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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
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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.

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).


edited by
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

@Digvijay Pandey

2

Please find the attachment

0
yes correct

little difference among two options

but that is why R1 is answer
0
and it is asking about equivalence relation

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

0

@Digvijay Pandey

a=gag−1  

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

 

0
How we can substitute the g inplace of g^-1?
0
i edited the answer, please check now.
0
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).

[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
b as r2 need not be reflexive.

And r1 is reflexive symmetric and transitive.
0
Can you please explain how R1 can be Equivalence?
0
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.
0

$g^{-1}ag$

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.

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

 

 

 

 

 

 


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