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  "body": "\n## Proof\n\nWe show that each inner automorphism is an automorphism, and then apply the definition of a characteristic subgroup.\n\n1. Let $G$ be a group with identity $e$, and let $H$ be a [[characteristic subgroup|01CS8Q2APC]] of $G$.\n2. For $g \\in G$, let $c_g \\colon G \\to G$, $c_g(x) = gxg^{-1}$, be the [[inner automorphism|0H47DMFN3S]] given by $g$.\n3. 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|0HV9Y7WG5R]].\n4. For $g, h \\in G$, $(gh)(h^{-1}g^{-1}) = e$, so $(gh)^{-1} = h^{-1}g^{-1}$ by the [[uniqueness of inverses|0D9NDGE9KP]]. Hence $c_g(c_h(x)) = (gh)x(gh)^{-1} = c_{gh}(x)$.\n5. 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|01VG2A0BHW]].\n6. By steps 3 and 5, $c_g$ is an [[automorphism|0CWXCWP0Y0]] of $G$.\n7. Since $H$ is characteristic, $c_g(H) = H$. Let $h \\in H$. Then $ghg^{-1} = c_g(h) \\in c_g(H) = H$.\n8. Step 7 holds for all $g \\in G$ and all $h \\in H$. So $H$ is a [[normal subgroup|0W77CVBCHQ]] of $G$.\n\n## Idea\n\nNormality asks a subgroup to be fixed by the conjugations, and the characteristic property asks it to be fixed by all automorphisms. The conjugations are automorphisms, so the second condition includes the first.\n",
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