*Proof.*
and all its nontrivial quotients are
-groups and thus have nontrivial centers. Thus if
,
. Since
is finite,
.Â â—»

*Proof.* Suppose
. Assume that
. Let
be the canonical epimorphism. Similarly for
. Verify that the canonical epimorphism
is the composition
where
and
is an isomorphism. Consequently,
The inductive step is proved so
for all
. Since
are nilpotent,
,
. Thus
.Â â—»

*Proof.* Let
and let
be the largest index such that
. Choose
.
. Thus
. This implies
. Thus
.Â â—»

*Proof.* If
is the direct product of its Sylow
-subgroups, then it is nilpotent by our previous results.
If
is nilpotent and
is a Sylow
-subgroup of
for some prime
, then either
or
is a proper subgroup. In the latter case,
is a proper subgroup of
. Since
,
. Thus
and hence the unique Sylow
-subgroup of
. Let
where
are distinct primes and let
be the corresponding proper normal Sylow subgroups of
. For
,
. Thus for any
,
. It is easy to see that
is an inner direct product.Â â—»

is abelian iff .

*Proof.* Note that for every endomorphism
,
. Hence,
. Since
and hence
is abelian. If
is abelian,
for all
, whence
. Therefore
. If
so
.Â â—»

*Proof.* By definition,
is abelian.
and
. For some
,
. Therefore
is abelian and hence
.
. Continuing,
. Hence
is solvable.Â â—»

*Proof.* (i) If
is a homomorphism [resp. epimorphism] then
[resp.
] for all
. Suppose
is an epimorphism and
is solvable. For some
,
, whence
is solvable. Similarly for the subgroup.

(ii) Let be the canonical epimorphism. . Hence . By (i), is solvable. Hence