Exercise 30, § 2

A topological group $G$ is said to have no arbitrarily small subgroups if there is a neighbourhood $V$ of $e$ in $G$ such that $\{e\}$ is the only subgroup of $G$ contained in $V$.

$a$ Let $G$ be a topological group, $H$ a normal subgroup of $G$. Show that if $H$ and $G/H$ have no arbitrarily small subgroups, then neither does $G$.

$b$ Deduce from $a$ that if $H_1, H_1$ are two normal subgroups of a topological group $G$ such that $G/H_1$ and $G/H_2$ have no arbitrarily small subgroups, then neither does $G/(H_1 \cap H_2)$.