Exercise 28, § 2

a) Let $S$ be a stable subset of a topological group $G$. Show that $\dot{S}$ and $\bar{S}$ are stable and that $S \dot{S} \subset \dot{S}$ and $S \bar{S} \subset \bar{S}$. If also $S$ is open and $e \in \bar{S}$, then $S = \dot{\bar{S}}$ (if $x \in \bar{S} - S$, show that $x$ is a frontier point of $S$).

b) From now on in this exercise, $G$ is a commutative topological group, written additively. For every non-empty subset $A$ of $G$, let $s(A)$ denote the set of all $x \in G$ such that $x + A \in A$, i.e. the intersection of the sets $A - y$ as $y$ runs through $A$; let $b(A) = s(A) \cap s(-A)$. The set $s(A)$ is a stable subset of $G$, and $b(A)$ is a subgroup of $G$ consisting of all $x \in G$ such that $x + A = A$. Show that if $F$ is any non-empty closed subset of $G$, then $s(F)$ is closed. For every non-empty subset $A$ of $G$, we have $s(A) \subset s(\overline{A})$ and $s(A) \subset s(\dot{A})$; and if $A = \dot{\overline{A}}$, then $s(\dot{\overline{A}}) = s(\overline{A})$.

c) Let $\mathscr{S}$ be the set of all stable open subsets $S$ of $G$ such that $0 \notin S$, and suppose that $\mathscr{S} \neq \emptyset$. Then the set $\mathscr{S}$, ordered by inclusion, is inductive; let $\mathfrak{M} \neq \emptyset$ be the set of its maximal elements. Show that if $M \in \mathfrak{M}$, then $M = \overline{M}$ [use $a$] and that, for each integer $n > 0$, $M$ is equal to the set of all $x \in G$ such that $nx \in M$. In addition, the subgroup $b(M)$ is equal to the complement of $M \cup (-M)$ in $G$ (if $x \notin M$ and $-x \notin M$, consider the union of the sets $kx + M$ for integers $k \geq 0$); deduce that $G = M - M$.

A closed subgroup $B$ of $G$ is said to be residual if there exists $M \in \mathfrak{M}$ such that $B = b(M)$.

$d$ The intersection of all the residual subgroups of $G$ is called the radical of $G$ and is denoted by $T_G$. If $T_G = G$ (resp. $T_G = \{0\}$) then $G$ is said to be a radical group (resp. a group without radical). For $x$ to belong to $T_G$ it is necessary and sufficient that every stable open subset of $G$ which contains $x$ should also contain $0$. Deduce that if $G$ is discrete, then $T_G$ is the torsion subgroup of $G$. If $f$ is any continuous homomorphism of $G$ into a commutative topological group $G'$, then $f(T_G) \subset T_{G'}$. If $H$ is a subgroup of $G$ contained in $T_G$, then $T_{G/H} = T_G / H$. The subgroup $T_G$ is the smallest of the subgroups $H$ of $G$ such that $G/H$ is without radical.

$e$ Show that, among the subgroups $H$ of $G$ such that $T_H = H$, there is a largest, $H_0 \subset T_G$, which is closed.

$f$ If $T_G \neq G$, then $T_G$ is open if and only if there is a residual open subgroup in $G$ [show that if $b(M)$ is open for some $M \in \mathfrak{M}$, then $T_G \supset b(M)$, by using $e$].