Exercise 6, § 2

Let $G$ be a topological group and let $H$ be a closed normal subgroup of $G$, isomorphic to $\mathbf{R}^n$ ($n \geqslant 1$) and such that $G/H$ is abelian.

a) Suppose, moreover, that $H$ contains no connected subgroup normal in $G$ except for $H$ itself and $\{ e \}$, and that $H$ is not contained in the centre of $G$. Given $x_0 \in G$ which does not commute with every element of $H$, show that, for every $v \in H$, there exists a unique $u \in H$ such that $x_0^{-1} u x_0 u^{-1} = v$. (Observe that in the additive notation in $H$ the equation becomes $u - x_0^{-1} u x_0 = -v$ and that the continuous endomorphism $u \to u - x_0^{-1} u x_0$ of $H$ is a linear mapping of $\mathbf{R}^n$ into itself. Show that its kernel is a normal subgroup of $G$.) Furthermore, if $f(u)$ is the unique $u \in H$ such that $x_0^{-1} u x_0 u^{-1} = v$, then $f$ is a bicontinuous automorphism of $H$.

b) Under the hypotheses of a), show that the continuous mapping $g : y \to (f(x_0^{-1}yx_0y^{-1}))^{-1}y$ of $G$ into itself has as image the subgroup $L$ of $G$ consisting of all elements which commute with $x_0$ and that the relation $g(y) = g(y')$ is equivalent to $y{y'}^{-1} \in H$. Deduce that there exists a continuous bijection $h : G/H \to L$ whose inverse is the restriction to $L$ of the canonical mapping $p : G \to G/H$, and deduce that $G$ is isomorphic to the topological semi-direct product of $H$ and $L$.