Relation equivalence


#1

Which of the following equivalent relation of a group G?
R 1 : ∀ a , b ∈ G , a R 1 b if only ∃ g ∈ G : a = g − 1 bg
R 2 : ∀ a , b ∈ G , a R 2 b if only a = b –1

  1. Both R 1 and R 2
  2. R 1
  3. R 2
  4. None of these

#2
  1. R1
    because it satsify the reflexive,symmetric and transitive equation

#4

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