$R_1 : \forall a,b \in G, aR_1b$ iff $\exists g \in G $ such that $a = g^{-1}bg$
let g and h are inverse for each other.
Reflexive:
$aR_1a$
$a = g^{−1}ag $
$ga = gg^{-1}ag$ //Left multiplication by $g$
$gag^{-1} = agg^{-1}$ //Right multiplication by $g^{-1}$
$gag^{-1} = a$
$a = gag^{-1}$
$a = h^{-1}ah$ ( $∃h∈G$ )
we have $aR_1a$. So the relation is reflexive.
Alternative :- in the group, there should exist identity element, So we can take it as some g.
Symmetric:
let (a,b) exist, then
$a = g^{−1}bg $
$gag^{-1} = b $
$h^{-1}ah = b $ ( $∃h∈G$ )
∴ (b,a) should be in the relation.
Hence the relation is symmetric.
Transitive:
If (a,b) and (b,c) present, then
$a = x^{−1} b x $ and $b = y^{−1} c y $
$a = \underline{x^{−1} \;y^{−1}} \;\;c \;\;\underline{y\; x} $
What is the inverse of $ x^{−1} y^{−1} $ ?
$ x^{−1} y^{−1} $ . _________ = Identity.
to cancel the term $ y^{−1} $, we must multiply with it's inverse.
to cancel the term $ x^{−1} $, we must multiply with it's inverse.
So, inverse of $ x^{−1} y^{−1} $ is $ y . x $
∴$a = p^{−1}\; c \;p \; \text{, where p is y.x} $
∴ (a,c) should be in the relation.
Hence the relation is transitive.
$R_1$ is equivalence relation.
$R_2 : \forall a,b \in G, aR_2b$ iff $a =b^{-1} $
Reflexive:
for including a pair of (a,a), we need to $a=a^{−1}$ .
For an arbitrary group this may not be true. So the relation is not reflexive.
$R_2$ is not equivalence relation.
Correct Answer (B).