Sylow tower not implies subgroups of all orders dividing the group order

Statement
It is possible to have a finite group $$G$$ such that $$G$$ possesses a fact about::Sylow tower, but there is a natural number $$d$$ dividing the order of $$G$$ such that $$G$$ has no subgroup of order $$d$$.

Example of the alternating group
The alternating group of degree four has a Sylow tower. Explicitly, if $$G$$ is the alternating group on $$\{ 1,2,3,4 \}$$, the Sylow tower of $$G$$ is given by:

$$\{ \} \le  \{ (1,2)(3,4), (1,3)(2,4), (1,4)(2,3),  \} \le G$$.

However, the group has no subgroup of order six. To see this, note that any subgroup of order six must contain an element of order two and an element of order three. But picking any element of order two and any element of order three generates the whole group.