Let G be a simply connected finite-dimensional solvable Lie group over R or C and M an integral subgroup of G. Then M is a Lie subgroup of G and is simply connected.
We continue to use the notation n,g,gi,xi,ϕ of Proposition 20 but impose on the xi the following supplementary condition: let ip>ip−1>⋯>i1 be the integers i such that L(M)∩gi=L(M)∩gi−1; then we take
xik∈L(M)∩gik
\dagger Such a sequence exists by Chapter I, § 5, Proposition 2.
for k=1,2,…,p. By induction on n, it is easily seen that (xip,xip−1,…,xi1) is a basis of L(M). Let N be a simply connected Lie group such that there exists an isomorphism h of L(N) onto L(M). Let
yp=h−1(xip),…,y1=h−1(xi1). By Proposition 20, the mapping
(λ1,λ2,…,λp)↦(expλ1y1)(expλ2y2)⋯(expλpyp) is an isomorphism of the manifold
Kp onto the manifold
N. There exists a Lie group morphism
τ of
N into
G such that
h=L(τ) and
τ(N)=M (§ 6, no. 2, Corollary 1 to Proposition 1). Hence
M is the set of elements of
G of the form
τ((expλ1y1)⋯(expλpyp))=exp(λ1L(τ)y1)⋯exp(λpL(τ)yp)=exp(λ1xi1)⋯exp(λpxip). Thus
M=ϕ(T) where
T is a vector subspace of
Kn.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.