Gate2019 cs Q20

0. Let G be an arbitrary group. Consider the following relations on G:
R₁: ∀a, b ∈ G, aR₁b if and only if ∃g ∈ G such that a = g⁻¹bg
R₂: ∀a, b ∈ G, aR₂b if and only if a = b⁻¹
Which of the above is/are equivalence relation/relations?

Answer
Comment
Array

Option : B
Explanation :
Given R₁ is a equivalence relation, because it satisfied reflexive, symmetric, and transitive conditions:
- Reflexive: a = g^–1ag can be satisfied by putting g = e, identity “e” always exists in a group.
- Symmetric:
  aRb ⇒ a = g^–1bg for some g
  ⇒ b = gag^–1 = (g^–1)^–1ag^–1
  g^–1 always exists for every g ∈ G.
- Transitive:
  aRb and bRc ⇒ a = g1^–1bg1
  and b = g2^–1 cg2 for some g1g2 ∈ G.
  Now a = g1^–1 g2^–1 cg2g1 = (g2g1)^–1 cg2g1
  g1 ∈ G and g2 ∈ G ⇒ g2g1 ∈ G since group is closed so aRb and aRb ⇒ aRc
R₂ is not equivalence because it does not satisfied reflexive condition of equivalence relation:
aR₂a ⇒ a = a^–1 ∀a which not be true in a group.
So, option (B) is correct.

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