statements:
1. A group of order 289 is abelian.
2. In S_{3} there are four elements satisfying x^{2}= e and six elements satisfying y^{3} = e.
3. Every proper subgroup of S_{3} is cyclic.
4. (Z_{30}, has 4 subgroups of order 15 .
Correct statements are:
every group of order p^{2} (p > prime) is always abelian group.
(1) true
In S_{3} → total no. of elements are 3 of order 2 i.e. x^{2} = e and identity element also satisfying this condition total elements are 4. but only 3 elements are exists which satisfies the condition y_{3} = e.
(3) A_{3} is proper subgroup of S_{3} 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(S_{3}) = 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
D_{4} = {1, r, r^{2}, r^{3} , s, sr, sr^{2}, sr^{3}}
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(K_{4}) = 4
K_{4} = {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(Z_{5}) = 5
Z_{5} = {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.
Let G be a group order 6, and H be a subgroup of G such that 1 < H < 6. Which one of the following options 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
If (G, ⋅) is a group such that (ab)^{1} = a^{1} b^{1}, ∀ a, b ∈ G, then G is a/an
A group is said to be abelian if (a*b) = (b*a) ∀a,b ∈G
Since (G, ⋅) is a group
∴ (ab)^{1 }= (b^{1}a^{1}) (1)
Given: (ab)^{–1} = a^{–1}b^{–1 }(2)
From (1) and (2)
(a*b) = (b*a)
Therefore it is abelian
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