Let G=HK, where H, K are characteristic subgroups of G with trivial intersection, i.e. . Then,
.
Proof:
Now suppose , where
and
are characteristic subgroups of
with
. Define
by
is a homomorphism, and bijective since
. Thus
and similarly,
so that
is well-defined.
Note that so
is a homomorphism.
Suppose . Then
. Then for
,
so that
. Thus
is injective.
For any , define
. Then
So is a homomorphism.
If , then
, so that
. Then since
, so
, so that
. Thus
and
is injective.
Any can be written as
since
is bijective. Similarly, any
can be written as
. Then
so
is surjective.
Thus . Note that
since
. Similarly,
. So
and
is surjective.
Hence is an isomorphism.