Question 20

Examining $R_1$ (Conjugacy relation):

• Reflexivity: For any $a \in G$, choose $g = e$ (identity element). Then $a = e^{-1}ae$, so $a R_1 a$. (Reflexive)
• Symmetry: If $a R_1 b$, then $\exists g \in G$ such that $a = g^{-1}bg$. Multiplying by $g$ on the left and $g^{-1}$ on the right gives $b = g a g^{-1} = (g^{-1})^{-1} a (g^{-1})$. Choosing $h = g^{-1} \in G$, we have $b = h^{-1}ah$, so $b R_1 a$. (Symmetric)
• Transitivity: If $a R_1 b$ and $b R_1 c$, then $a = g_1^{-1} b g_1$ and $b = g_2^{-1} c g_2$. Substituting gives $a = g_1^{-1}(g_2^{-1} c g_2)g_1 = (g_2 g_1)^{-1} c (g_2 g_1)$. Since $g_2 g_1 \in G$, $a R_1 c$. (Transitive)
  Thus, $R_1$ is an equivalence relation.

Examining $R_2$:
$a R_2 b \iff a = b^{-1}$. For $R_2$ to be reflexive, every element must satisfy $a R_2 a \iff a = a^{-1} \iff a^2 = e$. In an arbitrary group, not every element is self-inverse (for example, in $(\mathbb{Z}_4, +)$, $1^{-1} = 3 \ne 1$). Thus, $R_2$ is not reflexive in general, so it is NOT an equivalence relation.

Hence, only $R_1$ is an equivalence relation.