Exercise 17, § 4

Let $G$ be a locally compact group with no arbitrarily small subgroups [$\S$ 2, Exercise 30].

a) Show that every point of $G$ has a countable fundamental system of neighbourhoods (Exercise 15 b)).

b) There is a compact neighbourhood $V$ of $e$ such that for any two points $x, y$ of $V$, the relation $x^2 = y^2$ implies $x = y$. (We may suppose $G$ not commutative. If the result were false, there would exist two sequences $(x_n), (y_n)$ of points of $G$, tending to $e$, such that $x_n^2 = y_n^2$ and $x_n y_n^{-1} = a_n \neq e$. Let $U$ be a compact neighbourhood of $e$ which contains no subgroup of $G$ other than $\{e\}$, and let $p_n$ be the smallest integer $p > 0$ such that $a_n^{p+1} \notin U$: show (by passing to a subsequence if necessary) that we may suppose that $a = \lim_{n \to \infty} a_n^{p_n}$ exists, is not equal to $e$ and belongs to $U$. Show that $a^{-1} = a$ and thus obtain a contradiction).

c) Let $U$ be a compact symmetric neighbourhood of $e$ which does not contain any subgroup of $G$ other than $\{e\}$, and let $V$ be a neighbourhood of $e$. Show that there exists a number $c(V) > 0$ such that, for each pair of integers $p > 0, q > 0$ such that $p \leq c(V)q$, and each element $x \in G$ such that $x, x^2, \ldots, x^q$ belong to $U$, we have $x^p \in V$. [Argue by contradiction, by supposing that there exist two sequences of integers $(p_n), (q_n)$ such that $\lim_{n \to \infty} (p_n/q_n) = 0$ and for each $n$ an element $a_n \in G$ such that $a_n^h \in U$ for $1 \leq h \leq q_n$, but $a_n^{p_n} \notin V$; we may also suppose that the sequence $(a_n^{p_n})$ has a limit $a \neq e$ lying in $U$. Show that we should then have $a^m \in U$ for every integer $m > 0$ and hence arrive at a contradiction].