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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 by Veteran (416k points)
edited by | 2.3k views

4 Answers

+5 votes
  1. $\forall a,b \in G, aR_1b$ iff $\exists g \in G $ such that $a = g^{-1}bg$

    Reflexive:
    $aR_1a$
    $a = g^{−1}ag $
    $ga = gg^{-1}ag$ //Left multiplication by $g$
    $gag^{-1} = agg^{-1}$ //Right multiplication by $g^{-1}$
    $gag^{-1} = a$
    $a = gag^{-1}$

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

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

    Transitive:
    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} $
    Reflexive:

    $aR_2a$,
    $a=a^{−1}$ . this may not be true. So the relation is not reflexive.

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

by Veteran (60k points)
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

+1

Please find the attachment

0
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
0

@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

+2 votes
b as r2 need not be reflexive.

And r1 is reflexive symmetric and transitive.
by (51 points)
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
B should be the ans
by Junior (591 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).

[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
by (73 points)
edited by
Answer:

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