inner automorphism

Let $G$ be a group. For every $x\in G$, we define amapping

$${\varphi }_x:G\to G,y\mapsto xy{x}^{-1},y\in G,$$

called conjugation by $x$.It is easy to show the conjugation map is in fact, a group automorphism.

An automorphism of $G$ that corresponds to conjugation by some$x\in G$ is called inner. An automorphism that isn’t inner is calledan outer automorphism.

The composition operation gives the set of all automorphisms of $G$the structure of a group, $\mathrm{Aut}(G)$. The innerautomorphisms also form a group, $\mathrm{Inn}(G)$, which is anormal subgroup of $\mathrm{Aut}(G)$. Indeed, if ${\varphi }_x,x\in G$ is an inner automorphism and $\pi :G\to G$ an arbitraryautomorphism, then

$$\pi \circ {\varphi }_x\circ {\pi }^{-1}={\varphi }_{\pi (x)}.$$

Let us also note that the mapping

$$x\mapsto {\varphi }_x,x\in G$$

is a surjective group homomorphism with kernel$\mathrm{Z}(G)$, the centre subgroup. Consequently,$\mathrm{Inn}(G)$ is naturally isomorphic to the quotient of$G/\mathrm{Z}(G)$.

Note: the above definitions and assertions hold, mutatis mutandi, if we definethe conjugation action of $x\in G$ on $B$ to be the right action

$$y\mapsto x^{-1}yx,y\in G,$$

rather than the left action given above.