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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
edited by
B should be the ans

5 Answers

29 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.


$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.


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.


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}\; b \;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} $


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

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?
i edited the answer, please check now.
Here  if  reflexive  property  holds  then,  can  we  say  directly  that  R1  is equivalence  relation  because  smallest  equivalence  relation  contains  all diagonal  elements  that  satisfies  reflexive  property ?
5 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

edited by
2 votes
b as r2 need not be reflexive.

And r1 is reflexive symmetric and transitive.
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
=> for R1 :  Given that a =(inverse of g)*b*g  means a=b*e , and so a=b

(bcoz we know that (inverse of an element)*(element)=e ).

→ So statement boils down to  R1: ∀a,b∈G,aR1b iff a=b

        So if G={1,2} , Relation R1 = {(1,1),(2,2)} which is EQUIVALENCE RELATION.


=> for R2 : Given condition is that a = (inverse of b).

          Again if G={1,2} , Relation R1 should contain atleast {(1,1),(2,2)} for it to be reflexive which is not always possible bcoz a != (inverse of a) for all ‘a’ . So can’t be reflexive, so can’t be equivalence.


Group is given not abilean group so you cannot interchange a and g then write equation as:

$bg^{-1}g$ = $b.e$ for R1   

0 votes

For $R_1$:

As it’s given that $G$ be any arbitrary group , then it’s a must that every element has an inverse .

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

Let the inverse of $g$ be $\alpha$ , So we can say $g^{-1}$ = $\alpha$ and $\alpha ^{-1} = g $

Therefore, our equation looks like this : $a = \alpha bg$

And also there exists an identity element $’e’$ , as it’s a group.


Reflexivity :

We have to just prove that every element is related to itself , and also $a.e = e.a = a$ , here $e$ is an identity element

Now we can say if $(a,a)$ has to belong to the relation then it must satisfy :

$a = g^{-1}ag$ , and we can assume that element $’g’$ to be equivalent to our identity element $’e’$.

Then it becomes $a = e^{-1}ae$

$a = eae$  ($e^{-1}=e$)

$a = ae$

$a = a$

Hence , our relation is reflexive

Symmetric :

For relation to be symmetric $aRb$ and $bRa$ for all $a,b \in G$

So we just have to prove $b = k^{-1}ak  $ such that there $\exists k$ and $k \in G$ 

$a = g^{-1}bg$

(Left multiplication with g)

$ga = gg^{-1} b g$

$ga = ebg$ (As $ gg^{-1} = e$)

(Right multiplication with $g^{-1}$)

$gag^{-1} = bgg^{-1}$

$gag^{-1} = be$

$gag^{-1} = b$

$b = gag^{-1}$

And from our assumption we know $g^{-1}=\alpha$ and $\alpha^{-1}=g$

Therefore , we can write it as :

$b = \alpha^{-1}a\alpha$

And this satisfies the above equation as $\alpha$ do exists and belong to G. And , hence our relation is symmetric


We have to prove $aRb$ and $bRa$ , then $aRc$

Let there be to element $g$ and $k$ such that :

$a = g^{-1}bg$ and $b = k^{-1}ck$

From above we can say : $b = \alpha^{-1}a\alpha$


$\alpha^{-1}a\alpha = k^{-1}ck$

(Left multiplication with $\alpha$)

$a\alpha = \alpha k^{-1}ck$

$a\alpha =  g^{-1}k^{-1}ck$ (As $\alpha = g^{-1}$ )

(Right Multiplication with $\alpha^{-1}$)

$a =g^{-1}k^{-1}ck\alpha $

$a =g^{-1}k^{-1}ckg $  (As $\alpha^{-1}=g$)

Now what’s the inverse of $g^{-1}k^{-1}$ ? it would be $kg$

Because :

$g^{-1}k^{-1}*$ ____= $e$

So we’ve to operate it with $kg$ in order to get e

$g^{-1}k^{-1}kg = e$

$g^{-1}eg = e$ (As, $k^{-1}k = e$) 

$g^{-1}eg = e$

$g^{-1}g = e$

$e = e$

(Credits to Sir Shaik Masthan for this)

And, let $h = kg$ 

So $h^{-1} = g^{-1}k^{-1}$

Therefore our equation would look like this :

$a =h^{-1}ch $

Hence , relation is also transitive .

Therefore, $R_1$ is an equivalence relation.

For $R_2$:

$a = b^{-1}$ , doesn’t tells us anything.

 $b^{-1}$ might be equal to $a^{-1}$ or not.

Hence it’s not a reflexive relation, therefore not an equivalence relation







edited by

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