Finite supersolvable implies subgroups of all orders dividing the group order

Statement
Suppose $$G$$ is a finite supersolvable group: a finite group that is also supersolvable. Then, if $$d$$ is a positive divisor of the order of $$G$$, $$G$$ has a subgroup of order $$d$$.

Similar facts

 * Finite nilpotent implies every normal subgroup contains normal subgroups of all orders dividing its order
 * Finite nilpotent implies every normal subgroup is part of a chief series

Converse
The converse of this statement is false -- we can have a non-supersolvable group that has subgroups of all orders dividing the group order. In fact, every finite solvable group can be embedded inside a group with the property. See every finite solvable group is a subgroup of a finite group having subgroups of all orders dividing the group order.

The smallest examples of non-supersolvable groups with subgroups of all orders dividing their order are symmetric group:S4 and direct product of A4 and Z2, both of order 24.

Facts used

 * 1) uses::Supersolvability is subgroup-closed
 * 2) uses::Supersolvability is quotient-closed
 * 3) uses::Characteristic of normal implies normal
 * 4) uses::Normal Hall implies permutably complemented: This is the existence part of the Schur-Zassenhaus theorem.

Proof
Given: A finite supersolvable group $$G$$ of order $$n$$.

To prove: For any positive divisor $$d$$ of $$n$$, $$G$$ has a subgroup of order $$d$$.

Proof: We prove this by induction on the order of $$G$$. Note that by facts (1) and (2), every proper subgroup and every proper quotient of $$G$$ is supersolvable, and hence, by the inductive assumption, has subgroups of all orders dividing its order.


 * 1) $$G$$ has a cyclic normal subgroup $$N$$ of prime order $$p$$, for some $$p|n$$: By the definition of supersolvability, $$G$$ has a nontrivial cyclic normal subgroup, say $$C$$. $$C$$, being cyclic, has characteristic subgroups of all orders dividing the order of $$C$$. In particular, $$C$$ has a cyclic subgroup of order $$p$$ that is characteristic in it. By fact (3), this subgroup is normal in $$G$$, so we get a normal subgroup $$N$$ of order $$p$$ in $$G$$.
 * 2) If $$p|d$$, then $$G$$ has a subgroup of order $$d$$: $$G/N$$ has a subgroup of order $$d/p$$, and the inverse image of this via the quotient map is a subgroup of order $$d$$ in $$G$$.
 * 3) If $$p$$ does not divide $$d$$, then by the previous step, $$G$$ still has a subgroup $$K$$ of order $$pd$$ (it may be that $$K = G$$).
 * 4) Continuing from the previous step, $$K$$ contains a subgroup of order $$d$$. In particular, $$G$$ has a subgroup of order $$d$$: $$N$$ is a normal Sylow subgroup of $$K$$ (since $$p$$ does not divide $$d$$). Hence, fact (4) tells us that $$K$$ contains a complement to $$N$$, which must therefore have order $$d$$.

Thus, $$G$$ has a subgroup of order $$d$$ in both cases.