Let L(G) be given a norm compatible with its topology and such that $\| [x, y] \| \leq \| x \| \| y \|$ for all x, y in L(G). Let G_1 be the Lie group defined by L(G). By § 4, no. 2, Theorem 2, G and G_1 are locally isomorphic. Then it suffices to apply § 4, no. 2, Lemma 3 (iii).

The first assertion has been proved in § 4, no. 2, Lemma 3. The second is a special case of § 4, no. 2, Theorem 2.

The first assertion follows from Proposition 1 and § 4, no. 2, Theorem 3. The second is a special case of § 4, no. 2, Theorem 3.

(i) There exist an open subgroup G' of G and a Lie group morphism $\phi$ of G' into H such that $h = L(\phi)$.

(ii) Let G_1, G_2 be open subgroups of G and $\phi_i$ a morphism of G_i into H such that $h = L(\phi_i)$. Then $\phi_1$ and $\phi_2$ coincide on an open subgroup of G.

By Proposition 1, this follows from § 4, no. 1, Theorem 1.

(i) There exist an open subgroup G' of G and a normal Lie subgroup H of G' such that L(H) = h.

(ii) h is an ideal of L(H).

If there exist G' and H with the properties of (i), then L(G') = L(G) and L(H) is an ideal of L(G') by § 3, no. 12, Proposition 47.

Suppose that h is an ideal of L(G). There exists a Lie group F such that L(F) = L(G)/h (Theorem 1). Let h be the canonical morphism of L(G) onto L(F). By Theorem 3 (i), there exist an open subgroup G' of G and a Lie group morphism $\phi$ of G' into F such that $L(\phi) = h$. By § 3, no. 8, the kernel H of $\phi$ is a Lie subgroup of G' and $L(H) = \mathrm{Ker}\ L(\phi) = \mathrm{Ker}\ h = h$. Finally, H is normal in G' since $H = \mathrm{Ker}\ \phi$.

Let $L(G)$ be given a norm compatible with its topology and such that $\| [x, y] \| \leq \| x \| \| y \|$ for $x, y$ in $L(G)$. Let $G_1$ be the Lie group defined by $L(G)$. Let $\psi = \mathrm{Id}_{G_1}$, which is an exponential mapping of $G_1$. For all $\mu > 0$, let $L_\mu$ be the set of $x \in L(G)$ such that $\| x \| < \mu$. Then, for $\mu$ sufficiently small, $L_\mu$ is an open subgroup of the additive group $L(G)$, $\psi(L_\mu)$ is an open subgroup of $G_1$ (§ 4, no. 2, Lemma 3), $\psi|L_\mu$ is an isomorphism of analytic manifolds of $L_\mu$ onto $\psi(L_\mu)$ and $\psi(nx) = \psi(x)^n$ for all $x \in L_\mu$ and all $n \in \mathbf{Z}$. The $L_\mu$ form a fundamental system of neighbourhoods of 0 in $L(G)$. By Theorem 1, there exist $\mu$ and an open subgroup $G'$ of G such that $\psi(L_\mu)$ and $G'$ are isomorphic, whence the proposition.

(1)

(2)

These are special cases of Proposition 4 of § 4, no. 3.

By replacing G by a group isomorphic to an open subgroup of G, the problem is reduced to the case where G is an open subset of K', with identity element 0 and where the coordinates of the product x.y and the inverse x^{[-1]} are given by formulae

(3)

(4)

where the series on the right hand side are convergent for x, y in G (§ 5, no. 1). Let λ ∈ K* and let the group law be transported from G to G' = λG by the homothety of ratio λ. For x', y' in G', the product x'.y' and the inverse x'^{[-1]} evaluated in G' have coordinates

where

As the series (3) and (4) are convergent, we see that, for |λ| sufficiently large, for all α, β, i,

that is c'_{\alpha \beta i} ∈ A and d'_{\alpha i} ∈ A; and on the other hand G' ⊃ m × m × ... × m. Then m × m × ... × m is an open subgroup of G' and is a standard group.

(i) If $a$ is a non-zero ideal of $A$ contained in $m$, $G(a)$ is an open normal subgroup of $G$.

(ii) The $G(a_\lambda)$, for $\lambda > 0$, form a fundamental system of neighbourhoods of $e$ in $G$.

(iii) Suppose that $a_\lambda \subset a$ for $\lambda \geq \lambda_0$ and let the $G(a)/G(a_\lambda)$, for $\lambda \geq \lambda_0$, be given the discrete topology. Then the topological group $G(a)$ is the inverse limit of the groups $G(a)/G(a_\lambda)$.

