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.

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

Let $\mathscr{O}$ be the set of pairs $(X, f)$, where X is a subgroup of B containing C and $f$ a morphism of X into A extending $\varphi$. Let us order $\mathscr{O}$ by the relation “ $X\subset X'$ and $f'$ extends $f$ ”. One verifies that $\mathscr{O}$ is inductive. Let $(X, f)$ be a maximal element of $\mathscr{O}$ (E, III, p. 20 , Theorem 2). If $X\not= B$, let us take an element $b$ of B-X and let $X'$ be the subgroup generated by X and $b$. The commutativity of B shows that $X'$ is the set of elements $b^nx$ for $n\in \mathbf{Z}$ and $x\in X$. Suppose first that $b^n\notin X$ for every integer $n\not= 0$ and define $f'$ of $X'$ into A by taking an arbitrary element $y\in A$ and putting $f'(b^nx) =y^nf(x)$ for every $n\in \mathbf{Z}$ and every $x\in X$. Since A is commutative, $f'$ is a morphism, and it extends $f$. Suppose now that there exists $n\not= 0$ such that $b^n\in X$ and let $m >0$ be such that $m\mathbf{Z}=\{n\in \mathbf{Z}|b^n\in X\}$. Since A is divisible, there exists an element $y\in A$ such that $y^m=f(b^m)$. One then extends $f$ to a morphism of $X'$ into A by $f'(b^nx) =y^nf(x)$ for $n\in \{0,1, . . . , m-1\}$ and $x\in X$. In both cases, $(X, f)$ would not be maximal. Hence X = B and the lemma is proved.

(i) G is generated by a compact subset;

(ii) there exist positive integers $p$ and $q$ and a compact group K such that G is isomorphic to $\mathbf{R}^p\times \mathbf{Z}^q\times K$;

(iii) there exists an integer $n\geqslant 0$ such that $\widehat{G}$ is locally isomorphic to $\mathbf{R}^n$;

(iv) there exist positive integers $p$ and $q$ and a discrete group D such that $\widehat{G}$ is isomorphic to $\mathbf{R}^p\times \mathbf{T}^q\times D$.

(i) $=\Rightarrow$ (iii) : if G has property (i), there exists an integer $n\geqslant 0$ and a subgroup D of G isomorphic to $\mathbf{Z}^n$ such that $G/D$ is compact (lemma 2). Then $D^{\bot}$, which identifies with the dual of $G/D$, is discrete (Theorem 4 of II, p. 226 and prop. 18 of II, p. 233). Hence $\widehat{G}$ is locally isomorphic to $\widehat{G}/D^{\bot}$, that is to say to $\widehat{D}$, which is isomorphic to $\mathbf{T}^n$ (Theorem 4 of II, p. 226 and prop. 18 of II, p. 233). Now $\mathbf{T}^n$ is locally isomorphic to $\mathbf{R}^n$.

(iii) $=\Rightarrow$ (iv) : if $\widehat{G}$ is locally isomorphic to $\mathbf{R}^n$, there exists an integer $p$ such that $0\leqslant p\leqslant n$ such that the identity component $\widehat{G}_0$ of $\widehat{G}$ is an open subgroup isomorphic to $\mathbf{R}^p\times \mathbf{T}^{n-p}$ (TG, VII, p. 13, Theorem 1). In particular, $\widehat{G}_0$ is a divisible group. We then apply lemma 3 to the identity mapping of the subgroup $\widehat{G}_0$ of the group $\widehat{G}$ into the divisible group $\widehat{G}_0$. There therefore exists a morphism $\pi$ of $\widehat{G}$ into $\widehat{G}_0$ which is the identity mapping on $\widehat{G}_0$. Consequently, we have $\pi \circ \pi =\pi$, and $\pi$ is a projector. It is continuous, since its restriction to the open subgroup $\widehat{G}_0$ is so. It follows that $\widehat{G}$ is the direct product of $\widehat{G}_0$ and the subgroup $\overset{-1}{\pi}(e)$, which is discrete since it is isomorphic to $\widehat{G}/\widehat{G}_0$ (TG, III, p. 47, cor.)

(iv) $=\Rightarrow$ (ii) : follows from prop. 2 of II, p. 206, from prop. 18 of II, p. 233 and from corollary 3 of II, p. 236.

