GATE CSE 2019 | Question: 10

Question

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

• $R_1: \forall a , b \in G, \: a R_1 b \text{ if and only if } \exists g \in G \text{ such that } a = g^{-1}bg$
• $R_2: \forall a , b \in G, \: a R_2 b \text{ if and only if } a= b^{-1}$

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

• A. $R_1$ and $R_2$
• B. $R_1$ only
• C. $R_2$ only
• D. Neither $R_1$ nor $R_2$

Comments

B should be the ans

Answer 1

=> for R1 :  Given that a =(inverse of g)*b*g  means a=b*e , and so a=b

(bcoz we know that (inverse of an element)*(element)=e ).

→ So statement boils down to  R1: ∀a,b∈G,aR1b iff a=b

        So if G={1,2} , Relation R1 = {(1,1),(2,2)} which is EQUIVALENCE RELATION.

 

=> for R2 : Given condition is that a = (inverse of b).

          Again if G={1,2} , Relation R1 should contain atleast {(1,1),(2,2)} for it to be reflexive which is not always possible bcoz a != (inverse of a) for all ‘a’ . So can’t be reflexive, so can’t be equivalence.

Comments

Amcodes

$g^{-1}ag$

Group is given not abilean group so you cannot interchange a and g then write equation as:

$bg^{-1}g$ = $b.e$ for R1