# Subgroups of all orders dividing the group order not implies Sylow tower

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 group $G$ such that $G$ has subgroups of order $d$ for every natural number $d$ dividing the order of $G$, but such that $G$ does not possess a Sylow tower (?).

## Proof

### Example of the symmetric group

Further information: symmetric group:S4, subgroup structure of symmetric group:S4

The symmetric group of degree four has subgroups of every order dividing its order. There are cyclic subgroups of order $1,2,3,4$; there is a dihedral Sylow subgroup of order $8$; there is a symmetric group on three elements of order $6$, and there is an alternating group of order $12$.

On the other hand, this group does not possess a Sylow tower, because neither its $2$-Sylow subgroups nor its $3$-Sylow subgroups are normal.