[proofplan] We encode conjugation by elements of $G$ as a single group homomorphism from $G$ to $\operatorname{Aut}(G)$. Its image is exactly $\operatorname{Inn}(G)$ by definition of inner automorphism. Its kernel is exactly the centre $Z(G)$, because conjugation by $g$ fixes every element precisely when $g$ commutes with every element. The result then follows from the [First Isomorphism Theorem for Groups](/theorems/???). [/proofplan] [step:Show that conjugation by each element is an automorphism of $G$] For each $g\in G$, define the conjugation map \begin{align*} c_g: G &\to G \\ x &\mapsto gxg^{-1}. \end{align*} We first verify that $c_g\in \operatorname{Aut}(G)$. For $x,y\in G$, associativity in $G$ gives \begin{align*} c_g(xy) &= gxyg^{-1} \\ &= gx(g^{-1}g)yg^{-1} \\ &= (gxg^{-1})(gyg^{-1}) \\ &= c_g(x)c_g(y), \end{align*} so $c_g$ is a group homomorphism. Define \begin{align*} c_{g^{-1}}: G &\to G \\ x &\mapsto g^{-1}xg. \end{align*} For every $x\in G$, \begin{align*} (c_g\circ c_{g^{-1}})(x) &= g(g^{-1}xg)g^{-1} = x, \end{align*} and \begin{align*} (c_{g^{-1}}\circ c_g)(x) &= g^{-1}(gxg^{-1})g = x. \end{align*} Thus $c_{g^{-1}}$ is the inverse map of $c_g$, so $c_g$ is bijective. Since $c_g$ is both a group homomorphism and a bijection, $c_g\in \operatorname{Aut}(G)$. [/step] [step:Package conjugation into a homomorphism whose image is $\operatorname{Inn}(G)$] Define the map \begin{align*} \Phi: G &\to \operatorname{Aut}(G) \\ g &\mapsto c_g. \end{align*} The previous step shows that $\Phi$ is well-defined. For $g,h\in G$ and $x\in G$, we compute \begin{align*} (\Phi(g)\circ \Phi(h))(x) &= c_g(c_h(x)) \\ &= g(hxh^{-1})g^{-1} \\ &= (gh)x(h^{-1}g^{-1}) \\ &= (gh)x(gh)^{-1} \\ &= c_{gh}(x) \\ &= \Phi(gh)(x). \end{align*} Since the two automorphisms $\Phi(g)\circ \Phi(h)$ and $\Phi(gh)$ agree on every $x\in G$, they are equal. Hence \begin{align*} \Phi(gh)=\Phi(g)\circ \Phi(h), \end{align*} so $\Phi$ is a group homomorphism from $G$ to $\operatorname{Aut}(G)$. By definition, $\operatorname{Inn}(G)$ is the set of automorphisms $c_g$ with $g\in G$. Therefore \begin{align*} \operatorname{im}\Phi=\operatorname{Inn}(G). \end{align*} [guided] The goal is to convert the family of all conjugation maps into one ordinary group homomorphism, because kernels and images of homomorphisms are exactly what quotient theorems understand. For each $g\in G$, we already defined \begin{align*} c_g: G &\to G \\ x &\mapsto gxg^{-1}. \end{align*} Now define \begin{align*} \Phi: G &\to \operatorname{Aut}(G) \\ g &\mapsto c_g. \end{align*} This is well-defined because the previous step proved that each $c_g$ is an automorphism of $G$. We verify the homomorphism law. Let $g,h\in G$. The product in $\operatorname{Aut}(G)$ is composition of maps, so we must compare $\Phi(gh)$ with $\Phi(g)\circ \Phi(h)$. For every $x\in G$, \begin{align*} (\Phi(g)\circ \Phi(h))(x) &= c_g(c_h(x)) \\ &= g(hxh^{-1})g^{-1} \\ &= (gh)x(h^{-1}g^{-1}) \\ &= (gh)x(gh)^{-1} \\ &= c_{gh}(x) \\ &= \Phi(gh)(x). \end{align*} The equality $(gh)^{-1}=h^{-1}g^{-1}$ is the standard inverse rule in a group. Since the two maps $\Phi(g)\circ \Phi(h)$ and $\Phi(gh)$ have the same value at every element $x\in G$, they are the same automorphism. Therefore \begin{align*} \Phi(gh)=\Phi(g)\circ \Phi(h), \end{align*} so $\Phi$ is a group homomorphism. Finally, the image of $\Phi$ is exactly the subgroup of inner automorphisms. Indeed, an element of $\operatorname{im}\Phi$ has the form $\Phi(g)=c_g$ for some $g\in G$, and this is an inner automorphism. Conversely, every inner automorphism has the form $c_g$ for some $g\in G$, so it lies in $\operatorname{im}\Phi$. Hence \begin{align*} \operatorname{im}\Phi=\operatorname{Inn}(G). \end{align*} [/guided] [/step] [step:Identify the kernel of the conjugation homomorphism with the centre] The kernel of $\Phi$ is \begin{align*} \ker \Phi=\{g\in G:\Phi(g)=\operatorname{id}_G\}, \end{align*} where \begin{align*} \operatorname{id}_G: G &\to G \\ x &\mapsto x \end{align*} is the identity automorphism of $G$. Let $g\in \ker\Phi$. Then $c_g=\operatorname{id}_G$, so for every $x\in G$, \begin{align*} gxg^{-1}=x. \end{align*} Multiplying this equality on the right by $g$ gives \begin{align*} gx=xg \end{align*} for every $x\in G$. Hence $g\in Z(G)$. Conversely, let $g\in Z(G)$. Then $gx=xg$ for every $x\in G$. Multiplying the equality $gx=xg$ on the right by $g^{-1}$ gives \begin{align*} gxg^{-1}=x \end{align*} for every $x\in G$. Therefore $c_g=\operatorname{id}_G$, so $\Phi(g)=\operatorname{id}_G$ and $g\in \ker\Phi$. Thus \begin{align*} \ker\Phi=Z(G). \end{align*} [/step] [step:Apply the first isomorphism theorem to obtain the quotient] The map $\Phi:G\to \operatorname{Aut}(G)$ is a group homomorphism, its image is $\operatorname{Inn}(G)$, and its kernel is $Z(G)$. Since kernels of group homomorphisms are normal subgroups, $Z(G)\trianglelefteq G$, so the quotient group $G/Z(G)$ is defined. Applying the [First Isomorphism Theorem for Groups](/theorems/???) to $\Phi$ gives \begin{align*} G/\ker\Phi \cong \operatorname{im}\Phi. \end{align*} Substituting $\ker\Phi=Z(G)$ and $\operatorname{im}\Phi=\operatorname{Inn}(G)$ yields \begin{align*} G/Z(G)\cong \operatorname{Inn}(G). \end{align*} This is the desired isomorphism. [/step]