(i) The mapping t↦Lt (resp. t↦Rt) is an isomorphism of the vector space U(G) onto the vector field of left (resp. right) invariant distributions on G. (ii) For t,t′ in U(G), Lt∗t′=Lt∘Lt′, Rt∗t′=Rt′∘Rt, Lt∘Rt′=Rt′∘Lt (with the abuse of notation of no. 5). (iii) If θ is the mapping g↦g−1 of G onto G, then θ(Lt)=Rt∨. (iv) If t∈U(G) and g∈G, then (Lt)g=(Rεg∗t∗εg−1)g.
In G every right translation commutes with every left translation. By Proposition 21 of no. 5, Lt is therefore left invariant. As (Lt)e=t, the mapping t↦Lt is injective. Let Δ be a field of left invariant distributions on G; let t=Δe; then Δ and Lt have the same value at e and are left invariant and hence Δ=Lt. This proves (i) for Lt and the argument is similar for Rt. The formulae Lt∗t′=Lt∘Lt′, Rt∗t′=Rt′∘Rt follow from (21) and (18). Let t∈Us(G), t′∈Us′(G), f∈Cr(Ω,F), where Ω is open in G and s+s′≤r; then
LtRt′f=Lt(t′∨∗f)=(t′∨∗f)∗t=t′∨∗(f∗t∨)=Rt′Ltf(Proposition 20) and hence
Lt∘Rt′=Rt′∘Lt. As
θ is an isomorphism of
G onto
G∨,
θ(Lt) is a field of right invariant distributions on
G; its value at
e is
θ(t)=t∨; hence
θ(Lt)=Rt∨. Finally,
(Lt)g=εg∗t=(εg∗t∗εg−1)∗εg=(Rεg∗t∗εg−1)g.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.