Supplementary Exercises*: Topological Groups

• Alternatively, it is enough to require that $x\cdot y^{-1}:G\times G\rightarrow G$ is continuous.
• A subgroup $H$ of the topological group $G$ is a topological group.
  • Moreover, the closure $\overline{H}$ is also a subgroup, and, hence, a topological group.
• A topological group satisfies the regularity axiom: a closed subset $A\subset G$ and a point $x\in G$ can be separated by two disjoint open neighborhoods.
  • Hence, $G$ is also a Hausdorff space.
  • Note that we require $\{e\}$ to be closed in $G$ which is essential here, otherwise we would need to consider $G/\overline{\{e\}}$ , see the definition below.
• $f_{\alpha}(x)=\alpha\cdot x$ and $g_{\alpha}(x)=x\cdot\alpha$ are homeomorphisms. Hence, $G$ is a homogeneous space, i.e. for every $x,y\in G$ there is an automorphism $h$ of $G$ such that $h(x)=y$ .

• Note that we need additional requirements on $H$ for $G/H$ to be a topological group, namely, $H$ must be a normal subgroup and a closed subset of $G$ .
  • If $H$ is a normal subgroup, then $xH\cdot yH=(x\cdot y)H$ is well-defined, and $xH\cdot(yH)^{-1}$ is continuous.
  • If $H$ is closed, then $G/H$ is a $T_{1}$ -space, moreover, $G/H$ satisfies the regularity axiom, hence, $G/H$ is also a Hausdorff space.
• But regardless of what $H$ is, the following is true for $G/H$ as the quotient topological space:
  • The quotient map $p:G\rightarrow G/H$ is open.
  • $f_{\alpha}$ induces an automorphism of $G/H$ carrying $xH$ to $(\alpha\cdot x)H$ . Hence, $G/H$ is a homogeneous space.