The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
+4 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 $R_2$
asked in Set Theory & Algebra by Veteran (400k points)
edited by | 2.2k views

4 Answers

+5 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).

answered by Veteran (59.7k points)
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
+2 votes
b as r2 need not be reflexive.

And r1 is reflexive symmetric and transitive.
answered by (51 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 ?
0 votes
B should be the ans
answered by Junior (585 points)
0 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
answered by (73 points)
edited by

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
49,434 questions
53,630 answers
70,898 users