Exercise 5, § 2

A semi-topological group $G$ is quasi-topological if, for each $a \in G$, the mapping $x \to xax^{-1}$ of $G$ into $G$ is continuous. Every commutative semi-topological group is quasi-topological.

a) Show that the normalizer of a closed subgroup of a quasi-topological group is closed.

b) Let $G$ be a quasi-topological group which is generated by each neighbourhood of $e$ (resp. which is connected). Show that every discrete (resp. totally disconnected) normal subgroup $D$ of $G$ is contained in the centre of $G$ (if $a \in D$, show that there is a neighbourhood $V$ of $e$ such that $xax^{-1} = a$ for each $x \in V$).

c) Let $G$ be the group of matrices $\begin{pmatrix} 1 & 0 \\ y & x \end{pmatrix}$, where $x$ and $y$ are real numbers and $x > 0$. If we identify $G$ with a subset of $\mathbf{R}^2$ by means of the mapping $(x, y) \to \begin{pmatrix} 1 & 0 \\ y & x \end{pmatrix}$, then the topology $\mathfrak{T}_0$ induced on $G$ by the topology of $\mathbf{R}^2$ is compatible with the group structure of $G$. Let $H$ be the normal subgroup of $G$ consisting of all matrices $\begin{pmatrix} 1 & 0 \\ y & 1 \end{pmatrix}$, and let $H^*$ be the complement of the identity element in $H$. Define a topology $\mathfrak{T}$, finer than $\mathfrak{T}_0$, by taking as a fundamental system of neighbourhoods of $e$ in $G$ the intersections of neighbourhoods of $e$ in the topology $\mathfrak{T}_0$ with the complement of $H^*$ in $G$. With respect to this topology $\mathfrak{T}$, show that $G$ is a connected semi-topological group which is not quasi-topological and which has $H$ as a discrete subgroup not contained in the centre. \