(iv) Let $a, b$ be non-zero ideals of $A$ contained in $m$ such that $a \supset b \supset a^2$. The mapping $x \mapsto (x_1 \bmod b, \ldots, x_r \bmod b)$ of $G(a)$ into $(a/b) \times \cdots \times (a/b)$ defines on passing to the quotient an isomorphism of the group $G(a)/G(b)$ onto the additive group $(a/b) \times \cdots \times (a/b)$.

If $x \in G$ and $y \in G(a)$, the coordinates of $x$ and $x.y$ are equal modulo $a$. Hence, for $x', x''$ in $G$ and $y', y''$ in $G(a)$, the coordinates of $x'.x''$ and $(x'.y').(x''.y'')$ are equal modulo $a$. This proves (i).

(ii) is obvious.

(iii) follows from the above and General Topology, Chapter III, § 7, Proposition 2.

If $x \in G(a)$ and $y \in G(a)$, the coordinates of $x.y$ are congruent to those of $x + y$ modulo $G(a^2)$ by formula (4) of § 5. This proves (iv).

(i) If $a = m^a$ and $b = m^b$ with $b \geq a \geq 1$, $G(a)/G(b)$ is a $p$-group of cardinal $q^{r(b-a)}$.

(ii) $G(a)$ is an inverse limit of $p$-groups.

The number of elements in $G(a)/G(b)$ is $(\mathrm{Card}(a/b))^r$; if $b = a + 1$, $a/b$ is a 1-dimensional vector space over $A/m$, whence (i) in this case; the general case follows by induction on $b - a$. Assertion (ii) follows from (i) and Proposition 5 (iii).

If $x \in G(a)$ and $y \in G(b)$, then $x^{[-1]}y^{[-1]}x.y, x.y.x^{[-1]}y^{[-1]}$ and $[x,y]$ belong to $G(c')$ and are congruent modulo $G(c')$.

By § 5, no. 2, Proposition 1, there exist $c_{\alpha\beta} \in A'$ such that

If $x = 0$ or $y = 0$, then $x^{[-1]}y^{[-1]}x.y - [x,y] = 0$; hence $c_{0\beta} = c_{\alpha 0} = 0$. On the other hand, the conditions imply and hence $x^{[-1]}y^{[-1]}x.y - [x,y] \in G(c')$. We see similarly that Finally, by § 5, formula (13), $[x,y] \in G(ab) \subset G(c)$.

By Chapter II, § 4, no. 4, we can form the group $\mathrm{gr}(G)$ associated with the group $G$, with the central filtration $(G(a_\lambda))$. Writing $G_\lambda = G(a_\lambda)/G(a_\lambda^+)$ for all $\lambda > 0$, we obtain $\mathrm{gr}(G) = \bigoplus_{\lambda > 0} G_\lambda$. Recall (loc. cit., Proposition 1) that the commutator in $G$ allows us to define a bracket in $\mathrm{gr}(G)$ with which $\mathrm{gr}(G)$ is a Lie algebra, as follows: if $\bar{x} \in G_\lambda$ and $\bar{y} \in G_\mu$, choose a representative $x$ of $\bar{x}$ in $G(a_\lambda)$ and a representative $y$ of $\bar{y}$ in $G(a_\mu)$; then $[\bar{x}, \bar{y}]$ is the class of $x^{[-1]}y^{[-1]}x.y \in G(a_{\lambda+\mu})$ in $G_{\lambda+\mu}$. By Proposition 6, applied with $a = a_\lambda, b = a_\mu, c = a_{\lambda+\mu}, c' = a_{\lambda+\mu}^+$, we see that $[\bar{x}, \bar{y}]$ is also the class of $[x, y]$ in $G_{\lambda+\mu}$. Thus, when $G$ is considered as a Lie subalgebra of $L(G) = K^r$, filtered by the $G(a_\lambda)$, the associated graded Lie algebra (Chapter II, § 4, no. 3) is equal to $\mathrm{gr}(G)$.

By definition of standard groups, $h_n$ is equal on the whole of $G$ to the sum of an integral series with coefficients in $A'$. By § 5, formula (4), this series is of the form

Hence, for $x \in G$, where we write $S(x) = x + \sum_{|\alpha| \geq 2} a_\alpha n^{|\alpha|-2} x^\alpha$. This series $S(x)$ defines an analytic mapping, also denoted by $S$, of $G$ into $G$. By Algebra, Chapter IV, § 6, Proposition 8, there exists an integral series $S'$ in $r$ variables with coefficients in $A'$ such that $S'(S(X)) = S(S'(X)) = X$. Hence $S$ is an isomorphism of the analytic manifold $G$ onto itself and, for every non-zero ideal $b$ of $A$ contained in $m$, $S(G(b)) \subset G(b)$, $S'(G(b)) \subset G(b)$ and therefore As $h_n(y) = n^2 S \left( \frac{1}{n} y \right)$ for $y \in nG$, we see that $h_n|nG(b)$ is an isomorphism of the analytic manifold $nG(b)$ onto the analytic manifold $n^2 G(b)$. But, as $n \notin a$, $|n| > |\lambda|$ for all $\lambda \in a$, hence $n^{-1} a \subset m$ and hence $a$ is of the form $nb$ where $b$ is a non-zero ideal of $A$ contained in $m$.

