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# 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.

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

## Mathematics Subject Classification

20A05*no label found*

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## Comments

## hi

dont know if this is terribly important but:

G/Z(G) ~= Inn(G)

by sending G into Inn(G)

cheers,

vitriol

## Re: hi

( Don't know if -this- is terribly important but ) since the note by vitriol has justly been added into the inner automorphism main text shouldn't that justify the deletion of his note? It has been there ever since 2002.

-nilo de roock