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  "body": "\n## Proof\n\nWe apply the first isomorphism theorem to the map $g \\mapsto c_g$.\n\n1. Let $G$ be a group with identity $e$. For $g \\in G$, let $c_g \\colon G \\to G$, $c_g(x) = gxg^{-1}$, be the [[inner automorphism|0H47DMFN3S]] given by $g$.\n2. For $g, h \\in G$, $(gh)(h^{-1}g^{-1}) = e$. By the [[uniqueness of inverses|0D9NDGE9KP]], $(gh)^{-1} = h^{-1}g^{-1}$.\n3. For $g, h, x \\in G$, step 2 gives $c_g(c_h(x)) = ghxh^{-1}g^{-1} = (gh)x(gh)^{-1} = c_{gh}(x)$. So $c_g \\circ c_h = c_{gh}$.\n4. The map $c_e$ is the identity of $G$. By step 3, $c_g \\circ c_{g^{-1}} = c_{g^{-1}} \\circ c_g = c_e$, so $c_g$ is [[bijective|01VG2A0BHW]].\n5. 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]].\n6. By steps 4 and 5, each $c_g$ lies in the [[automorphism group|0CWXCWP0Y0]] $\\operatorname{Aut}(G)$, whose operation is composition.\n7. By step 3, the map $c \\colon G \\to \\operatorname{Aut}(G)$, $g \\mapsto c_g$, is a group homomorphism. Its image is the set $\\operatorname{Inn}(G)$ of all inner automorphisms.\n8. The identity of $\\operatorname{Aut}(G)$ is $c_e$, the identity map. So $g$ lies in the [[kernel|0Z424HK8CQ]] of $c$ exactly when $gxg^{-1} = x$ for all $x \\in G$.\n9. Multiplying on the right by $g$ or by $g^{-1}$ shows that $gxg^{-1} = x$ holds exactly when $gx = xg$.\n10. By steps 8 and 9, $\\ker c = \\{g \\in G : gx = xg \\text{ for all } x \\in G\\}$. This is the [[centre|0S7Y9Q990Q]] $Z(G)$.\n11. By the [[first isomorphism theorem|0CXXJNG5B6]], the map $gZ(G) \\mapsto c_g$ is an isomorphism from $G/Z(G)$ onto $\\operatorname{im} c = \\operatorname{Inn}(G)$.\n\n## Idea\n\nConjugation turns each element $g$ into an automorphism $c_g$, and $c_g c_h = c_{gh}$, so this is a homomorphism onto $\\operatorname{Inn}(G)$. The elements that give the trivial conjugation are those that commute with everything, which is the centre. Dividing out the kernel gives the isomorphism.\n",
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