(ii) $=\Rightarrow$ (i) : for every compact group K, the group $\mathbf{R}^p\times \mathbf{Z}^q\times K$ is generated by the compact set $[0,1]^p\times \{0,1\}^q\times K$.

a) There exists a compact subgroup K of G and positive integers $p$ and $q$ such that G is isomorphic to $\mathbf{R}^p\times \mathbf{Z}^q\times K$.

b) Conversely, let K be a compact group, let $p$ and $q$ be positive integers, and let G be a group isomorphic to $\mathbf{R}^p\times \mathbf{Z}^q\times K$. Then K is the unique maximal compact subgroup of G, and the integers $(p, q)$ are determined in a unique way by G.

Assertion a) follows from prop. 1. Let then K be a compact group, and let $p,q$ be positive integers. Suppose that G is isomorphic to the group $\mathbf{R}^p\times \mathbf{Z}^q\times K$, and identify G with this group. By the canonical projection of G onto $\mathbf{R}^p\times \mathbf{Z}^q$, the image of every compact subgroup of G is a compact subgroup of $\mathbf{R}^p\times \mathbf{Z}^q$, and is therefore reduced to the identity element. Hence $K'\subset K$ and K is the largest compact subgroup of G. The subgroup $\mathbf{R}^p\times K$ is also unique since $\mathbf{R}^p$ is the identity component of $G/K$. In view of TG, VII, p. 13, cor. 3, the integer $p$ is determined in a unique way by G. Since $G/(\mathbf{R}^p\times K)$ is isomorphic to $\mathbf{Z}^q$, the integer $q$ is likewise determined in a unique way by G.

(i) The groups G and $\widehat{G}$ are generated by compact subsets ;

(ii) There exist positive integers $n$ and $m$ such that G is locally isomorphic to $\mathbf{R}^m$ and $\widehat{G}$ to $\mathbf{R}^n$;

(iii) There exist positive integers $p,q$ and $r$ and a finite group A such that G is isomorphic to $\mathbf{R}^p\times \mathbf{T}^q\times \mathbf{Z}^r\times A$ ;

(iv) There exist positive integers $p,q$ and $r$ and a finite group A such that $\widehat{G}$ is isomorphic to a product $\mathbf{R}^p\times \mathbf{Z}^q\times \mathbf{T}^r\times A$.

One has (i) $\Leftrightarrow$ (ii) according to prop. 1, and (iii) $\Leftrightarrow$ (iv) by duality, hence (iii) $\Rightarrow$ (i). Finally, if (i) is true, then $\widehat{G}$ is isomorphic to $\mathbf{R}^p\times \mathbf{T}^r\times D$ where D is discrete (prop. 1). The group D is generated by a compact subset, hence finite, of D. Consequently, there exists $q\geqslant 0$ such that D is isomorphic to $\mathbf{Z}^q\times A$, where A is a finite group (A, VII, p. 22, th. 3).

a) the group G identifies with the projective limit of the $G/H_i$;

b) for every $i$, there exists an integer $q\geqslant 0$ and a finite group A such that $G/H_i$ is isomorphic to $\mathbf{T}^q\times A$.

Indeed, $\widehat{G}$ is discrete (prop. 18 of II, p. 233), hence union of an increasing filtrant family $(D_i)_{i\in I}$ of subgroups of finite type. Put $H_i= D^{\bot}_i$ ; the group G identifies with the projective limit of the $G/H_i($II, p. 234, cor. 3), and $(H_i)$ is a decreasing filtrant family.

Let $i\in I$. There exists an integer $q\geqslant 0$ and a finite group A such that the group $D_i$ is isomorphic to $\mathbf{Z}^q\times A$ (A, VII, p. 22, th. 3), hence $G/H_i$ is isomorphic to $\mathbf{T}^q\times \widehat{A}($cf. corollary 1 of II, p. 232).

According to (ii) of prop. 1 of II, p. 246, the corollary follows from prop. 2.

b) The group G is union of an increasing filtrant family of open subgroups, each of which is a projective limit of groups isomorphic to groups of the form $\mathbf{R}^p\times \mathbf{T}^q\times \mathbf{Z}^r\times A$, where A is a finite group and $p,q,r$ are positive integers.

Prove b). For every compact neighbourhood V of $e$, denote by $G_V$ the subgroup of G generated by V. It is open, and by the corollary of Prop. 2, the group $G_V$ is the projective limit of groups of the form $\mathbf{R}^p\times \mathbf{T}^q\times \mathbf{Z}^r\times A$, where A is a compact group and $p,q$ and $r$ are in $\mathbf{N}$. When V runs through the compact neighbourhoods of $e$, these subgroups $G_V$ form a filtered family (since $G_V$ and $G_W$ are contained in $G_{V\cup W}$ for all compact neighbourhoods V and W of $e$). Finally, the group G is the union of the subgroups $G_V$.

