Let G be an arbitrary group. Consider the following relations on G: R1 : ∀a,bϵG,aR1b if and only if ∃gϵG such that a=g−1bg R2: ∀a,bϵG,aR2b if and only if a=b−1
Which of the above is/are equivalence relation/relations?
A
R1 and R2
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B
R1 only
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C
R2 only
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D
Neither R1 nor R2
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Solution
The correct option is BR1 only R1:∀a,bϵG,aR1b if and only if ∃gϵG such that a=g−1bg
Reflexive: a=g−1ag can be satisfied by putting g=e, identity "e" always exists in a group.
So reflexive.
Symmetric: aRb⇒a=g−1bg for some g⇒b=gag−1=(g−1)−1ag−1 g−1 always exists for every gϵG
So symmetric
Transitive: aRb and bRc⇒a=g−11bg1 at b = g−12cg2 for some g1g2ϵG Now a=g−11g−12cg2g1=(g2g1)−1cg2g1g1ϵGandg2ϵG⇒g2g1ϵG since group closed so aRb and aRb⇒aRc hence transitive
Clearly R1 is equivalence relation. R2 is not equivalence it need not even be reflexive since aR2a⇒a=a−1∀a which not be true it is group. R1 is equivalence relation is the correct answer.