This follows immediately from Proposition 8.

There exists an open subgroup of $G$ which is isomorphic to a standard group (no. 3, Theorem 4). Assertion (i) then follows from Lemma 1.

Let $U$ be an open subgroup of $L(G)$ and $\phi : U \to \phi(U)$ an exponential mapping of $G$ with the properties of Proposition 3 of no. 2. $U$ can be assumed to be small enough for $\phi(U) \subset G_f$. Let $x \in G_f$. There exists $m \in \mathbf{Z} - \{0\}$ such that $x^m \in \phi(U)$. The element $\frac{1}{m} \phi^{-1}(x^m)$ does not depend on the choice of $m$. For let $m' \in \mathbf{Z}$ be such that $x^{m'} \in \phi(U)$. Then $x^{m m'} \in \phi(U)$ and

whence our assertion. Let $\psi(x) = \frac{1}{m} \phi^{-1}(x^m)$. Then $\psi|\phi(U) = \phi^{-1}$. On the other hand, if $n \in \mathbf{Z}$, then

Hence $\psi$ has properties (a) and (b) of the proposition. In a neighbourhood of $x$, $\psi$ is composed of the mappings $x \mapsto x^m$, $y \mapsto \phi^{-1}(y)$ and $z \mapsto \frac{1}{m} z$; hence $\psi$ is analytic on $G_f$.

Finally, let $\psi'$ be a mapping of $G_f$ into $L(G)$ and $V'$ a neighbourhood of $e$ in $G_f$ such that $\psi'(x^n) = n \psi'(x)$ for $x \in G_f$ and $n \in \mathbf{Z}$ and such that $\psi'|V'$ is the inverse mapping of an injective exponential mapping. Then $\psi$ and $\psi'$ coincide on a neighbourhood $W$ of $e$. If $x \in G_f$, there exists $n \in \mathbf{Z}$ such that $x^n \in W$. Then

and hence $\psi = \psi'$.

The fact that $xy \in G_f$ follows from Lemma 1. Let $U$ be an open subgroup of the additive group $L(G)$ and $\phi : U \to \phi(U)$ an exponential mapping of $G$ with the properties of Proposition 3 of no. 2; $U$ can be assumed to be small enough for $\log|\psi(U)|$ to be the inverse mapping of $\phi$. For $n \in \mathbf{Z} - \{0\}$ suitably chosen, $x^n \in \phi(U)$, $y^n \in \phi(U)$. Let $u = \log x^n$, $v = \log y^n$, whence $x^n = \phi(u)$, $y^n = \phi(v)$. By formula (2), $[u, v] = 0$. The Hausdorff formula proves then that $\phi(\lambda(u + v)) = \phi(\lambda u) \phi(\lambda v)$ for $|\lambda|$ sufficiently small; hence, for every sufficiently large integer $i$,

that is or

If there exists an integer $n > 0$ such that $x^n = e$, it follows that

whence $\log x = 0$. If $\log x = 0$, let $V$ be a neighbourhood of $e$ in $G_f$ such that $\log|V|$ is the inverse mapping of an injective exponential mapping. There exists an integer $n > 0$ such that $x^n \in V$; the equality $\log x^n = 0$ implies $x^n = e$.

If G is standard, it suffices to use Lemma 1. Suppose that G is compact. Let $x \in G$ and V be a neighbourhood of e in G. Let y be a limit point of the sequence $(x^n)_{n \geq 0}$. For all $n > 0$, there exist two integers $n_1, n_2$ such that $n_1 \geq 2n_2 \geq n$ and $x^{n_1} \in yV, x^{n_2} \in yV$, whence $x^{n_1 - n_2} \in V^{-1}V$ and $n_1 - n_2 \geq n$. Hence $x \in G_f$.

