We can prove this by showing that (1) every
a∈A belongs to an orbit, and (2) orbits are
disjoint.
(1) Since
1∈G, then
a=1∙a is in the orbit
G∙a
(2) As follows
First note that for any
g0∈G and
a∈A we have that
G∙(g0∙a)=G∙a. This is because
G∙(g0∙a)={g∙(g0∙a):g∈G}={gg0∙a:g∈G}={g′∙a:g′∈g0G}=(g0G)∙a. Note
g0G is a
left coset of
G; since
g0∈G then
g0G=G. Thus
(g0G)∙a=G∙a.
Now note that any orbit
G∙a is decided fully by a single element. If an orbit
G∙a contains some element
x∈A, then exists some
g0∈G where
x=g0∙a. Thus
G∙x=G∙(g0∙a)=G∙a.
Now take two orbits
G∙a1 and
G∙a2. If some element
x∈A belongs to both, then
G∙a1=G∙x=G∙a2, so they are the same. Thus orbits are
disjoint.