Let G be an arbitrary group. Consider the following relations on G: R_1: \forall a, b \in G, a R_{1} b if and only if \exists g \in G such that a=g^{-1} b g {R}_{2}: \forall a, {b} \in G, a R_{2} b if and only if {a}= {b}^{-1} Which of the above is/are equivalence relation/relations?

GATE Computer Science (CS) 2019 Solved Papers

Question : 20 of 65

Marks: +1, -0

Let G be an arbitrary group. Consider the following relations on G:

R_1:

∀ a, b ∈ G, a R_{1} b

if and only if

∃ g ∈ G

such that

a=g^{-1} b g

{R}_{2}: ∀ a, {b} ∈ G, a R_{2} b

if and only if

{a}= {b}^{-1}

Which of the above is/are equivalence relation/relations?

Solution:

{R}_{1}: ∀ {a}, {b} ∈ {G}, {a} {R}_{1} {b}

if and only if

∃ {g} ∈

G such that

a= {g}^{-1} {bg}

Reflexive:

{a}= {g}^{-1} {ag}

can be satisfied by putting

{g}= {e}

, identity "e" always exists in a group.
So reflexive
Symmetric:

{aRb} ⇒ {a}= {g}^{-1} {bg}

for some

{g}

⇒ {b}= {gag}^{-1}=( {g}^{-1})^{-1} {ag}^{-1}

{g}^{-1}

always exists for every

{g} ∈ {G}

.
So symmetric
Transitive:

{aRb}

and

{bRc} ⇒ {a}= {g} 1^{-1} {bg}_{1}

and

{b}= {g}_{2}^{-1} {cg}_{2}

for some

{g}_{1} {g}_{2} ∈ {G}

.
Now

{a}= {g}_{1}^{-1} {g}_{2}^{-1} {cg}_{2} {g}_{1}=( {g}_{2} {g}_{1})^{-1} {cg}_{2} {g}_{1}

{g}_{1} ∈ {G}

and

{g}_{2} ∈ {G} ⇒ {g}_{2} {g}_{1} ∈ {G}

since group is closed

{so} {aRb}

and

{aRb} ⇒ {aRc}

hence transitive
Clearly

{R}_{1}

is equivalence relation.

{R}_{2}

is not equivalence it need not even be reflexive, since

{aR}_{2} {a} ⇒ {a}= {a}^{-1} ∀ {a}

which not be true in a group.

{R}_{1}

is equivalence relation is the correct answer

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