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?

Let us check the three properties of equivalence relations (reflexivity, symmetry, and transitivity) for both $R_1$ and $R_2$:

1. For $R_1$ (Conjugacy relation):

   • Reflexive: For any $a \in G$, choose $g = e$ (identity element). Then $a = e^{-1}ae = a$, so $a \, R_1 \, a$.

   • Symmetric: If $a \, R_1 \, b$, then $a = g^{-1}bg$ for some $g \in G$. Multiplying by $g$ on the left and $g^{-1}$ on the right gives $b = gag^{-1} = (g^{-1})^{-1} a (g^{-1})$. Letting $h = g^{-1} \in G$, we have $b = h^{-1}ah$, so $b \, R_1 \, a$.

   • Transitive: If $a \, R_1 \, b$ and $b \, R_1 \, c$, then $a = g_1^{-1}bg_1$ and $b = g_2^{-1}cg_2$. Substituting gives $a = g_1^{-1}(g_2^{-1}cg_2)g_1 = (g_2g_1)^{-1}c(g_2g_1)$. Since $g_2g_1 \in G$, $a \, R_1 \, c$.

   • Therefore, $R_1$ is an equivalence relation.

2. For $R_2$:

   • Reflexive: For $R_2$ to be reflexive, we must have $a \, R_2 \, a \iff a = a^{-1}$ (or $a^2 = e$) for all $a \in G$. In an arbitrary group, not every element is self-inverse (e.g., in $(\mathbb{Z}_3, +)$, $1^{-1} = 2 \neq 1$). Hence, $R_2$ is not reflexive in general.

   • Therefore, $R_2$ is not an equivalence relation.

Hence, $R_1$ only is an equivalence relation (B).