Let G be an arbitrary group- Consider the following relations on G- R1 - - a- b - G- a R1 b if and only if - g - G such that a -g-1bgR2- - a- b - G- a R2 b if and only if a-b-1Which of the above is-are equivalence relation-relations-

R_1: ∀ a, b ϵ G, a R_1 b if and only if ∃ g ϵ G such that a =g^ 1bg nbsp; Reflexive: a = g^ 1ag can be satisfied by putting g=e, identity "e" always exists in ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Reflexive Relations

Let G be an a...

Question

Let $G$ be an arbitrary group. Consider the following relations on $G$ :
$R_{1}$ : $\forall a, b ϵ G, a R_{1} b$ if and only if $\exists g ϵ G$ such that $a = g^{- 1} b g$
$R_{2}$ : $\forall a, b ϵ G, a R_{2} b$ if and only if $a = b^{- 1}$

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

R_{1}

and

R_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R_{1}

only

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

R_{2}

only

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Neither

R_{1}

nor

R_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $R_{1}$ only
$R_{1} : \forall a, b ϵ G, a R_{1} b$ if and only if $\exists g ϵ G$ such that $a = g^{- 1} b g$
Reflexive: $a = g^{- 1} a g$ can be satisfied by putting $g = e$ , identity $" e "$ always exists in a group.
So reflexive.
Symmetric: $a R b \Rightarrow a = g^{- 1} b g$ for some $g \Rightarrow b = g a g^{- 1} = (g^{- 1})^{- 1} a g^{- 1}$
$g^{- 1}$ always exists for every $g ϵ G$
So symmetric
Transitive: $a R b$ and $b R c \Rightarrow a = g_{1}^{- 1} b g_{1}$ at b = $g_{2}^{- 1} c g_{2}$ for some $g_{1} g_{2} ϵ G$ Now $a = g_{1}^{- 1} g_{2}^{- 1} c g_{2} g_{1} = (g_{2} g_{1})^{- 1} c g_{2} g_{1}$ $g_{1} ϵ G$ and $g_{2} ϵ G \Rightarrow g_{2} g_{1} ϵ G$ since group closed so $a R b$ and $a R b \Rightarrow a R c$ hence transitive
Clearly $R_{1}$ is equivalence relation.
$R_{2}$ is not equivalence it need not even be reflexive since $a R_{2} a \Rightarrow a = a^{- 1} \forall a$ which not be true it is group.
$R_{1}$ is equivalence relation is the correct answer.

Suggest Corrections

Similar questions

Q. Let

R_{1}, R_{2}

are relation defined on

Z

such that

a R_{1} b ⟺ (a - b)

is divisible by

3

and

a R_{2} b ⟺ (a - b)

is divisible by

4

. Then which of the two relation

(R_{1} \cup R_{2}), (R_{1} \cap R_{2})

is an equivalence relation?

Q. Show that the relation 'a R b if and only if

a - b

is an even integer defined on the Z of integers is an equivalence relation.

Prove that on the set of integers, the relation R defined as aRb if and only if a=±b is an equivalence relation

Let n be a fixed positive integer. Define a relation R in Z as follows $\forall a, b \in Z$ , aRb if and only if a - b is divisible by n. Show that R is an equivalence relation.

Q. Show that the relation R on the set N×N
(a,b)R(c,d) if and only if a+d = b+c is an equivalence relation

Join BYJU'S Learning Program

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?