Example

(1) Let $G$ be a group and $E$ a homogeneous principal $G$-set; let $\Gamma$ denote the automorphism group of $G$. Let $A$ be the set of permutations $f$ of $E$ with the following property: There exists $\gamma \in \Gamma$ such that $f$ is a $\gamma$-morphism of $E$ into $E$ (that is, $f(gb) = \gamma(g)f(b)$ for $b \in E$ and $g \in G$). The above formula $f(gb) = \gamma(g)f(b)$ shows that if $f \in A$ there exists a unique $\gamma \in \Gamma$ such that $f$ is a $\gamma$-morphism, we shall denote it by $p(f)$. Let $f, f'$ be in $A$, $\gamma = p(f), \gamma' = p(f')$. Then, for all $b \in E$ and all $g \in G$,

which proves that $f' \circ f \in A$ and that $p(f' \circ f) = p(f')p(f)$. On the other hand, $f(\gamma^{-1}(g)f^{-1}(b)) = gb$, whence $f^{-1}(gb) = \gamma^{-1}(g)f^{-1}(b)$ and $f^{-1} \in A$. Thus $A$ is a subgroup of $\mathscr{S}_E$ and $p$ is a homomorphism of $A$ into $\Gamma$. The kernel of $p$ is the set $\mathrm{Aut}_G(E)$ of automorphisms of the $G$-set $E$. We fix $a \in E$. We have defined in § 5, no. 6 an isomorphism $\psi_a$ of $G^0$ onto $\mathrm{Aut}_G(E)$ such that $\psi_a(x)(ga) = gxa$ for all $g, x$ in $G$. On the other hand, for $\gamma \in \Gamma$, let $s_a(\gamma)$ be the permutation of $E$ defined by $s_a(\gamma)(ga) = \gamma(g)a$ for all $g \in G$; it is immediately verified that $s_a$ is a homomorphism of $\Gamma$ into $A$ such that $p \circ s_a = \mathrm{Id}_{\tau}$. Thus $G^0 \xrightarrow{\psi_a} A \xrightarrow{p} \Gamma$ is an extension of $\Gamma$ by $G^0$ and $s_a$ is a section of this extension. This extension and this section define an operation of $\Gamma$ on $G^0$, $s_a(\Gamma)$ acting on $\psi_a(G^0)$ by inner automorphisms; we write this operation exponentially. We show that this operation is the natural operation ($§ 3$, no. 1, Example 3): for $x, g$ in $G$ and $\gamma \in \Gamma$, whence $^\gamma x = \gamma(x)$. Proposition 4 then shows that $A$ is isomorphic to the semidirect product of $\Gamma = \mathrm{Aut}(G)$ by $G^0$ under the natural operation of $\mathrm{Aut}(G)$ on $G^0$. Note that the isomorphism which we have constructed depends in general on the choice of the element $a \in E$. (2) Let $A$ be a commutative ring. The upper triangular group $T(n, A)$ is the semi-direct product of the diagonal subgroup $D(n, A)$ by the upper strict triangular group $T_1(n, A)$.