The Gateway to Computer Science Excellence
+11 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 $R_2$
in Set Theory & Algebra by Veteran (431k points)
edited by | 3.5k views
B should be the ans

3 Answers

+12 votes
  1. $\forall a,b \in G, aR_1b$ iff $\exists g \in G $ such that $a = g^{-1}bg$

    $a = g^{−1}ag $
    $ga = gg^{-1}ag$ //Left multiplication by $g$
    $gag^{-1} = agg^{-1}$ //Right multiplication by $g^{-1}$
    $gag^{-1} = a$
    $a = gag^{-1}$

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

    If $aR_1b$, then $∃g∈G$ such that $gag^{−1}=b$ then $a=g^{−1}bg$, so $bR_1a$. Hence the relation is symmetric.

    If $aR_1b$ and $bR_1c$ there are $g, h∈G$ such that $b=gag^{−1}$ and $c=hbh^{−1}$. Then $c=hga(hg)^{−1}$ so $aR_1c$. Hence the relation is transitive.

    $R_1$ is equivalence relation.
  2. $\forall a,b \in G, aR_2b$ iff $\exists g \in G $ such that $a =b^{-1} $

    $a=a^{−1}$ . this may not be true. So the relation is not reflexive.

$R_2$ is not equivalence relation. Answer (B).

by Veteran (60.9k points) 1 flag:
✌ Edit necessary (toxicdesire “Point 2.) there exists g \in G shouldn't be there.”)

edited by

How you are saying gag^-1 = g^-1ag in your first line in proving reflexive..

In a group a commutative property may or may not is not written that it is abelian group

@Digvijay Pandey


Please find the attachment

yes correct

little difference among two options

but that is why R1 is answer
and it is asking about equivalence relation

So, reflexive, symmetric, transitivity must have to satisfy

@Sankha Narayan Bose Since one of the conditions to form a group is that every element should have their inverse. So g and g^-1 will be both in group G. So if you substitute g with g^-1 the relation still satisfies


@Digvijay Pandey


how is it implying that g $\epsilon$ G?


How we can substitute the g inplace of g^-1?
+4 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).



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


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
by (113 points)
edited by
+2 votes
b as r2 need not be reflexive.

And r1 is reflexive symmetric and transitive.
by (47 points)
Can you please explain how R1 can be Equivalence?
I dont understand here :

Say G = (<0,1,2>,+3) and set A = {0,1,2}

here so Option 1 : (1,1) pair wont be present because :

for 1 R 1 there should exist a g such that         1 = g^-1 .1.g        now there will be no such g for which it exist then how the first option is true ?

What am i doing wrong ?

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,737 questions
57,313 answers
105,046 users