Proof that characteristic subgroups are normal

Proof

We show that each inner automorphism is an automorphism, and then apply the definition of a characteristic subgroup.

1. Let $G$ be a group with identity $e$, and let $H$ be a characteristic subgroup of $G$.
2. For $g \in G$, let $c_g \colon G \to G$, $c_g(x) = gxg^{-1}$, be the inner automorphism given by $g$.
3. For $x, y \in G$, $c_g(x) c_g(y) = gxg^{-1}gyg^{-1} = gxyg^{-1} = c_g(xy)$. So $c_g$ is a group homomorphism.
4. For $g, h \in G$, $(gh)(h^{-1}g^{-1}) = e$, so $(gh)^{-1} = h^{-1}g^{-1}$ by the uniqueness of inverses. Hence $c_g(c_h(x)) = (gh)x(gh)^{-1} = c_{gh}(x)$.
5. By step 4, $c_g \circ c_{g^{-1}} = c_{g^{-1}} \circ c_g = c_e$, which is the identity map of $G$. So $c_g$ is bijective.
6. By steps 3 and 5, $c_g$ is an automorphism of $G$.
7. Since $H$ is characteristic, $c_g(H) = H$. Let $h \in H$. Then $ghg^{-1} = c_g(h) \in c_g(H) = H$.
8. Step 7 holds for all $g \in G$ and all $h \in H$. So $H$ is a normal subgroup of $G$.