Proof. and all its nontrivial quotients are -groups and thus have nontrivial centers. Thus if , . Since is finite, . ◻
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 . ◻
Proof. Let and let be the largest index such that . Choose . . Thus . This implies . Thus . ◻
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. ◻
is abelian iff .
Proof. Note that for every endomorphism , . Hence, . Since and hence is abelian. If is abelian, for all , whence . Therefore . If so . ◻
Proof. By definition, is abelian. and . For some , . Therefore is abelian and hence . . Continuing, . Hence is solvable. ◻
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