retagged by
17,148 views
37 votes
37 votes

Let $G$ be an arbitrary group. Consider the following relations on $G$:

  • $R_1: \forall a , b \in G, \: a R_1 b \text{ if and only if } \exists g \in G \text{ such that } a = g^{-1}bg$
  • $R_2: \forall a , b \in G, \: a R_2 b \text{ if and only if } a= b^{-1}$

Which of the above is/are equivalence relation/relations?

  1. $R_1$ and $R_2$
  2. $R_1$ only
  3. $R_2$ only
  4. Neither $R_1$ nor $R_2$
retagged by

9 Answers

53 votes
53 votes

$R_1 : \forall a,b \in G, aR_1b$ iff $\exists g \in G $ such that $a = g^{-1}bg$

let g and h are inverse for each other.

Reflexive:
$aR_1a$

$a = g^{−1}ag $

$ga = gg^{-1}ag$ //Left multiplication by $g$

$gag^{-1} = agg^{-1}$ //Right multiplication by $g^{-1}$

$gag^{-1} = a$

$a = gag^{-1}$

$a = h^{-1}ah$  ( $∃h∈G$ )

we have $aR_1a$. So the relation is reflexive.

Alternative :- in the group, there should exist identity element, So we can take it as some g.

Symmetric:

let (a,b) exist, then

$a = g^{−1}bg $

$gag^{-1} = b $

$h^{-1}ah = b $  ( $∃h∈G$ )

∴ (b,a) should be in the relation.

Hence the relation is symmetric.

Transitive:

If (a,b) and (b,c) present, then

$a = x^{−1} b x $ and $b = y^{−1} c y $

$a = \underline{x^{−1} \;y^{−1}} \;\;c \;\;\underline{y\; x} $

What is the inverse of $ x^{−1} y^{−1} $ ?

$ x^{−1} y^{−1} $ . _________ = Identity.

to cancel the term $ y^{−1} $, we must multiply with it's inverse.

to cancel the term $ x^{−1} $, we must multiply with it's inverse.

So, inverse of $ x^{−1} y^{−1} $ is $ y . x $

∴$a = p^{−1}\; c \;p \; \text{, where p is y.x} $

∴ (a,c) should be in the relation.

Hence the relation is transitive.

$R_1$ is equivalence relation.

 

$R_2 : \forall a,b \in G, aR_2b$ iff $a =b^{-1} $


Reflexive:

for including a pair of (a,a), we need to $a=a^{−1}$ .

For an arbitrary group this may not be true. So the relation is not reflexive.

$R_2$ is not equivalence relation.

 

Correct Answer (B).

edited by
7 votes
7 votes
R2 is not Reflexive.

-R1 is Reflexive.means

we have to prove a= g^-1 .a .g for some g belonging to group

Since , identity element belong to every group.

For g=identity element(e).

[e^-1=e

e.a=a.e=a]

So, g^-1 .a . g = e^-1 .a . e = e . a = a

Hence, for all a there exist g( an identity element) such that a= g^-1 . a .g

- R1 is symmetric

Given: a= g^-1 .b . g and g belongs to the group.

To Prove: b = k^-1 . a . k for k belonging to group.

Premultiplying with g

g.a = g.g^-1.b.g= b.g

Postmultiplying with g^-1

g .a. g^-1= b.g.g^-1= b

implies

b= g.a.g^-1

g can be written as (g^-1)^-1

b=(g^-1)^-1 . a . g^-1

for k= g^-1

Since, g belongs to group g^-1 also belongs to group and hence k belongs to group

b= k^-1 . a . k

Hence, if there exist g such that a=g^-1.b.g

then there also exist g'(which is g^-1) such that b= g'^-1.a.g'

 

-Dont know about transitive
edited by
2 votes
2 votes
C option

R1 reflexive  symmetric and transitive

R1is reflexive:

aRa because e −1ae = a.

R1 is symmetric:

if aRb, i.e. if b = g −1ag for some g, then a = gbg−1 = (g −1 ) −1 bg−1 and bRa.

R1 is transitive: if aRb, i.e. b = g −1ag, and bRc, i.e. c = h −1 bh, then c = h −1 g −1agh = (gh) −1a(gh)
edited by
2 votes
2 votes
b as r2 need not be reflexive.

And r1 is reflexive symmetric and transitive.
Answer:

Related questions

32 votes
32 votes
14 answers
2
Arjun asked Feb 7, 2019
20,977 views
Let $G$ be an undirected complete graph on $n$ vertices, where $n 2$. Then, the number of different Hamiltonian cycles in $G$ is equal to$n!$$(n-1)!$$1$$\frac{(n-1)!}{2}...
3 votes
3 votes
3 answers
3
2 votes
2 votes
3 answers
4
Sayan Bose asked May 7, 2019
2,041 views
Let $G =\{a_1,a_2, \dots ,a_{12}\}$ be an Abelian group of order $12$ . Then the order of the element $ ( \prod_{i=1}^{12} a_i)$ is$1$$2$$6$$12$