Let H be an open subgroup of G generated by a compact neighbourhood of $e$. There exists a compact group K and positive integers $p$ and $q$ such that H is isomorphic to $\mathbf{R}^p\times \mathbf{Z}^q\times K$ (Prop. 1 of II, p. 246); identify H with this product. The canonical surjection of H onto the divisible group $\mathbf{R}^p$ extends to a morphism $\pi$ of G onto $\mathbf{R}^p$ (Lemma 3 of II, p. 245). It is a projector $\pi$ of G onto $\mathbf{R}^p$, which is continuous since its restriction to the open subgroup H is continuous. Thus G is the direct product of $\mathbf{R}^p$ and the kernel L of $\pi$ (TG, III, p. 47, cor.). We have $\mathbf{Z}^q\times K = H\cap L$, hence $\mathbf{Z}^q\times K$ is an open subgroup of L. Thus K is a compact open subgroup of L, and consequently $L/K$ is discrete.

The set $B_G$ is a subgroup of G since the product of two compact subsets of G is a compact subset of G.

Let H be an open subgroup of G generated by a compact neighbourhood of $e$. There exist positive integers $p$ and $q$ and a compact group K such that the subgroup H is isomorphic to $\mathbf{R}^p\times \mathbf{Z}^q\times K$ (Prop. 1 of II, p. 246). If these groups are identified, we see that $B_G\cap H = K$ is closed in H. Since the family of open subgroups H generated by compact neighbourhoods is an open covering of G (for example, $x$ belongs to the subgroup generated by $U\cap \{x\}$ for every fixed compact neighbourhood U of $e$), it follows that $B_G$ is closed.

We now calculate $B^{\bot}_G$. By Prop. 3, a), there exist a positive integer $n\geqslant 0$ and a group L admitting a compact open subgroup such that G can be identified with $\mathbf{R}^n\times L$. Then $B_G$ is identified with $\{0\} \times B_L$, and $B^{\bot}_G$ with $\mathbf{R}^n\times B^{\bot}_L$. We are therefore reduced to the case where G = L admits a compact open subgroup K.

On then has $K\subset B_G$; if one identifies $G/\widehat{K}$ with $K^{\bot}$ (Theorem 4 of II, p. 226), the orthogonal $(B_G/K)^{\bot}$ of $B_G/K$ in $\widehat{G}/K$ is identified with $B^{\bot}_G$. But on the other hand $B_G/K = B_{G/K}$, and since $K^{\bot}$ is an open subgroup of $\widehat{G}$ (Cor. 2 of II, p. 233), the connected component $\widehat{G}_0$ of $\widehat{G}$ is also the connected component of $K^{\bot}$ (TG, III, p. 35, Prop. 14). Thus the assertion for the discrete group $G/K$ is equivalent to that for G.

Finally, suppose that G is discrete. The group $\widehat{G}$ is then compact (Prop. 18 of II, p. 233). The connected component $(\widehat{G})_0$ is the intersection of the open subgroups of $\widehat{G}$ (TG, III, p. 35, Prop. 14) and a subgroup of $\widehat{G}$ is open if and only if it is closed and of finite index, or equivalently if its orthogonal is finite (Corollary 2 of II, p. 233); Corollary 4 of II, p. 228 shows that $(\widehat{G})^{\bot}_0$ is the union of the finite subgroups of G, which is none other than $B_G$ since G is discrete. We conclude by duality.

(i) The group G is connected;

(ii) The group $\widehat{G}$ is without torsion;

(iii) The group G is divisible.

This follows from Prop. 4 and Cor. 7 of II, p. 229 since $B_G= G$.

The group G is totally discontinuous if and only if its connected component is reduced to $\{e\}$; Prop. 4 shows that this condition is equivalent to $B_{\widehat{G}}=\widehat{G}$. Since the group $\widehat{G}$ is discrete (Prop. 18 of II, p. 233), this amounts to saying that every element of $\widehat{G}$ generates a finite group, hence that $\widehat{G}$ is torsion.

Indeed, there exists a compact connected group K and an integer $n\geqslant 0$ such that G is isomorphic to $\mathbf{R}^n\times K$ (II, p. 248, Prop. 3, where one must have L = K). Corollary 1 shows that K is divisible, and therefore G is divisible.