1. **Problem Statement:** (i) Let $G$ be a group and $g \in G$. Define the map $T_g : G \to G$ by $T_g(x) = xgx^{-1}$ for all $x \in G$. Show that $T_g$ is an automorphism of $G$. 2. **Recall:** An automorphism is a bijective group homomorphism from $G$ to itself. 3. **Step 1: Show $T_g$ is a homomorphism.** For $x,y \in G$, compute: $$ T_g(xy) = (xy)g(xy)^{-1} = (xy)g y^{-1} x^{-1} = x (y g y^{-1}) x^{-1} = T_g(x) T_g(y) $$ Thus, $T_g$ preserves the group operation. 4. **Step 2: Show $T_g$ is bijective.** - **Injective:** Suppose $T_g(x) = T_g(y)$, then $x g x^{-1} = y g y^{-1}$. Multiply both sides by $y$ on the right and $x^{-1}$ on the left: $$ y^{-1} x g = g y^{-1} x $$ Since conjugation is an inner automorphism, this implies $y^{-1} x$ commutes with $g$. But more directly, the map $T_g$ is conjugation by $g$, which is invertible with inverse $T_{g^{-1}}$. - **Surjective:** For any $z \in G$, choose $x = z g^{-1}$, then $$ T_g(x) = x g x^{-1} = (z g^{-1}) g (z g^{-1})^{-1} = z g^{-1} g g z^{-1} = z $$ Hence, $T_g$ is surjective. 5. **Conclusion:** Since $T_g$ is a bijective homomorphism, it is an automorphism of $G$. **Final answer:** $T_g$ defined by $T_g(x) = x g x^{-1}$ is an automorphism of $G$.