Let (ϕi:Hi→Gi)i∈I be a family of homomorphisms of groups with operators and let Φ=∏i∈Iϕi. Then:
(a) Ker(Φ)=∏i∈IKer(ϕi); in particular, if all the ϕi *are injective, Φ is injective.
(b) Im(Φ)=∏i∈IIm(ϕi); in particular, if all the ϕi *are surjective, Φ is surjective.
This is immediate.
In particular, let (Gi)i∈I be a family of groups with operators and, for all i, let Hi be a stable (resp. normal stable) subgroup of Gi. The product ∏i∈IHi is a stable (resp. normal stable) subgroup of ∏i∈IGi and the canonical mapping of ∏i∈IGi onto ∏i∈I(Gi/Hi) defines when passing to the quotient an isomorphism of
(i∈I∏Gi)/(i∈I∏Hi) onto
∏i∈I(Gi/Hi). For example, let
J be a subset of
I. The subgroup
GJ of
∏i∈IGi consisting of the
(xi)i∈I such that
xi=ei for
i∈/J is a normal stable subgroup. The mapping
ιJ which associates with
x=(xj)j∈J the element
y=(yi)i∈I such that
yi=ei for
i∈/J and
yi=xi for
i∈J, is an isomorphism of
∏j∈JGj onto
GJ. The mapping
prI−J defines when passing to the quotient an isomorphism
θJ of the quotient group
G/GJ onto
∏i∈I−JGi. The composition
prJ∘ιJ is the identity mapping of
∏j∈JGj.
G/GJ is often identified with
∏i∈I−JGi because of
θJ and
∏i∈JGi with
GJ because of
ιJ.
If J1 and J2 are disjoint subsets of I, it follows from the definitions that every element of GJ1 commutes with every element of GJ2.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.
Nothing.