If is finite, prove that (1) is nilpotent if and only if it has a normal subgroup of each order dividing , and (2) is cyclic if and only if it has a unique subgroup of each order dividing .
We begin with some lemmas.
Lemma 1: Let be a group, and let be characteristic subgroups of . Then is characteristic in . Proof: is a subgroup because each is normal. If is an automorphism of , then .
Lemma 2: Let be a finite group. If is a normal Sylow subgroup, then is characteristic. Proof: is the unique subgroup of order .
Now to the main result; first we prove (1).
We proceed by induction on the breadth (number of prime divisors with multiplicity) of .
For the base case, if is a finite nilpotent group of breadth 1, then is an abelian simple group. Thus has a normal subgroup of order 1 and , the only divisors of .
For the inductive step, suppose that every finite nilpotent group of breadth has a normal subgroup of each order dividing , and let be a finite nilpotent group of breadth . Now is nontrivial, so there exists an element of prime order, and is normal. We showed in a lemma to the previous exercise that is nilpotent, and has breadth . By the inductive hypothesis, has a normal subgroup of each order dividing . By the Lattice Isomorphism Theorem, a subgroup is normal if and only if for some normal subgroup . Thus we see that has a normal subgroup of order for all dividing such that . Now let divide , with not dividing , and let be a normal subgroup of order . Since , is a finite nilpotent group. By Theorem 3, is the direct product of its Sylow subgroups. Write , where is the Sylow subgroup of and the direct product of the remaining Sylow subgroups. By Lemma 2, each factor of is characteristic in , so that is characteristic in . Note that , thus is normal and has order .
Suppose has a normal subgroup of each order dividing . In particular, has a normal (thus unique) Sylow -subgroup for each prime . By Theorem 6.3, is nilpotent.
Now we prove (2).
The direction was proved in Theorem 2.7.
Suppose is finite and has a unique subgroup of order for each properly dividing . In particular, all the Sylow subgroups of are unique, hence normal; say there are of these. Suppose some Sylow -subgroup of is not cyclic; then by this previous exercise, has a subgroup congruent to , and thus has distinct subgroups of order , a contradiction. Thus each Sylow subgroup of is cyclic; say . Then , so that is cyclic.