Lemma 2

If G is generated by a compact subset V, there exists an integer $n\geqslant 0$ and a discrete subgroup D of G isomorphic to $\mathbf{Z}^n$ such that $G/D$ is compact.

After replacing V by $V\cup V^{-1}$, we may suppose that V is symmetric; the assumption then means that G is the union of the sets $V^n$ where $n\in \mathbf{N}$.

Since $V^2$ is compact, there exist an integer $k\geqslant 1$ and elements $x_1, . . . , x_k\in G$ such that $V^2\subset \bigcup_{1\leqslant i\leqslant k}x_iV$. Let $D_0$ be the subgroup of G generated by the family $(x_i)_{1\leqslant i\leqslant k}$. We have $V^2\subset D_0V$, whence by induction $V^n\subset D_0V$ for every integer $n\geqslant 1$, and therefore $G = D_0V$ since V generates G. Let then J be a subset of $\{1,2, . . . , k\}$ such that the subgroup D generated by the family $(x_i)_{i\in J}$ is topologically isomorphic to $\mathbf{Z}^{Card(J)}$, and maximal for this property. Let us show that $G/D$ is compact.

Let $p$ be the canonical surjection of G onto $G/D$. Let $i\in \{1,2, . . . , k\}-J$. If the subgroup $H_i$ of $G/D$ generated by $p(x_i)$ is topologically isomorphic to $\mathbf{Z}$, the subgroup of G generated by D and $x_i$ is discrete and the mapping $(d, n)\mapsto dx^n_i$ is an isomorphism of $D\times \mathbf{Z}$ onto this subgroup, contrary to the maximality of J. Lemma 1 therefore implies that $\overline{H}_i$ is compact. Hence $G/D = (\prod_{i\notin J}\overline{H}_i)p(V)$ is compact