Lemma 1

Let H be a locally compact group, and let R be one of the groups $\mathbf{R}$ or $\mathbf{Z}$. Let $\varphi$ be a continuous morphism from R into H. If $\varphi$ is not a topological isomorphism of R onto a subgroup of H, then the image of R in H is relatively compact.

Let I be the image of $\varphi$. After replacing H by $\overline{I}$, we may suppose that I is dense in H. We must then prove that H is compact if $\varphi$ is not a topological isomorphism of R onto I.

Suppose that there exists a neighbourhood V of $e$ in H and an integer $M>0$ such that, for every $t >M$ in R, one has $\varphi (t)\notin V$. Then $\varphi$ is injective: if $\varphi (u) =e$, one has $\varphi (nu) =e$ for every integer $n\geqslant 1$, hence the set $\mathbf{N}u$ is bounded in R, which means that $u= 0$. The restriction of $\varphi$ to $[-M,M]\cap R$ is therefore a homeomorphism onto its image, which contains $V\cap I$. The restriction of $\varphi^{-1}$ to $V\cap I$ being continuous, it follows that $\varphi$ is a topological isomorphism of R onto I.

Suppose now that $\varphi$ is not a topological isomorphism of R onto I. Let W be a relatively compact open neighbourhood of $e$ in H, and let V be a symmetric neighbourhood of $e$ such that $V^2\subset W$. For every $x\in$ H = I, there exists an element $s\in$ R such that $x\in \varphi (s)V$. According to the preceding paragraph and the assumption on $\varphi$, there exists $t\in$ R such that $t >|s|$ and $\varphi (t)\in V$. We then have $x\in \varphi (t+s)\varphi (t)^{-1}V\subset \varphi (t+s)W$, and $t+s >0$. Consequently, the open sets $\varphi (u)W$ for $u >0$ form an open covering of H. Since W is relatively compact, there exist an integer $n\geqslant 1$ and elements $u_1, . . . , u_n$ of R, strictly positive, such that $W\subset \bigcup_{1\leqslant i\leqslant n}\varphi (u_i)W$. Let U be the largest of the $u_i$.

Let $x\in H$ and let $s=$ inf$\{t\in R|t\geqslant 0,\varphi (t)x^{-1}\in \overline{W}\}$. Since $\overline{W}$ is compact, we then have $\varphi (s)x^{-1}\in \overline{W}$. There exists an integer $i$ such that $\varphi (s)x^{-1}\in \varphi (u_i)W$, whence $\varphi (s-u_i)x^{-1}\in \overline{W}$. The definition of $s$ implies $s-u_i<0$, whence $s\leqslant U$. It follows that $H =\varphi ([0,U]\cap R)W$ is compact.