20 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?

- $R_1$ and $R_2$
- $R_1$ only
- $R_2$ only
- Neither $R_1$ nor $R_2$

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.

__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}\; 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} $

__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).**

0

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 hold.it is not written that it is abelian group

0

and it is asking about equivalence relation

So, reflexive, symmetric, transitivity must have to satisfy

So, reflexive, symmetric, transitivity must have to satisfy

5

@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

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).

[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

-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

2 votes

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.

(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.

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.

Now,

**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

**Transitivity:**

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 : $b = \alpha^{-1}a\alpha$

So

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

(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