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 . ◻
Proof. Since is nonabelian, is not trivial. Since and is simple, . Thus for all whence is not solvable. Thus is not solvable. ◻
Proof. (i) Since , conjugation by is an automorphism of . Since is characteristic in , for all . Hence .
(ii) Let be a normal Sylow -subgroup of . Then . Let be an endomorphism. For , . Suppose . Then so implying . Thus .
(iii) It is easy to see that is fully invariant in . Whence . Thus or . Since is solvable, and is a nontrivial abelian group. Let be a nontrivial Sylow -subgroup of for some prime . Since is abelian, is normal in and hence fully-invariant in . Consequently, . Since is minimal and is nontrivial, . ◻
Proof. The proof proceeds by induction on , with orders less than or equal to 5 being trivial. There are two cases:
There is a nontrivial proper normal subgroup of whose order is not divisible by .
Every proper nontrivial normal subgroup of has order divisible by .
Case 1. (i) . is a solvable group of order with . By induction, contains a subgroup of order where . Then . A is solvable and contains a subgroup of order .
(ii) Suppose are subgroups of of order . Since , is a subgroup whose order necessarily divides . whence . Since , . By Lagrange’s theorem, and . Thus implies . Therefore . Similarly, . Thus and are subgroups of of order . By induction, they are conjugate for some . . Thus . and are subgroups of of order and are therefore conjugate in by induction. Hence is conjugate to in .
(iii) If a subgroup of has order , then has order dividing . Since , . implies . By induction, there is a subgroup of of order containing . Clearly . Since , is contained in a subgroup of of order by induction.
Case 2. If is a minimal normal subgroup, then for some prime . Since and , and is a Sylow -subgroup of . is the only minimal subgroup of . Every nontrivial normal subgroup of contains .
(i) Let be a normal subgroup of such that is a minimal normal subgroup of . for some prime where . . Let be a Sylow -subgroup of and . We shall show . Since , . Clearly so that whence . Since , every conjugate of in lies in . Since is a Sylow subgroup of , all these subgroups are conjugate in . Let . Let be the number of conjugates of in . Since , and We shall show that so that and hence . We first show then show . Let and . Since . Since is abelian, we only need to show . since . But . Thus . It is easy to see that is a characteristic subgroup of . Since . If , then . This with implies . is a normal Sylow -subgroup of and is thus fully invariant in , and hence normal in . This implies , a contradiction. Thus .
(ii) Let be as in (i) and suppose is a subgroup of of order . Note and . Since , . Hence . Thus implying that . by the second Sylow theorem, is conjugate to in . Furthermore, is normal in and hence . Conjugate subgroups have conjugate normalizers. Hence and are conjugate in . Thus . But so . Thus and are conjugate.
(iii) Let , where and . Let and be as in (i). Then and . Also, , and . Hence and . Let then by applying (ii) to , and are conjugate. . Since , and . ◻