----- > [!proposition] Proposition. ([[inner automorphism]]) > Let $G$ be a [[group]] and let $g\in G$. The map $c_{g}:G \to G$ corresponding to [[conjugate|conjugation]] by $g$, $c_{g}(h):=ghg^{-1},$ > is an [[automorphism]] of $G$, called an **inner automorphism** of $G$. > > The function $G \xrightarrow{} \text{Aut}(G)$ defined by $g \mapsto c_{g}$ is a [[group homomorphism]] whose image, denoted $\text{Inn}(G) \leq \text{Aut}(G)$, is [[group homomorphism|trivial]] iff it is [[cyclic group|cyclic]] and iff $G$ is [[abelian group|abelian]]. ^3761ad > [!proof]- Proof. ([[inner automorphism]]) > First we show that $c_{g}$ is a [[group homomorphism]] from $G\to G$. Let $h_{1},h_{2} \in G$, then $c_{g}(h_{1}h_{2})=gh_{1}h_{2}g^{-1}=gh_{1}g^{-1}gh_{2}g^{-1}=c_{g}(h_{1})c_{g}(h_{2}).$ Next we show that $c_{g}$ is a [[bijection]]. Let $h \in \ker c_{g}$. Then $ghg^{-1} \in \ker c_{g}$ by [[kernel iff normal subgroup]]. So $g^{-1}=hg^{-1}$, implying $h=e$. By [[group homomorphism is injective iff kernel is trivial]], $c_{g}$ is an [[injection]]; since it is an [[injection]] between finite sets of equal [[cardinality]] it is also a [[surjection]] and thus a [[bijection]]. > Now we verify $g \xmapsto{\phi} c_{g}$ is a [[group homomorphism]] between $G$ and its [[automorphism|automorphism group]]: $\begin{align} \phi(g_{1}g_{2})(h)= & c_{g_{1}g_{2}[](group%20homomorphism%20is%20injective%20iff%20kernel%20is%20trivial%20iff%20is%20a%20monomorphism.md)c_{g_{2}}(h) g_{1}^{-1} \\ = & c_{g_{1}}(c_{g_{2}}(h))) \\ = & \phi(g_{1}) \circ \phi(g_{2}) (h). \end{align}$ If $G$ is [[abelian group|abelian]] then $\phi(g_{})(h)=c_{g}(h)= ghg^{-1}=h$ and hence $c_{g}=\id$. If $G$ is not [[abelian group|abelian]] then there exist $g_{1},g_{2} \in G$ such that $g_{1}g_{2} \neq g_{2}g_{1}$, implying$\phi(g_{1})(g_{2})=c_{g_{1}}(g_{2})=g_{1}g_{2}g_{1}^{-1} \neq g_{2}$ for if $g_{1}g_{2}g_{1}^{-1}=g_{2}$ then $g_{1}g_{2}=g_{2}g_{1}$ by rearrangement. > Finally, we check that $\text{Inn}(G)$ is [[cyclic group|cyclic]] iff it is trivial. Suppose it is cyclic: $\text{Inn}(G)=\langle \phi \rangle=\langle c_{a} \rangle$ for some $a \in G$. Then for any $g \in G$, $c_{g}(h)=(c_{a}(h))^{m}=(aha^{-1})^{m}=a^{m}ha^{-m}$ for some $m$. Then in particular $c_{g}(a)=a^{m} a a^{-m}=a \text{ and }c_{g}(a)=gag^{-1}$ implying $a=gag^{-1}$ and hence $ag=ga$ for all $g \in G$. But this means that given any pair $g,h \in G$ we have $ghg^{-1}=c_{g}(h)=a^{m}ha^{-m}=a^{m+1}ha^{-m+1}=\dots = a^{2m}h=h$ which implies $gh=hg$, thereby showing $G$ is [[abelian group|abelian]]. Now we have shown $\text{cyclic} \implies \text{abelian} \iff \text{trivial}$. For the final implication, it suffices to note $\text{trivial} \implies \text{cyclic}$. ^403867 ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```