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

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This article gives the statement and possibly, proof, of a non-implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., group having subgroups of all orders dividing the group order) need not satisfy the second group property (i.e., group having a Sylow tower)
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Statement

It is possible to have a finite group G such that G possesses a Sylow tower (?), but there is a natural number d dividing the order of G such that G has no subgroup of order d.

Proof

Example of the alternating group

Further information: alternating group:A4

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.