Let G be an arbitrary group. Consider the following relations on G: R1: ∀a, b ϵ G, a R1b if and only if ∃g ϵ G such that a = g-1bg R2: ∀a, b ϵ G, a R1b if and only if a = b-1 Which of the above is/are equivalence relation/relations?

Let G be an arbitrary group. Consider the following relations on G: R1: ∀a, b ϵ G, a R1b if and only if ∃g ϵ G such that a = g-1bg R2: ∀a, b ϵ G, a R1b if and only if a = b-1 Which of the above is/are equivalence relation/relations? Correct Answer R₁ only

∀ a, b ∈G, aR₁b iff ∃g ∈ G such that a=g⁻¹bg

Reflexive:

aR₁a

a=g⁻¹ag

ga=gg⁻¹ag //Left multiplication by g

gag⁻¹=agg⁻¹ //Right multiplication by g

gag⁻¹=a

a = gag⁻¹

we have aR₁a. So, the relation is reflexive.

Symmetric:

If aR₁b, then ∃g ∈ G such that gag⁻¹ = b then a=g⁻¹bg, so bR₁a. Hence the relation is symmetric.

Transitive:

If aR₁b and bR₁c there are g, h ∈ G such that b=gag⁻¹ and c=hbh⁻¹. Then c = hga(hg)⁻¹ so aR1c.

Hence the relation is transitive.

R1 is an equivalence relation.

∀ a, b ∈ G, aR₂b iff ∃g ∈ G such that a=b⁻¹

Reflexive:

aR₂a,

a=a⁻¹. this may not be true. So, the relation is not reflexive.

R₂ is not an equivalence relation.