InnerAutomorphism 2002-07-04 02:40:08 2007-07-31 12:47:20 Definition conjugation outer outer automorphism automorphism group \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \newcommand{\reals}{\mathbb{R}} \newcommand{\natnums}{\mathbb{N}} \newcommand{\cnums}{\mathbb{C}} \newcommand{\znums}{\mathbb{Z}} \newcommand{\lp}{\left(} \newcommand{\rp}{\right)} \newcommand{\lb}{\left[} \newcommand{\rb}{\right]} \newcommand{\supth}{^{\text{th}}} \newtheorem{proposition}{Proposition} \newtheorem{definition}[proposition]{Definition} \newtheorem{theorem}[proposition]{Theorem} Let $G$ be a group. For every $x\in G$, we define a mapping $$\phi_x:G\rightarrow G,\quad y\mapsto x y x^{-1},\quad y\in G,$$ 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 \emph{inner}. An automorphism that isn't inner is called an \emph{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 \[ y\mapsto x^{-1} y x,\quad y\in G,\] rather than the left action given above.