# inner automorphism

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

 $\phi_{x}:G\rightarrow G,\quad y\mapsto xyx^{-1},\quad y\in G,$

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

The composition operation gives the set of all automorphisms of $G$ the structure  of a group, $\operatorname{Aut}(G)$. The inner automorphisms also form a group, $\operatorname{Inn}(G)$, which is a normal subgroup  of $\operatorname{Aut}(G)$. Indeed, if $\phi_{x},\;x\in G$ is an inner automorphism and $\pi:G\rightarrow G$ an arbitrary automorphism, then

 $\pi\circ\phi_{x}\circ\pi^{-1}=\phi_{\pi(x)}.$

Let us also note that the mapping

 $x\mapsto\phi_{x},\quad x\in G$

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

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

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

rather than the left action given above.

Title inner automorphism InnerAutomorphism 2013-03-22 12:49:53 2013-03-22 12:49:53 rmilson (146) rmilson (146) 12 rmilson (146) Definition msc 20A05 inner conjugation outer outer automorphism automorphism group