Let $x \in G$. If x belongs to a compact subgroup of G, then $x \in G_f$ (Proposition 13). Suppose that $x \in G_f$. As K is locally compact, there exists an open subgroup $G_1$ of G which is compact. Then there exists an integer $m > 0$ such that $x^m \in G_1$. The closed subgroup $G_2$ generated by $x^m$ is contained in $G_1$ and is therefore compact. Then x commutes with the elements of $G_2$ and hence $G_2 \cup xG_2 \cup \cdots \cup x^{m-1}G_2$ is a compact subgroup of G which contains x.

(i) The domain of absolute convergence (Differentiable and Analytic Manifolds, R, 4.1.3) of E contains the set $\Delta$ of $x \in G$ such that $\omega(x) > \frac{1}{p-1}$. Let $E'$ denote the mapping defined on $\Delta$ by this series. Then $E'$ is an exponential mapping of G and is an isomorphism of the manifold $\Delta$ onto itself.

(ii) The domain of absolute convergence of L contains G. Let $L'$ denote the mapping on G defined by this series. Then $L'$ is the logarithmic mapping of G and the restriction of $L'$ to $\Delta$ is the inverse mapping of $E'$.

(iii) The mapping $E'$ is an isomorphism of $\Delta$, with the Hausdorff law, onto the subgroup $\Delta$ of G.

Using the notation of § 5, nos. 3 and 4, $E = \sum_{m \geq 1} \frac{\psi_{m,m}}{m!}$ ($§ 5$, no. 4, Proposition 3). As the coefficients $c_{\alpha \beta \gamma}$ belong to A, $\| \psi_{m,m} \| \leq 1$ (Differentiable and Analytic Manifolds, R, Appendix) $K^r$ is assumed to have the norm

By Chapter II, § 8, no. 1, Lemma 1, $v(m!) \leq \frac{m-1}{p-1}$. If $\omega(x) > \frac{1}{p-1}$, we see that $m \omega(x) - v(m!)$ tends to $+\infty$ with $m$, whence

which tends to 0 as $m$ tends to $+\infty$ and Therefore $\Delta$ is contained in the domain of absolute convergence of $E$ and $E'(\Delta) \subset \Delta$. Clearly $E'$ is an exponential mapping.

If $L_m$ denotes the homogeneous component of $L$ of degree $m$, Proposition 3 of § 5, no. 4, proves that each coefficient of $L_m$ is of the form

with $a_1, a_2, \ldots, a_m$ in $A$; but and Therefore, if $\omega(x) > 0$, $\|L_m\| \cdot \|x\|^m$ tends to 0 as $m$ tends to $+\infty$, so that $G$ is contained in the domain of absolute convergence of $L$. On the other hand, if $\omega(x) > \frac{m}{p-1}$, then $\omega(L_m(x)) > \frac{m}{p-1} - \frac{m-1}{p-1} = \frac{1}{p-1}$ for $m \geq 1$ and hence $L'(\Delta) \subset \Delta$.

As the formal power series $L(E(X))$ and $E(L(X))$ are equal to $X$, no. 4.1.5 of Differentiable and Analytic Manifolds, R, proves that

for $x \in \Delta$. Hence $E'$ is an isomorphism of the manifold $\Delta$ onto itself and the inverse isomorphism is the restriction of $L'$ to $\Delta$.

$L(X^{[n]}) = nL(X)$ for $n$ an integer $> 0$ (cf. § 5, no. 4). As $G$ is contained in the domain of absolute convergence of $L$ and $X^{[n]}$, therefore $L'(x^n) = nL'(x)$ for all $x \in G$. The relation $L'|_{\Delta} = {E'}^{-1}$ implies that $L'(x^n) = \log x^n$ for $n$ sufficiently large. Hence $L'(x) = \log x$. We have thus proved (i) and (ii).

Let $H = \sum_{r,s \geq 0} H_{r,s}$ be the Hausdorff formal power series and $h$ the Hausdorff function relative to $L(G)$. The domain of absolute convergence of $\tilde{H}$ contains $\Delta \times \Delta$ and $h$ is defined on $\Delta \times \Delta$ (Chapter II, § 8, Proposition 2). Then

for $x, y$ sufficiently close to 0 ($§ 4$, Theorem 4 (v)). Hence, in the notation of no. 3, Definition 1, the formal power series $F(E(X), E(Y))$ and $E(H(X, Y))$ are equal. Let $x, y$ be elements of $\Delta$. Then by Chapter II, § 8, formula (14). By Differentiable and Analytic Manifolds, R, 4.1.5, $E'(x)E'(y)$ is obtained by substituting $x$ for $X$ and $y$ for $Y$ in and $E'(h(x, y))$ is obtained by substituting $x$ for $X$ and $y$ for $Y$ in $E(H(X, Y))$. Hence $E'(x)E'(y) = E'(h(x, y))$.